2=|G(L/\Q)|$, so $L$ is not normal over $\Q$. Explicitly, the polynomial $f(t)=t^3-3\in\Q[t]$ has a root in $L$ but does not split in $L$. \end{solution} \begin{exercise}\exlabel{ex-galois-iii} Put $L=\Q(\sqrt[4]{3},i)$. Find all the automorphisms of $L$, and show that $L$ is normal over $\Q$. \end{exercise} \begin{solution} Put $\al=\sqrt[4]{3}$ and \[ f(t) = (t-\al)(t+\al)(t-i\al)(t+i\al). \] We find that $(t-\al)(t+\al)=t^2-\sqrt{3}$, but $(t-i\al)(t+i\al)=t^2+\sqrt{3}$, so $f(t)=t^4-3$. It follows easily that $L=\Q(\al,i)$ is a splitting field for $f(t)$ over $\Q$, so $L$ is normal over $\Q$. The set $R=\{\al,i\al,-\al,-i\al\}$ of roots is the set of vertices of a square in the complex plane. We claim that the group $G(L/\Q)$ is just the dihedral group of rotations and reflections of this square. Indeed, complex conjugation gives an automorphism $\sg$ which reflects the square across the real axis. Next, we can use Eisenstein's criterion at the prime $3$ to see that $f(t)$ is irreducible, so $G(L/\Q)$ acts transitively on $R$. It follows that there is an automorphism $\phi$ with $\phi(\al)=i\al$. Now $\phi(i)$ must be a square root of $-1$, so $\phi(i)=\pm i$. If $\phi(i)=i$ then we put $\rho=\phi$, otherwise we put $\rho=\phi\sg$. Either way we find that $\rho(i)=i$ and $\rho(\al)=i\al$. This implies that $\rho(i^m\al)=i^{m+1}\al$ for all $m$, so $\rho$ is a quarter turn of the square. This means that $\rho$ and $\sg$ generate $D_8$, so $|G(L/\Q)|\geq|D_8|=8$. On the other hand, the set \[ B = \{1,\al,\al^2,\al^3,i,i\al,i\al^2,i\al^3\} \] clearly spans $L$ over $\Q$, so $[L:\Q]\leq |B|=8$, and for any extension we have $|G(L/\Q)|\leq [L:\Q]$. It follows that all these inequalities must be equalities, so $G(L/\Q)=D_8$ and $B$ is a basis. \begin{center} \begin{tikzpicture}[scale=1.5] \draw[blue] (1.32,0) -- (0,1.32) -- (-1.32,0) -- (0,-1.32) -- cycle; \fill ( 0.00, 0.00) circle(0.05); \fill ( 1.32, 0.00) circle(0.05); \fill ( 0.00, 1.32) circle(0.05); \fill (-1.32, 0.00) circle(0.05); \fill ( 0.00,-1.32) circle(0.05); \draw ( 1.54, 0.00) node{$\al$}; \draw ( 0.00, 1.54) node{$i\al$}; \draw (-1.59, 0.00) node{$-\al$}; \draw ( 0.00,-1.54) node{$-i\al$}; \draw[red,ultra thick,<->] (1.7,-0.3) -- (1.7,0.3); \draw[red,ultra thick,->] (0,0) +(0:0.3) arc(0:90:0.3); \draw[red] (1.85,0) node{$\sg$}; \draw[red] (0.4,0.4) node{$\rho$}; \end{tikzpicture} \end{center} \end{solution} \begin{exercise}\exlabel{ex-galois-iv} Consider the polynomial $f(x)=x^4+x^2+4$. This is irreducible over $\Q$; you can either prove that, or just assume it and continue with the rest of the question. Put \[ \al=\sqrt{-\frac{1}{2}+\frac{1}{2}\sqrt{-15}}. \] \begin{itemize} \item[(a)] Show that the roots of $f(x)$ are $\pm\al$ and $\pm\frac{2}{\al}$, so $\Q(\al)$ is a splitting field for $f(x)$. \item[(b)] Compute $G(\Q(\al)/\Q)$. What well-known group is it? \end{itemize} \end{exercise} \begin{solution} We will do~(a) and~(b) first, and then check that $f(x)$ is irreducible. \begin{itemize} \item[(a)] From the definition we have $2\al^2+1=\sqrt{-15}$, and squaring again gives $4\al^4+4\al^2+16=0$, so $f(\al)=0$. As $f(x)$ only involves even powers of $x$ we have $f(-x)=f(x)$ and so $f(-\al)=0$. Now \[ f(2/\al) = \frac{16}{\al^4} + \frac{4}{\al^2} + 4 = \frac{4}{\al^4}(4+\al^2+\al^4) = \frac{4}{\al^4}f(\al)= 0, \] and similarly $f(-2/\al)=0$. Numerically we have $\al\simeq 0.87+0.12i$, and from that one can check that $\al,-\al,2/\al$ and $-2/al$ are all distinct. We must therefore have \[ f(x) = (x-\al)(x+\al)(x-2/\al)(x+2/\al). \] \item[(b)] We have a normal extension of degree $4$, so the Galois group $G$ must have order $4$. We know that $G$ acts transitively on the roots, so there are automorphisms $\sg$ and $\rho$ with $\sg(\al)=-\al$ and $\rho(\al)=2/\al$. These satisfy $\sg^2(\al)=\sg(-\al)=-\sg(\al)=\al$ and $rho^2(\al)=\rho(2/\al)=2/\rho(\al)=\al$, so $\sg^2=\rho^2=1$. We also have $\sg(\rho(\al))=\rho(\sg(\al))=-2/\al$. It follows that \[ G = \{1,\sg,\rho,\sg\rho\}, \] and this is isomorphic to $C_2\tm C_2$. \end{itemize} We now prove that $f(x)$ is irreducible. It is clear that $f(x)>0$ for all $x\in\R$, so there are no roots in $\Q$. This means that the only way $f(x)$ could factor would be as the product of two quadratics, say $f(x)=(x^2+ax+b)(x^2+cx+d)$ for some $a,b,c,d\in\Q$. By looking at the term in $x^3$, we see that $c=-a$. After substituting this, expanding and comparing the remaining coefficients we obtain \begin{align*} b+d-a^2 &= 1 \\ a(d-b) &= 0 \\ bd &= 4. \end{align*} If $a=0$ we quickly obtain $b=(1\pm\sqrt{-3})/2$, which is impossible as $b\in\Q$. Thus $a\neq 0$, so the second equation above gives $d=b$, so the last equation gives $b=\pm 2$. The first equation then becomes $a^2=\pm 4-1$, which is impossible for $a\in\Q$. \end{solution} \begin{exercise}\exlabel{ex-galois-v} Put $f(x)=x^4+8x^2-2\in\Q[x]$, and $\al=\sqrt{3\sqrt{2}-4}$, and $M=\Q(\al,\sqrt{-2})$. \begin{itemize} \item[(a)] Show that $f(x)$ is irreducible over $\Q$. \item[(b)] Show that $f(x)$ has roots $\pm\al,\pm\sqrt{-2}/\al$, so that $M$ is a splitting field for $f(x)$. \item[(c)] Show that $\Q(\al)=M\cap\R\neq M$, and deduce that $[M:\Q]=8$. \item[(d)] Show that there exist automorphisms $\phi,\psi\in G(M/\Q)$ such that $\phi$ has order 4, $\psi$ has order 2, and $G(M/\Q)=\langle\phi,\psi\rangle$. \item[(e)] Write $\psi\phi\psi^{-1}$ in the form $\phi^i\psi^j$. To what well-known group is $G(M/\Q)$ isomorphic? \end{itemize} \end{exercise} \begin{solution}\ \\ \begin{itemize} \item[(a)] As $f(x)=x^4\pmod{2}$ and $f(0)\neq 0\pmod{4}$ we can use Eisenstein's criterion to see that $f(x)$ is irreducible. \item[(b)] Note that $\al^2+4=3\sqrt{2}=\sqrt{18}$, and squaring again shows that $\al^4+8\al^2+16=18$, so $f(\al)=0$. As $f(x)$ only involves even powers of $x$ we have $f(-x)=f(x)$ and so $f(-\al)=0$. Now put $\bt=\sqrt{-3\sqrt{2}-4}$; the same argument shows that $f(\pm\bt)=0$. We also have $(\al\bt)^2=(3\sqrt{2}-4)(-3\sqrt{2}-4)=-2$, so $\bt=\pm\sqrt{-2}/\al$. (With the standard conventions for square roots we have $\al>0$, and $\bt$ and $\sqrt{-2}$ are positive multiples of $i$, and it follows that $\bt=\sqrt{-2}/\al$.) It follows that the roots of $f(x)$ are as described, so the splitting field is $\Q(\al,\bt)=\Q(\al,\al\bt)=\Q(\al,\sqrt{-2})=M$ as claimed. \item[(c)] We have $3\sqrt{2}-4\simeq 0.24>0$ so $\al$ is real, so $\Q(\al)\sse M\cap\R$. As $f(x)$ is irreducible, it must be the minimal polynomial for $\al$, and so $[\Q(\al):\Q]=\deg(f(x))=4$. As $\Q(\al)\sse\R$ and $\sqrt{-2}$ is purely imaginary we see that $1,\sqrt{-2}$ is a basis for $M$ over $\Q(\al)$, so $M\cap\R=\Q(\al)$ and $[M:\Q]=[M:\Q(\al)][\Q(\al):\Q]=2\tm 4=8$. \item[(d)] First let $\psi\:M\to M$ be given by complex conjugation, so $\psi(\sqrt{-2})=-\sqrt{-2}$ and $\psi(\al)=\al$. It is clear that $\psi^2=1$. Next, the Galois group of the splitting field of an irreducible polynomial always acts transitively on the roots, so we can find $\sg\in G(M/\Q)$ with $\sg(\al)=\sqrt{-2}/\al$. Now $\sg$ must permute the roots of $x^2+2$, so $\sg(\sqrt{-2})=\pm\sqrt{-2}$. If the sign is positive we put $\phi=\sg\psi$, otherwise we put $\phi=\sg$. In either case we then have $\phi(\al)=\sqrt{-2}/\al=\bt$ and $\phi(\sqrt{-2})=-\sqrt{-2}$. This means that \[ \phi^2(\al)=\phi(\sqrt{-2}/\al)=\phi(\sqrt{-2})/\phi(\al) = -\sqrt{-2}/(\sqrt{-2}/\al) = -\al \] and $\phi^2(\sqrt{-2})=\sqrt{-2}$. It follows in turn that $\phi^4=1$. We now have various different automorphisms, whose effect we can tabulate as follows: \[ \renewcommand{\arraystretch}{1.5} \begin{array}{|c||c|c|c|c|c|c|c|c|} \hline & 1 & \phi & \phi^2 & \phi^3 & \psi & \phi\psi & \phi^2\psi & \phi^3\psi \\ \hline \al & \al & \bt & -\al & -\bt & \al & \bt & -\al & -\bt \\ \hline \bt & \bt & -\al & -\bt & \al & -\bt & \al & \bt & -\al \\ \hline \sqrt{-2} & \sqrt{-2} & -\sqrt{-2} & \sqrt{-2} & -\sqrt{-2} & -\sqrt{-2} & \sqrt{-2} & -\sqrt{-2} & \sqrt{-2}. \\ \hline \end{array} \] We see that the eight automorphisms listed are all different, but $|G(M/\Q)|=[M:\Q]=8$, so we have found all the automorphisms. \item[(e)] We can read off from the above table that $\psi\phi\psi^{-1}=\phi^3=\phi^{-1}$. This means that $G(M/\Q)$ is the dihedral group $D_8$, with $\phi$ corresponding to a rotation through $\pi/2$, and $\psi$ to a reflection. \end{itemize} \end{solution} \section{Cyclotomic extensions} \label{sec-cyclotomic} \begin{definition}\lbl{defn-cyclotomic} For any $n>0$ we put \[ \mu_n = \{z\in\C\st z^n=1\} = \{\exp(2\pi ik/n) \st k=0,1,\dotsc,n-1\}. \] If $z\in\mu_n$ for some $n$, then the \emph{order} of $z$ is the smallest $d>0$ for which $z^d=1$; this is a divisor of $n$. We write $\mu_n^\tm$ for the subset of $\mu_n$ consisting of numbers of order precisely $d$. We also define \[ \vph_n(t) = \prod_{z\in\mu_n^\tm}(t-z) \in \C[t], \] and call this the \emph{$n$'th cyclotomic polynomial}. We write $\Q(\mu_n)$ for the subfield of $\C$ generated by $\mu_n^\tm$, and call this the \emph{$n$'th cyclotomic field}. This is evidently a splitting field for $\vph_n(t)$. \end{definition} \begin{remark}\lbl{rem-cyclotomic} If $\xi=\exp(2\pi ik/n)\in\mu_n$, then $\xi=\zt_1^k$, where $\zt_1=\exp(2\pi i/n)\in\mu_n^\tm$. By definition we have $\zt_1\in\Q(\mu_n)$ and $\Q(\mu_n)$ is a subfield so it is closed under multiplication, so $\zt_1^k=\xi\in\Q(\mu_n)$. This shows that $\Q(\mu_n)$ does indeed contain $\mu_n$ as suggested by the notation. \end{remark} \begin{proposition}\lbl{prop-cyclotomic-product} The polynomial $\vph_n(t)$ is actually in $\Z[t]$, and satisfies \[ t^n-1 = \prod_{d|n} \vph_d(t). \] \end{proposition} \begin{proof} Firstly, for each divisor $d$ of $n$, we note that $\mu_d^\tm\sse\mu_d\sse\mu_n$. Every element $z\in\mu_n$ lies in precisely one of the sets $\mu_d^\tm$, so we see that \[ \prod_{z\in\mu_n}(t-z) = \prod_{d|n} \prod_{z\in\mu_d^\tm}(t-z) = \prod_{d|n} \vph_d(t). \] On the other hand, the elements of $\mu_n$ are precisely the roots of $x^n-1$, and there are $n$ of them, so we see from Proposition~\ref{prop-several-roots} that $t^n-1=\prod_{z\in\mu_n}(t-z)=\prod_{d|n}\vph_d(t)$ as claimed. We will now prove by induction that $\vph_n(t)\in\Z[t]$ for all $n$. To start the induction, note that $\mu_1^\tm=\{1\}$ so $\vph_1(t)=t-1\in\Z[t]$. Now suppose that $\vph_d(t)\in\Z[t]$ for all $d0$ and $k$ is coprime to $n$ then we can write $k=p_1p_2\dotsb p_r$ for some primes $p_1,\dotsc,p_r$ (not necessarily distinct) that do not divide $n$. We then see that $f(t)=\min(\zt^k,\Q)$ by an obvious extension of the argument for the case $k=pq$. Finally, if $k<0$ and $(k,n)=1$ then we can choose $j$ such that the number $k'=k+jn$ is positive (and still coprime to $n$). We then see that $f(t)=\min(\zt^{k'},\Q)$ but $\zt^{k'}=(\zt^n)^j\zt^k=\zt^k$ so $f(t)=\min(\zt^k,\Q)$ as claimed. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop-phi-irreducible}] Put $\zt=\exp(2\pi i/n)\in\mu_n^\tm$, and $f(t)=\min(\zt,\Q)$. As $\vph_n(\zt)=0$ we see that $f(t)$ divides $\vph_n(t)$. On the other hand, the roots of $\vph_n(t)$ are precisely the elements of $\mu_n^\tm$, or in other words the powers $\zt^k$ with $0\leq k 2$ we have $\sqrt{p}\in\Q(\mu_{4p})$. More precisely, if $\xi=\exp(\pi i/(2p))$ is the standard generator of $\mu_{4p}$ then \[ \sqrt{p} = \prod_{k=1}^{(p-1)/2} (\xi^{p-2k}-\xi^{p+2k}) = \xi^{(p-1)^2/4} \prod_{k=1}^{(p-1)/2} (1-\xi^{4k}). \] (Note here that $(p-1)/2$ and $(p-1)^2/4$ are integers, because $p$ is odd.) \end{proposition} \begin{example}\lbl{eg-five} Before discussing the general case we will look at the example where $p=5$. There we have $\xi=\exp(\pi i/10)$ and the claim is that \[ \sqrt{5} = (\xi^3-\xi^7)(\xi-\xi^9) = \xi^4(1-\xi^4)(1-\xi^8). \] Put \begin{align*} \lm &= (\xi^3-\xi^7)(\xi-\xi^9) \\ \mu &= \xi^4(1-\xi^4)(1-\xi^8), \end{align*} so the claim is that $\lm=\mu=\sqrt{p}$. If we start with $\lm$ and extract a factor of $\xi^3$ from the first bracket and a factor of $\xi$ from the second bracket then we end up with $\mu$, so $\lm=\mu$ as claimed. It will be convenient to rewrite $\mu$ in terms of $\zt=\xi^4=\exp(2\pi i/5)$, which is a primitive $5$th root of unity. This satisfies \[ (1+\zt+\zt^2+\zt^3+\zt^4)(1-\zt)=1-\zt^5=0 \] and $1-\zt\neq 0$ so we must have $1+\zt+\zt^2+\zt^3+\zt^4=0$. Note that \[ \mu = \zt(1-\zt)(1-\zt^2) = \zt-\zt^2-\zt^3+\zt^4. \] If we square this and collect terms in the most obvious way, we get \begin{align*} \mu^2 &= \zt^2+\zt^4+\zt^6+\zt^8 -2\zt^3-2\zt^4+2\zt^5+2\zt^5-2\zt^6-2\zt^7 \\ &= \zt^2-2\zt^3-\zt^4+4\zt^5-\zt^6-2\zt^7+\zt^8. \end{align*} If we now use the identity $\zt^5=1$ (so $\zt^6=\zt$ and so on) we get \begin{align*} \mu^2 &= \zt^2-2\zt^3-\zt^4+4-\zt-2\zt^2+\zt^3 \\ &= 4-\zt-\zt^2-\zt^3-\zt^4. \end{align*} Finally, we can combine this with the identity $1+\zt+\zt^2+\zt^3+\zt^4=0$ to get $\mu^2=5$, so $\mu=\pm\sqrt{5}$. It is not hard to check that the factors $\xi^3-\xi^7$ and $\xi-\xi^9$ are positive real numbers, so $\mu>0$, so $\mu=\sqrt{5}$; we will explain this in more detail when we discuss the general case. \end{example} \begin{lemma}\lbl{lem-root-p-xi} With $p$ and $\xi$ as above we have $\prod_{k=1}^{p-1}(1-\xi^{4k})=p$. \end{lemma} \begin{proof} The powers $\xi^{4k}$ for $k=0,\dotsc,p-1$ are precisely the $p$'th roots of unity, so we have \[ t^p-1 = \prod_{k=0}^{p-1} (t-\xi^{4k}). \] The $k=0$ term on the right hand side is just $t-1$. We can move this to the left hand side and use the standard geometric progression formula to get \[ 1 + t + \dotsb + t^{p-1} = \frac{t^p-1}{t-1} = \prod_{k=1}^{p-1} (t-\xi^{4k}). \] Now set $t=1$. On the left hand side we have $p$ terms which all become $1$, and on the right we have $\prod_{k=1}^{p-1}(1-\xi^{4k})$ so $\prod_{k=1}^{p-1}(1-\xi^{4k})=p$ as claimed. \end{proof} \begin{corollary}\lbl{cor-norm} We have $|\prod_{k=1}^{(p-1)/2}(1-\xi^{4k})|=\sqrt{p}$. \end{corollary} \begin{proof} Put $\kp=\prod_{k=1}^{(p-1)/2}(1-\xi^{4k})$. We then have $\ov{\kp}=\prod_{k=1}^{(p-1)/2}(1-\xi^{-4k})$, and $\xi^{4p}=1$ so we can rewrite $\xi^{-4k}$ as $\xi^{4p-4k}$ or as $\xi^{4(p-k)}$. Now, as $k$ runs from $1$ to $(p-1)/2$ we find that $p-k$ runs through the numbers from $(p+1)/2$ to $p-1$ (in reverse order), so the numbers $k$ and $p-k$ together cover all the numbers from $1$ to $p-1$ (each number exactly once). Thus \[ \kp\ov{\kp} = \prod_{k=1}^{(p-1)/2}(1-\xi^{4k}) \prod_{k=1}^{(p-1)/2}(1-\xi^{4(p-k)}) = \prod_{j=1}^{p-1}(1-\xi^{4j}) = p. \] (The last step here is just the previous lemma.) On the other hand, we have $\kp\ov{\kp}=|\kp|^2$, so $|\kp|=\sqrt{p}$ as claimed. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop-root-p}] Put \begin{align*} \lm &= \prod_{k=1}^{(p-1)/2} (\xi^{p-2k}-\xi^{p+2k}) \\ \mu &= \xi^{(p-1)^2/4} \prod_{k=1}^{(p-1)/2} (1-\xi^{4k}), \end{align*} so the claim is that $\lm=\mu=\sqrt{p}$. First, combine Corollary~\ref{cor-norm} with the fact that $|\xi|=1$ to get \[ |\mu| = |\xi|^{(p-1)^2/4} \left|\prod_{k=1}^{(p-1)/2}(1-\xi^{4k})\right| = 1^{(p-1)^2/4} \sqrt{p} = \sqrt{p}. \] Next, note that \[ \xi^{p-2k}(1-\xi^{4k}) = \xi^{p-2k} - \xi^{p+2k}. \] Take the product for $k=1,\dotsc,(p-1)/2$ to get \[ \xi^N \prod_{k=1}^{(p-1)/2} (1-\xi^{4k}) = \lm, \] where $N=\sum_{k=1}^{(p-1)/2}(p-2k)$. This is the sum of $(p-1)/2$ equally spaced terms from $p-2$ down to $1$, so the average term is $\half((p-2)+1)=(p-1)/2$ and the total is the number of terms times the average, which gives $N=(p-1)^2/4$. Given this, the displayed equation tells us that $\mu=\lm$, so $|\lm|=|\mu|=\sqrt{p}$. Next, note that $\xi^p=i$ and \begin{align*} \xi^{2k} &= \exp( k\pi i/p) = \cos(k\pi/p) + i\sin(k\pi/p) \\ \xi^{-2k} &= \exp(-k\pi i/p) = \cos(k\pi/p) - i\sin(k\pi/p) \end{align*} so \[ \xi^{p-2k}-\xi^{p+2k} = i(\cos(k\pi/p) - i\sin(k\pi/p)) - i(\cos(k\pi/p) + i\sin(k\pi/p)) = 2\sin(k\pi/p). \] Moreover, when $1\leq k\leq (p-1)/2$ we have $0 0$. It follows that $\lm$ is a positive real number, so $\lm=|\lm|=\sqrt{p}$. \end{proof} \begin{corollary}\lbl{cor-mquad-cyclotomic} For any field of the form $K=\Q(\sqrt{p_1},\dotsc,\sqrt{p_m})$ (where the $p_i$ are odd primes) there exists $N$ such that $K\sse\Q(\mu_N)$. \end{corollary} \begin{proof} Put $N=4\prod_ip_i$. For each $i$ we see that $4p_i$ divides $N$ and so $\sqrt{p_i}\in\Q(\mu_{4p_i})\sse\Q(\mu_N)$. It follows that $K\sse\Q(\mu_N)$ as claimed. \end{proof} %============================================================ %============================================================ \begin{center} \Large \textbf{Exercises} \end{center} \begin{exercise}\exlabel{ex-cyclotomic-twenty} Find the cyclotomic polynomial $\vph_{20}(x)$. \end{exercise} \begin{solution} Recall the key fact that \[ x^n-1 = \prod_{d|n} \vph_d(x). \] In particular, we have \begin{align*} x-1 &= \vph_1(x) \\ x^2-1 &= \vph_1(x)\vph_2(x) \\ x^4-1 &= \vph_1(x)\vph_2(x)\vph_4(x) \\ x^5-1 &= \vph_1(x)\vph_5(x) \\ x^{10}-1 &= \vph_1(x)\vph_{2}(x)\vph_5(x)\vph_{10}(x) \\ x^{20}-1 &= \vph_1(x)\vph_{2}(x)\vph_4(x) \vph_5(x)\vph_{10}(x)\vph_{20}(x). \end{align*} Dividing the second and third of these gives \[ \vph_4(x) = \frac{x^4-1}{x^2-1} = x^2+1. \] On the other hand, we can divide the last two equations to give \[ \vph_{20}(x)\vph_4(x) = \frac{x^{20}-1}{x^{10}-1} = x^{10}+1. \] Putting these together, we get \[ \vph_{20}(x) = \frac{x^{10}+1}{x^2+1} = x^8-x^6+x^4-x^2+1. \] (The calculation can also be arranged in various other ways, but this is probably the most efficient.) \end{solution} \begin{exercise}\exlabel{ex-phi-CC} What is $\vph_{200}(x)$? \end{exercise} \begin{solution} We have \begin{align*} x^{200}-1 &= \vph_{200}(x)\vph_{100}(x)\vph_{50}(x)\vph_{40}(x) \vph_{25}(x)\vph_{20}(x)\vph_{10}(x)\vph_8(x)\vph_5(x) \vph_4(x)\vph_2(x)\vph_1(x)\\ x^{100}-1 &= \vph_{100}(x)\vph_{50}(x)\vph_{25}(x)\vph_{20}(x) \vph_{10}(x)\vph_5(x)\vph_4(x)\vph_2(x)\vph_1(x)\\ x^{40}-1 &= \vph_{40}(x)\vph_{20}(x)\vph_{10}(x)\vph_8(x) \vph_5(x)\vph_4(x)\vph_2(x)\vph_1(x)\\ x^{20}-1 &= \vph_{20}(x)\vph_{10}(x)\vph_5(x)\vph_4(x)\vph_2(x)\vph_1(x) \end{align*} and it follows that \[ \vph_{200}(x) = \frac{(x^{200}-1)(x^{20}-1)}{(x^{100}-1)(x^{40}-1)} = \frac{x^{100}+1}{x^{20}+1} = x^{80}-x^{60}+x^{40}-x^{20}+1. \] \end{solution} \begin{exercise}\exlabel{ex-mu-seven} Explicitly compute a polynomial $f(t)\in\Q[t]$ of degree six with $e^{3\pi i/7}+1$ as a root. Prove that this polynomial is irreducible over $\Q$, using Eisenstein's criterion. \end{exercise} \begin{solution} Put $\zt=e^{3\pi i/7}=(e^{2\pi i/14})^3$ and $\al=\zt+1$. As $3$ and $14$ are coprime, we see that $\zt$ is a primitive 14th root of unity, and so is a root of the cyclotomic polynomial $\vph_{14}(t)$. We know that \begin{align*} t^{14}-1 &= \vph_{14}(t)\vph_7(t)\vph_2(t)\vph_1(t) \\ t^7 - 1 &= \vph_7(t)\vph_1(t) \\ t+1 &= \vph_2(t). \end{align*} We can divide the first of these by the second and the third to give \[ \vph_{14}(t) = \frac{t^7+1}{t+1} = t^6-t^5+t^4-t^3+t^2-t+1. \] Now put $f(t)=\vph_{14}(t-1)$. This is again a polynomial of degree $6$ over $\Q$, and we have $f(\al)=\vph_{14}(\al-1)=\vph_{14}(\zt)=0$. More explicitly, we can use the expression $\vph_{14}(t)=(t^7+1)/(t+1)$ to get \[ f(t) = \frac{(t-1)^7+1}{t-1+1} = ((t-1)^7+1)/t = \sum_{i=0}^6 (-1)^i\bcf{7}{i} t^{6-i} = t^6-7t^5+21t^4-35t^3+35t^2-21t+7. \] This reduces to $t^6$ modulo $7$, either by inspecting the coefficients directly, or by recalling that $(t-1)^7=t^7-1^7\pmod{7}$. Moreover, the constant term is $7$, which is not divisible by $7^2$. Thus Eisenstein's criterion is applicable, and we see that $f(t)$ is irreducible. \end{solution} \begin{exercise}\exlabel{ex-mu-fifteen} Describe the automorphisms of $\Q(\mu_{15})$. Find two cyclic subgroups $A$ and $B$ such that $G(\Q(\mu_{15})/\Q)=A\tm B$. \end{exercise} \begin{solution} Put $\zt=e^{2\pi i/15}$ and $K=\Q(\zt)=\Q(\mu_{15})$. The general theory tells us that for each integer $k$ that is coprime to $15$, there is a unique automorphism $\sg_k$ of $K$ with $\sg_k(\zt)=\zt^k$, and that the rule $k+15\Z\mapsto\sg_k$ gives a well-defined isomorphism $(\Z/15\Z)^\tm\to G(K/\Q)$. Every element of $\Z/15\Z$ has a unique representative lying between $-7$ and $7$, and the integers in that range that are coprime to $15$ form the set \[ U = \{-7,-4,-2,-1,1,2,4,7\}, \] so we can identify this set with $(\Z/15\Z)^\tm$. Put $A=\{1,-1\}$, which is a cyclic subgroup of $U$ of order $2$. Note that $2^3=8=-7\pmod{15}$ and $2^4=16=1\pmod{15}$. It follows that the set $B=\{1,2,4,-7\}$ is a cyclic subgroup of $U$ of order $4$, and we see directly that $U=A\tm B$. \end{solution} \begin{exercise}\exlabel{ex-cyclotomic-real} Let $\zt$ be a primitive $n$th root of unity, where $n\geq 3$, and write $\bt=\zt+\zt^{-1}$. \begin{itemize} \item[(a)] Show that $\zt$ satisfies a quadratic equation over $\Q(\bt)$ and deduce that $[\Q(\zt):\Q(\bt)]\leq 2$. \item[(b)] Show that $\Q(\bt)\subset\R$, and deduce that $\zt\not\in\Q(\bt)$. Deduce that $[\Q(\zt):\Q(\bt)]=2$. \item[(c)] Prove by induction that for all $m$, $\zt^m+\zt^{-m}\in\Q(\bt)$. \item[(d)] Express $\zt^5+\zt^{-5}$ as a polynomial in $\bt$. \end{itemize} [{\sl Hint for (c) and (d): if $\zt^m+\zt^{-m}=p_m(\bt)$, show that $\zt^{m+1}+\zt^{-m-1}=\bt p_m(\bt)-p_{m-1}(\bt)$.}] \end{exercise} \begin{solution}\ \\ \begin{itemize} \item[(a)] Put $f(x)=x^2-\bt x+1\in\Q(\bt)[x]$. As $\bt=\zt+\zt^{-1}$, we see that $\bt\zt=\zt^2+1$, so $f(\zt)=0$. Thus, $\zt$ satisfies a quadratic equation over $\Q(\bt)$, as claimed. The minimal polynomial $\min(\zt,\Q(\bt))$ must divide $f(x)$, so it has degree one (if $\zt\in\Q(\bt)$) or two (if $\zt\not\in\Q(\bt)$). Thus, we have $[\Q(\zt):\Q(\bt)]\leq 2$. \item[(b)] We next observe that $\zt^n=1$ so $|\zt|>0$ and $|\zt|^n=1$, so $|\zt|=1$. If $\zt$ is real this means that $\zt=\pm 1$, so $\zt^2=1$, but this contradicts the assumption that $\zt$ is a primitive $n$th root for some $n\geq 3$. Thus, we see that $\zt\not\in\R$. On the other hand, as $|\zt|=1$ we see that $\zt^{-1}=\ov{\zt}$, so $\bt=\zt+\ov{\zt}=2\text{Re}(\zt)\in\R$. It follows that $\Q(\bt)\sse\R$ and so $\zt\not\in\Q(\bt)$. In conjunction with~(a) this means that $[\Q(\zt):\Q(\bt)]=2$. \item[(c)] We claim that $\zt^m+\zt^{-m}=p_m(\bt)$ for some polynomial $p_m(x)$. Indeed, we can put $p_0(x)=2$ and $p_1(x)=x$, and then define $p_m(x)$ recursively for $m>1$ by $p_{k+1}(x)=x\,p_k(x)-p_{k-1}(x)$. We claim that $p_k(\bt)=\zt^k+\zt^{-k}$. This is clear for $k\in\{0,1\}$. If the claim holds for all $k\leq m$, we have \begin{align*} p_{m+1}(\bt) &= \bt p_m(\bt) - p_{m-1}(\bt) \\ &= (\zt+\zt^{-1})(\zt^m+\zt^{-m}) - (\zt^{m-1}+\zt^{1-m}) \\ &= (\zt^{m+1}+\zt^{1-m}+\zt^{m-1}+\zt^{-m-1}) - (\zt^{m-1}+\zt^{1-m}) \\ &= \zt^{m+1}+\zt^{-m-1}. \end{align*} The claim therefore holds for all $m$, by induction. \item[(d)] The first few steps of the recursive scheme are as follows: \begin{align*} p_0(x) &= 2 \\ p_1(x) &= x \\ p_2(x) &= x\,p_1(x) - p_0(x) = x^2-2 \\ p_3(x) &= x\,p_2(x) - p_1(x) = x^3-3x \\ p_4(x) &= x\,p_3(x) - p_2(x) = x^4-4x^2+2 \\ p_5(x) &= x\,p_4(x) - p_3(x) = x^5-5x^3+5x. \end{align*} Thus, we have $\zt^5+\zt^{-5}=\bt^5-5\bt^3+5\bt$. \end{itemize} \end{solution} \begin{exercise}\exlabel{ex-shift-irr} Show that if $f(t)\in K[t]$ and $a\in K$ and the polynomial $g(t)=f(t+a)$ is irreducible, then $f(t)$ itself is also irreducible. Apply this together with Eisenstein's criterion to give an alternative proof that $\vph_p(t)$ is irreducible (for any prime $p$). \end{exercise} \begin{solution} Suppose that $g(t)=f(t+a)$ is irreducible as above. Suppose we have a factorisation $f(t)=p(t)q(t)$, where $p(t)$ and $q(t)$ are nonconstant polynomials in $K[t]$. We then have nonconstant polynomials $r(t)=p(t+a)$ and $s(t)=q(t+a)$ with $g(t)=r(t)s(t)$. This is impossible, because $g(t)$ is assumed to be irreducible. This means that no such factorisation $f(t)=p(t)q(t)$ can exist, so $f(t)$ must be irreducible. Now take $f(t)=\vph_p(t)=(t^p-1)/(t-1)$ and $a=1$. We then have \[ g(t) = \frac{(t+1)^p-1}{(t+1)-1} = t^{-1}((t+1)^p-1) = \sum_{i=0}^{p-1}\bcf{p}{i+1}t^i. \] This is monic, and using Lemma~\ref{lem-F-additive} we see that $g(t)=t^{p-1}\pmod{p}$, so the coefficients of $t^0,\dotsc,t^{p-2}$ are all divisible by $p$. Moreover, the constant term is $g(0)=p$, which is not divisible by $p^2$. Eisenstein's criterion therefore tells us that $g(t)=f(t+1)$ is irreducible, so we can use the first paragraph above to see that $f(t)$ is also irreducible. \end{solution} \begin{exercise}\exlabel{ex-phi-two-power} Prove that $\vph_{2^{k+1}}(t)=t^{2^k}+1$. \end{exercise} \begin{solution} Put $s=t^{2^k}$. As the divisors of $2^k$ are just the powers $2^j$ for $j\leq k$, we have $s-1=\prod_{j=0}^k\vph_{2^j}(t)$. We also have $s^2=t^{2\tm 2^k}=t^{2^{k+1}}$, so $s^2-1=\prod_{j=0}^{k+1}\vph_{2^j}(t)$. By dividing these two equations we get $\vph_{2^{k+1}}(t)=(s^2-1)/(s-1)=s+1=t^{2^k}+1$ as claimed. Alternatively, if $\zt$ is a $2^{k+1}$th root of unity, then $\zt^{2^k}$ cannot be equal to $1$ (by primitivity) but $(\zt^{2^k})^2=\zt^{2^{k+1}}=1$. We must therefore have $\zt^{2^k}=-1$. It follows that the primitive $2^{k+1}$th roots of unity are precisely the same as the roots of $t^{2^k}+1$. This polynomial is monic and coprime with its derivative, so there are no repeated roots. It follows that $t^{2^k}+1$ is the product of $t-\zt$ as $\zt$ runs over the roots, which is $\vph_{2^{n+1}}(t)$. \end{solution} \begin{exercise}\exlabel{ex-phi-families}\ \\ \begin{itemize} \item[(a)] Prove that $\zt$ is a primitive $m$th root of unity if and only if $\ov{\zt}$ is a primitive $m$th root of unity. Deduce that if $m>2$ then $\vph_n(x)$ has even degree. \item[(b)] Let $n\geq 6$ be even, but not divisible by $4$. Prove that $\zt$ is a primitive $n$th root of 1 if and only if $-\zt$ is a primitive $(n/2)$th root of 1. Deduce that $\vph_n(x)=\vph_{n/2}(-x)$. \item[(c)] Suppose that $n$ is divisible by $p^2$ for some prime $p$. Show that $\zt$ is a primitive $n$th root of $1$ if and only if $\zt^p$ is a primitive $(n/p)$th root of $1$. Deduce that $\vph_n(x)=\vph_{n/p}(x^p)$. \item[(d)] Recall that $\vph_1(x)=x-1$, and that $\vph_p(x)=1+x+\dotsb+x^{p-1}$ when $p$ is prime. How many cyclotomic polynomials can you calculate using these facts together with~(b) and~(c)? \item[(e)] For small $n$ one observes that all coefficients in $\vph_n(x)$ are $0$, $1$ or $-1$, but this pattern does not persist for ever. Let $N$ be the smallest number such that $\vph_N(x)$ has a coefficient not in $\{0,1,-1\}$. What do~(b) and~(c) tell you about $N$? \item[(f)] Use~(e) to find $N$, with help from Maple if necessary. (Start by entering \verb+with(numtheory):+; then you can use the notation \verb+cyclotomic(n,x)+ for $\vph_n(x)$.) \end{itemize} \end{exercise} \begin{solution} We will write $\mu_k$ for the set of all $k$th roots of unity, and $\mu_k^\tm$ for the subset of primitive roots. \begin{itemize} \item[(a)] Note that $\zt^k=1$ if and only if $\ov{\zt}^k=1$, so $\zt$ and $\ov{\zt}$ have the same order. In other words, $\zt$ is a primitive $m$th root of unity if and only if $\ov{\zt}$ is a primitive $m$th root of unity. Now suppose that $m>2$. The only roots of unity on the real axis are $+1$ (of order $1$) and $-1$ (of order $2$), so all primitive $m$th roots of unity have nonzero imaginary part. Our first observation shows that the roots with positive imaginary part biject with those of negative imaginary part, so the total number of roots is even. This number is the same as the degree of $\vph_m(x)$. \item[(b)] We can write $n=2m$, where $m$ is odd. Suppose that $\zt\in\mu_n^\tm$, so $\zt^k=1$ if and only if $n|k$. This means that $\zt^m\neq 1$, but $(\zt^m)^2=\zt^n=1$, so we must have $\zt^m=-1$. This means that $(-\zt)^m=(-1)^m\zt^m=(-1)^{m+1}$, which is $1$ because $m$ is odd. On the other hand, if $(-\zt)^k=1$ then $\zt^{2k}=(-\zt)^{2k}=1^2=1$, so $2k$ must be divisible by $n=2m$, so $k$ must be divisible by $m$. This proves that $-\zt\in\mu_m^\tm$. Conversely, suppose that $-\zt\in\mu_m^\tm$. As $m$ is odd we then have $\zt^m=(-1)^m(-\zt)^m=-1$, and thus $\zt^n=(\zt^m)^2=1$, so $\zt\in\mu_n$. On the other hand, if $\zt^k=1$ then $(-\zt)^{2k}=(\zt^k)^2=1$, so $2k$ is divisible by $m$. As $m$ is odd this can only happen if $k$ is divisible by $m$, say $k=mj$. This means that $\zt^k=(\zt^m)^j=(-1)^j$, but we also assumed that $\zt^k=1$, so $j$ must be even. As $k=mj$ this means that $k$ is divisible by $2m=n$. This shows that $\zt\in\mu_n^\tm$. Next, $\vph_m(x)$ is the product of the terms $x-\zt$ for $\zt\in\mu_m^\tm$, so $\vph_m(-x)$ is the product of the corresponding terms $-x-\zt$. The number of terms here is $|\mu_m^\tm|$, which is even, by part~(a). It therefore does not matter if we change all the signs, so $\vph_m(x)$ is the product of the terms $x+\zt$. Now $x+\zt=x-(-\zt)$, and $\{-\zt\st\zt\in\mu_m^\tm\}=\mu_n^\tm$, so we see that $\vph_m(-x)=\vph_n(x)$. \item[(c)] We can write $n=p^2m$ for some $m$, so $n/p=mp$. Suppose that $\zt\in\mu_n^\tm$. Then $(\zt^p)^{mp}=\zt^n=1$. On the other hand, if $(\zt^p)^k=\zt^{pk}=1$, then $pk$ must be divisible by $p^2m$, so $k$ must be divisible by $pm$. It follows that $\zt^p\in\mu_{pm}^\tm$. Conversely, suppose that $\zt^p\in\mu_{mp}^\tm$. It is then clear that $\zt^n=(\zt^p)^{mp}=1$, so $\zt\in\mu_n$. On the other hand, suppose that $\zt^k=1$. Then $(\zt^p)^k=1$, so $k$ is divisible by $mp$, say $k=mpj$. Now the original relation $\zt^k=1$ can be written as $(\zt^p)^{mj}=1$, so $mj$ must be divisible by $mp$, say $mj=mpi$. It follows that $k=mpj=p.mj=mp^2i=ni$, so $k$ is divisible by $n$. This shows that $\zt\in\mu_n^\tm$ as claimed. Now note that $\vph_{n/p}(x^p)$ is the product of the terms $x^p-\xi$ for $\xi\in\mu^\tm_{n/p}$. Here $x^p-\xi$ can be rewritten as the product of the terms $x-\zt$, as $\zt$ runs over the $p$th roots of $\xi$. Thus, $\vph_{n/p}(x^p)$ is the product of all terms $x-\zt$ for which $\zt^p\in\mu_{n/p}^\tm$, or equivalently (by what we just proved) $\zt\in\mu_n^\tm$. This means that $\vph_{n/p}(x^p)=\vph_n(x)$. \item[(d)] If we start with $\vph_p(x)$ and apply~(c) repeatedly we can find $\vph_{p^k}(x)$ for all $k$ (and any prime $p$). If $p$ is odd we can then use~(b) to find $\vph_{2p^k}(x)$, and then we can use method~(c) at the prime $2$ to find $\vph_{4p^k}(x)$, $\vph_{8p^k}(x)$ and so on. Eventually this gives $\vph_{2^ip^j}(x)$ for all $i$ and $j$. If $p$ and $q$ are distinct odd primes, then we cannot find $\vph_{pq}(x)$ by this method. In particular, the first case that we do not cover is $\vph_{15}(x)$. However, if we compute $\vph_{pq}(x)$ by some other method then using~(b) and~(c) we can find $\vph_{2^ip^jq^k}(x)$. \item[(e)] Let $N$ be the smallest number such that $\vph_N(x)$ has a coefficient not in $\{0,1,-1\}$. If $N$ is divisible by $p^2$ for some prime $p$, then $\vph_N(x)=\vph_{N/p}(x^p)$ by~(c). Here $N/p 0 then print([n,sort(f)]); break; fi: od: \end{verbatim} \end{itemize} \end{solution} \begin{exercise}\exlabel{ex-phi-pq} Let $p$ and $q$ be distinct odd primes, and consider the power series \[ f(x) = \sum_{i=0}^{q-1}\sum_{j=0}^{p-1}\sum_{k=0}^\infty (x^{ip+jq+kpq}-x^{1+ip+jq+kpq}). \] Prove that $f(x)=\vph_{pq}(x)$ (so in particular, enough terms must cancel to make $f(x)$ a polynomial). \end{exercise} \begin{solution} We can reorganise the definition and use the geometric progression formula as follows: \begin{align*} f(x) &= (1-x)\left(\sum_{i=0}^{q-1}x^{ip}\right) \left(\sum_{j=0}^{p-1}x^{jq}\right) \left(\sum_{k=0}^\infty x^{kpq}\right) \\ &= (1-x)\frac{x^{pq}-1}{x^p-1} \frac{x^{pq}-1}{x^q-1}\frac{1}{1-x^{pq}} = \frac{(x-1)(x^{pq}-1)}{(x^p-1)(x^q-1)} \\ &= \frac{\vph_1(x)\vph_{pq}(x)\vph_p(x)\vph_q(x)\vph_1(x)} {\vph_p(x)\vph_1(x)\vph_q(x)\vph_1(x)} = \vph_{pq}(x). \end{align*} Now consider an arbitrary natural number $m$. The element $m/p\in\F_q$ is represented by some $i\in\{0,\dotsc,q-1\}$, and the element $m/q\in\F_p$ is represented by some $j\in\{0,\dotsc,p-1\}$. We find that $m-(ip+jq)$ is divisible by both $p$ and $q$, so $m=ip+jq+kpq$ for some $k\in\Z$. We define $\lm(m)$ to be $1$ if $k\geq 0$, and $0$ if $k<0$. Note that $ip+jq\leq(q-1)p+(p-1)q<2pq$, so $\lm(m)=1$ for $m\geq 2pq$. The definition of $f(x)$ can now be rewritten as \[ f(x) = \sum_{m=0}^\infty \lm(m)(x^m-x^{m+1}) = \sum_{m=0}^\infty (\lm(m)-\lm(m-1)) x^m. \] It follows that all the coefficients of $f(x)$ are in $\{0,1,-1\}$. We also see that for $m>2pq$ we have $\lm(m)-\lm(m-1)=1-1=0$, so $f(x)$ is a polynomial as expected. \end{solution} \begin{exercise}\exlabel{ex-fifth-root} Let $\zt$ be a primitive 5th root of unity, and let $\al$ denote the real 5th root of 2. You are given that $\Q(\zt,\al)$ is the splitting field of $x^5-2$ over $\Q$ and that $[\Q(\zt,\al):\Q]=20$. \begin{itemize} \item Specify the elements of $\Gal(\Q(\zt,\al)/\Q)$ by writing down how they act on $\zt$ and on $\al$. \item Show that there exist automorphisms $\phi$, $\psi\in\Gal(\Q(\zt,\al)/\Q)$ such that $\phi$ has order 4, $\psi$ has order 5, and $\Gal(\Q(\zt,\al)/\Q)=\langle\phi,\psi\rangle$. \item Write $\phi\psi\phi^{-1}$ in the form $\phi^i\psi^j$. \item Recall that if $\bt=\zt+\frac{1}{\zt}$, then $\Q(\bt)=\Q(\sqrt{5})$. Under the Galois correspondence, what should be the order of the corresponding subgroup $\Gal(\Q(\zt,\al)/\Q(\bt))$? \item Show that the group $\Gal(\Q(\zt,\al)/\Q(\bt))$ is $\langle\phi^2,\psi\rangle$. \end{itemize} \end{exercise} \begin{solution} \begin{itemize} \item Any automorphism is uniquely determined by its effect on $\al$ and on $\zt$. The image of $\al$ must be a root of $x^5-2$, so must be one of $\al$, $\zt\al$, $\zt^2\al$, $\zt^3\al$ or $\zt^4\al$. In the same way, the image of $\zt$ must be another primitive 5th root of unity, i.e., a root of $\vph_5$, so is one of $\zt$, $\zt^2$, $\zt^3$ or $\zt^4$. This gives 20 possible automorphisms, $\theta_{ij}$ say, defined by \begin{eqnarray*} \theta_{ij}(\zt)&=&\zt^i\\ \theta_{ij}(\al)&=&\zt^j\al \end{eqnarray*} for $i=1$, 2, 3 or 4 and $j=0$, 1, 2, 3 or 4. As the extension $\Q(\zt,\al)/\Q$ is Galois and has degree 20, these are all of the automorphisms. \item The automorphism $\psi$ which fixes $\zt$ and maps $\al$ to $\zt\al$ is clearly of order 5. The automorphism $\phi$ which fixes $\al$ and maps $\zt$ to $\zt^2$ is of order 4 because $\phi^2(\zt)=\phi(\zt^2)=\zt^4$, and so $\phi^4(\zt)=\phi^2(\zt^4)=(\zt^4)^4=\zt$. The group generated by $\phi$ and $\psi$ has as subgroups $\langle\phi\rangle$ and $\langle\psi\rangle$ so its order must be a multiple of 4 and of 5 by Lagrange's Theorem. It follows that this group must have order 20, so is the whole Galois group. \item We have: \begin{eqnarray*} &\phi\psi\phi^{-1}(\al)=\phi\psi(\al)=\phi(\zt\al)=\phi(\zt)\phi(\al)=\zt^2.\al\\ &\phi\psi\phi^{-1}(\zt)=\phi\psi(\zt^3)=\phi(\zt^3)=\zt \end{eqnarray*} It follows that $\phi\psi\phi^{-1}=\psi^2$. \item We see that $$\zt^2+\zt+1+\zt^{-1}+\zt^{-2}=0.$$ Rearranging, we get $$(\zt+\frac{1}{\zt})^2+(\zt+\frac{1}{\zt})-1=0.$$ It follows that $\bt$ is a root of $X^2+X-1$, and so $\bt=\frac{-1\pm\sqrt{5}}{2}$, from the quadratic formula. It is then easy to see that $\Q(\bt)=\Q(\sqrt{5})$. $[\Q(\bt):\Q]=2$, so the index of the corresponding subgroup of $\Gal(M/\Q)$ must be 2, so its order must be 10. \item The group $\langle\phi^2,\psi\rangle$ is of order 10 (it contains an element of order 2, and an element of order 5, so its order must be a multiple of 10~--~but it isn't the whole group, as it doesn't contain $\phi$). Let $G$ be the subgroup associated to $\Q(\bt)$. If we can show that $\bt$ is fixed by both $\phi^2$ and by $\psi$, we will know that $\langle\phi^2,\psi\rangle\subseteq G$. But by the previous part of the question, $|G|=10$, and so we have to have $G=\langle\phi^2,\psi\rangle$, as required. But this is easy to check: \begin{eqnarray*} &\phi^2(\bt)=\phi^2(\zt)+\frac{1}{\phi^2(\zt)}=\zt^{-1}+\frac{1}{\zt^{-1}}=\frac{1}{\zt}+\zt=\bt\\ &\psi(\bt)=\psi(\zt)+\frac{1}{\psi(\zt)}=\zt+\frac{1}{\zt}=\bt. \end{eqnarray*} \end{itemize} \end{solution} \begin{exercise}\exlabel{ex-forty-two} Let $L$ be the splitting field of $x^7-3$ over $\Q$. You know that $[L:\Q]=42$. Calculate the elements of $\Gal(L/\Q)$. Find $\psi$, $\phi\in\Gal(L/\Q)$ which satisfy: \begin{itemize} \item $\psi$ has order 7, $\phi$ has order 6 \item $\phi\psi\phi^{-1}=\psi^3$ \item $\Gal(L/\Q)=\langle\phi,\psi\rangle$ \end{itemize} \end{exercise} \begin{solution} \begin{itemize} \item $L=\Q(\al,\zt)$, where $\zt=e^{{2\pi i}/{7}}$ and $\al$ is the real 7th root of 3. Any automorphism must send $\zt$ to another primitive 7th root of unity, and send $\al$ to a 7th root of 3. There is an automorphism $\psi$ which fixes $\zt$ but maps $\al$ to $\zt\al$. Clearly $\psi$ is of order 7, as doing $\psi$ seven times fixes $\al$. Further, there is an automorphism $\phi$ which fixes $\al$ but sends $\zt$ to $\zt^3$. Applying $\phi$ successively to $\zt$ we see that $\zt$ is sent successively to $$\zt\mapsto\zt^3\mapsto\zt^2\mapsto\zt^6\mapsto\zt^4\mapsto\zt^5\mapsto\zt\mapsto\cdots$$ so $\phi$ has order 6. \item Further, $$\phi\psi\phi^{-1}(\al)=\phi\psi(\al)=\phi(\zt\al)=\phi(\zt)\phi(\al)=\zt^3\al=\psi^3(\al)$$ and $$\phi\psi\phi^{-1}(\zt)=\phi\psi(\zt^5)=\phi(\zt^5)=\zt=\psi^3(\zt)$$ Thus $\phi\psi\phi^{-1}=\psi^3$. \item Finally, it remains to see that $\phi$ and $\psi$ generate the whole Galois group. But the Galois group has order 42, and the subgroup generated by $\phi$ and $\psi$ has order which is a multiple of both 6 and 7, so it must be the whole group. \end{itemize} \end{solution} \section{Finite fields} \label{sec-finite-fields} We now divert temporarily from our main focus on fields of characteristic zero, and instead discuss finite fields. It turns out that the relevant theory is quite closely related to that of cyclotomic fields. \begin{theorem}\lbl{thm-finite-fields} \begin{itemize} \item[(a)] There is a finite field of order $n$ if and only if $n=p^r$ for some prime $p$ and $r>0$. \item[(b)] If $K$ is a field of order $p^r$ then $K$ has characteristic $p$, and $K^\tm\simeq C_{p^r-1}$. Moreover, the function $\sg(a)=a^p$ defines an automorphism of $K$, called the \emph{Frobenius automorphism}. \item[(c)] If $K$ and $L$ are fields of the same order then they are isomorphic. \item[(d)] If $|L|=p^{rs}$ then the set $K=\{a\in L\st a^{p^r}=a\}$ is a subfield of $L$, and is the unique subfield of order $p^r$. Moreover, this procedure gives all the subfields of $L$. \item[(e)] If $K$ and $L$ are as above, then $L$ is normal over $K$, and $G(L/K)$ is cyclic of order $s$, generated by $\sg^r$. \end{itemize} \end{theorem} The proof will be given at the end of this section; it will consist of collecting together a number of smaller results that we will prove separately. We first discuss a few examples. \begin{example}\lbl{eg-finite-misc} We have already seen the fields $\F_p=\Z/p\Z$ for $p$ prime, and Example~\ref{eg-F-four} exhibited a field $\F_4$ of order four. Now suppose that $p>2$, and consider the ring $\F_p[i]$ of ``mod $p$ complex numbers'', as discussed in Exercise~\ref{ex-Ri-field}. The elements of $\F_p[i]$ have the form $a+bi$, with $a,b\in\F_p$, and the multiplication rule is \[ (a+bi)(c+di) = (ac-bd)+(ad+bc)i. \] We saw in Exercise~\ref{ex-Ri-field} that $\F_3[i]$ is a field (of order $9$) but that $\F_2[i]$ and $\F_5[i]$ are not fields. More generally, we will see in Proposition~\ref{prop-Fpi} that $\F_p[i]$ is a field (of order $p^2$) if and only if $p=3\pmod{4}$. \end{example} \begin{lemma}\lbl{lem-finite-field-order} Let $K$ be a finite field. Then $K$ has characteristic $p>0$ for some prime $p$, and $|K|=p^r$ for some $r>0$. \end{lemma} \begin{proof} As $K$ is finite, the elements $n.1$ (for $n\in\N$) cannot all be different. It follows that there exist integers $n 0$, and so $\chr(K)$ is a prime $p$ by Proposition~\ref{prop-char}. It therefore follows from Proposition~\ref{prop-hom-char} that $K$ contains a copy of $\F_p$. Note that the whole of $K$ is certainly a finite spanning set for $K$ over $\F_p$, so $K$ is finite-dimensional over $\F_p$, with dimension $r$ say. This means that $K\simeq\F_p^r$ and so $|K|=p^r$. As $1\neq 0$ in $K$ (by one of the field axioms) we have $|K|>1$ and so $r>0$. \end{proof} \begin{lemma}\lbl{prop-frobenius-exists} Let $K$ be a finite field of order $p^r$. Then the function $\sg(a)=a^p$ defines an automorphism of $K$. \end{lemma} \begin{proof} It is clear that $\sg(0)=0$ and $\sg(1)=1$ and $\sg(ab)=\sg(a)\sg(b)$. We also see from Lemma~\ref{lem-F-additive} that $\sg(a+b)=\sg(a)+\sg(b)$. This means that $\sg$ is a homomorphism from $K$ to $K$. Now suppose that $K=\{a_1,\dotsc,a_{p^r}\}$. We see from Proposition~\ref{prop-hom-inj} that $\sg$ is injective, so the $p^r$ elements $\sg(a_1),\dotsc,\sg(a_{p^r})$ are all different, so between them they must cover all the $p^r$ elements of $K$. This means that $\sg$ is also surjective, so it is an isomorphism as required. \end{proof} \begin{remark}\lbl{rem-frobenius-powers} We observe that \begin{align*} \sg(a) &= a^p \\ \sg^2(a) &= \sg(\sg(a)) = (a^p)^p = a^{p\tm p} = a^{p^2} \\ \sg^3(a) &= \sg(\sg^2(a)) = (a^{p^2})^p = a^{p^2\tm p} = a^{p^3} \\ \sg^4(a) &= \sg(\sg^3(a)) = (a^{p^3})^p = a^{p^3\tm p} = a^{p^4} \end{align*} and so on; in general, $\sg^r(a)=a^{p^r}$. \end{remark} \begin{lemma}\lbl{lem-cyclotomic-coprime} Suppose that $p$ is prime and $r>0$ and put $q=p^r$. If $f(x),g(x)\in\F_p[x]$ and $x^q-x$ is divisible by $f(x)g(x)$, then $f(x)$ and $g(x)$ are coprime. \end{lemma} \begin{proof} We have $x-x^q=f(x)g(x)h(x)$ for some $h$. Taking derivatives gives \[ 1-qx^{q-1} = f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x). \] The left hand side is just $1$, because we are working mod $p$. We can rewrite the right hand side in terms of the polynomials $a(x)=g'(x)h(x)+g(x)h'(x)$ and $b(x)=f'(x)h(x)$ to get \[ 1= a(x)f(x)+b(x)g(x), \] showing that $f(x)$ and $g(x)$ are coprime. \end{proof} \begin{lemma}\lbl{lem-Fq-exists} Suppose again that $p$ is prime and $r>0$ and $q=p^r$. Let $f(x)\in\F_p[x]$ be an irreducible factor of the mod $p$ reduction of $\vph_{q-1}(x)$, and put $K=\F_p[x]/f(x)$. Then $K$ is a field with $|K|=q$, and $K^\tm$ is cyclic of order $q-1$. \end{lemma} \begin{proof} We write $\al$ for the image of $x$ in $K$, so $f(\al)=0$. As $f(x)$ is irreducible, we see from Corollary~\ref{cor-quotient-field} that $K$ is a field and $K=\F_p(\al)$. If $f(x)$ has degree $s$ we also see from Proposition~\ref{prop-quotient-basis} that $K\simeq\F_p^s$ as vector spaces over $\F_p$, so in particular $|K|=p^s$. As $f(x)\;|\;\vph_{q-1}(x)\;|\;x^{q-1}-1\;|\;x^q-x$, we see that $\al^q=\al$. Here $q=p^r$ and so one checks that $\sg^r(t)=t^{p^r}=t^q$, so we see that $\sg^r(\al)=\al$. Now put $K'=\{a\in K\st\sg^r(a)=a\}$. We see from Proposition~\ref{prop-fixed-subfield} that $K'$ is a subfield of $K=\F_p(\al)$, and it contains $\al$ so it must be all of $K$. This means that every element in $K$ is a root of the polynomial $g(x)=x^q-x$. However, $g(x)$ has degree $q$ and so cannot have more than $q$ roots in any field. We must therefore have $|K|\leq q$. We next consider the order of $\al$ in $K^\tm$. As explained above we have $f(x)\;|\;x^{q-1}-1$ and so $\al^{q-1}=1$, so the order of $\al$ divides $q-1$. Write $r$ for this order, and suppose (for a contradiction) that $r ->}[r] \dto & U \dto_\al \ar@{->>}[rr] & & U/U[n/d] \dto_\simeq^{\ov{\al}} \\ 1 \ar@{ >->}[r] & U & U[d] \ar@{ >->}[l] & \img(\al) \ar@{ >->}[l] } \] \end{proof} \begin{lemma}\lbl{lem-cyclic-test} Let $U$ be a finite abelian group such that $|U[d]|\leq d$ for all $d$. Then $U$ is cyclic. \end{lemma} \begin{proof} Put $n=|U|$ and let $C$ be a cyclic group of order $n$; we will compare $U$ with $C$. Put \[ U\ip{d} = \{x\in U\st x \text{ has exact order } d\}. \] Note that $x^d=1$ if and only if the exact order of $x$ is a divisor of $d$. Using this together with Lemma~\ref{lem-cyclic-test-aux} we see that $d=|U[d]|=\sum_{e|d}|U\ip{e}|$, so \[ |U\ip{d}| = d - \sum_{e|d,e 0$, so $|U\ip{n}|>0$. If $x$ is any element of $U\ip{n}$ then $x$ generates a cyclic subgroup of $U$ of order $n$, which must therefore be $U$ itself. Thus $U$ is cyclic as claimed. \end{proof} \begin{remark}\lbl{rem-classify} We have chosen to give a proof that does not depend on the classification of finite abelian groups. Readers who are familiar with that classification may prefer to proceed as follows. The general theory implies that there is a unique sequence $d_1,\dotsc,d_r$ of integers with $d_k>0$ and $d_1|d_2|\dotsb|d_r$, such that $U$ is isomorphic to $\prod_{k=1}^rC_{d_k}$. In particular, $U$ is cyclic if and only if $r=1$, so we must show that this is the case. As $d_1$ divides $d_k$ for all $k$, we see that each cyclic factor $C_{d_k}$ contains a copy of $C_{d_1}$, and thus $U[d_1]\simeq C_{d_1}^r$ and $|U[d_1]|=d_1^r$. By assumption we have $|U[d_1]|\leq d_1$, so we must have $r=1$ as required. \end{remark} \begin{corollary}\lbl{cor-units-cyclic} If $K$ is a finite field of order $q$ then $K^\tm$ is cyclic of order $q-1$. \end{corollary} \begin{proof} This is immediate from Proposition~\ref{prop-units-cyclic}. \end{proof} We now pause to justify the claims made in Example~\ref{eg-finite-misc}. \begin{proposition}\lbl{prop-Fpi} Let $p$ be a prime. Then $\F_p[i]$ is a field if and only if $p=3\pmod{4}$. \end{proposition} \begin{proof} We first dispose of the case $p=2$. There $p\neq 3\pmod{4}$, and the ring $\F_2[i]=\{0,1,i,1+i\}$ is not a field because the element $1+i$ has no inverse. From now on we assume that $p$ is odd, so either $p=1\pmod{4}$ or $p=3\pmod{4}$. We will say that $p$ is \emph{bad} if $\F_p[i]$ is not a field. We must show that $p$ is bad if and only if $p=1\pmod{4}$. We next claim that $p$ is bad if and only if there is an element $a\in\F_p$ with $a^2=-1$. Indeed, if there exists such an $a$ then we have $a+i,a-i\neq 0$ but $(a+i)(a-i)=a^2-i^2=(-1)-(-1)=0$ so $\F_p[i]$ is not a field, so $p$ is bad. On the other hand, if there is no such $a$ then the polynomial $f(x)=x^2+1$ has no roots in $\F_p[x]$ and so is irreducible (because any nontrivial factor would have to have degree one). It therefore follows that $\F_p[x]/f(x)$ is a field, which is easily seen to be isomorphic to $\F_p[i]$; so $p$ is good. Next, Corollary~\ref{cor-units-cyclic} tells us that $\F_p^\tm$ is a cyclic group of order $p-1$. If $p$ is bad then there is an element $a$ with $a^2=-1$ so the subgroup generated by $a$ is $\{1,a,-1,-a\}$, which has order $4$. It follows by Lagrange's theorem that $p-1$ is divisible by $4$, so $p=1\pmod{4}$. Conversely, suppose that $p=1\pmod{4}$, so $(p-1)/4$ is an integer. Choose a generator $b$ for the cyclic group $\F_p^\tm$, and put $a=b^{(p-1)/4}$. The powers $1,b,\dotsc,b^{p-2}$ are then distinct, so we see that $a^2=b^{(p-1)/2}\neq 1$ but $a^4=b^{p-1}=1$. This means that $(a^2+1)(a^2-1)=a^4-1=0$ but $a^2-1\neq 0$ so $a^2+1=0$, which implies that $p$ is bad. \end{proof} \begin{proposition}\lbl{prop-factor} Let $K$ be a finite field of order $q=p^r$. Then $\prod_{\al\in K}(x-\al)=x^q-x$. \end{proposition} \begin{proof} We have $|K^\tm|=q-1$, so for all $\al\in K^\tm$ we have $\al^{q-1}=1$. It follows that for all $\al\in K$ we have $\al^q-\al=0$. Thus the elements of $K$ give $q$ distinct roots of $x^q-x$, and it follows that $x^q-x=\prod_\al(x-\al)$. \end{proof} \begin{proposition}\lbl{prop-unique} Let $K$ and $L$ be fields of order $q=p^n$. Then $K\simeq L$. \end{proposition} \begin{proof} We have seen that $K^\tm$ is cyclic, generated by some element $\al$, say. We then have a surjective homomorphism $\ep\:\F_p[x]\xra{}K$ given by $x\mapsto\al$. Let $f(x)$ be the minimal polynomial of $\al$ over $\F_p$, or in other words, the monic generator of $\ker(\ep)$. Then $\al$ induces an isomorphism $\ov{\ep}\:\F_p[x]/f(x)\xra{}K$. Moreover, $f(x)$ is non-constant and divides $x^q-x$, which factors in $L[x]$ as $\prod_{\bt\in L}(x-\bt)$. It follows that $f(\bt)=0$ for some $\bt\in L$. We can therefore define $\phi\:\F_p[x]/f(x)\xra{}L$ by $\phi(x)=\bt$. Now the map $\psi=\phi\circ\ov{\ep}^{-1}\:K\xra{}L$ is a homomorphism of fields, and so is injective. As $|K|=|L|=q$, it follows that $\psi$ must be a bijection, and thus an isomorphism. \end{proof} \begin{proposition}\lbl{prop-finite-subfield} Let $L$ be a field of order $p^{rs}$. Then the subset $K=\{a\in L\st a^{p^r}=a\}$ is a subfield of $L$, and is the unique subfield of order $p^r$. Moreover, we have $[L:K]=s$. \end{proposition} \begin{proof} Using Remark~\ref{rem-frobenius-powers} we see that $K=\{a\in L\st\sg^r(a)=a\}=L^{\{\sg^r\}}$, which is a subfield by Proposition~\ref{prop-fixed-subfield}. Next, put $f_k(t)=t^{p^k}-t$, so that $K$ is the set of roots of $f_r(t)$ in $L$, whereas $f_{rs}(t)=\prod_{\al\in L}(t-\al)$. We claim that $f_r(t)$ divides $f_{rs}(t)$. To see this, consider the standard identity \[ u^m-1 = (u-1)(1+u+\dotsb+u^{m-1})=(u-1)\sum_{i=0}^{m-1}u^i. \] Put $u=t^{p^r-1}$ and \[ m = \frac{p^{rs}-1}{p^r-1} = 1+p^s+p^{2s}+\dotsb+p^{(r-1)s}\in\N \] so that $u^m=t^{p^{rs}-1}$; we find that $t^{p^{rs}-1}-1$ is divisible by $t^{p^r-1}-1$, and we can multiply by $t$ to see that $f_{rs}(t)$ is divisible by $f_r(t)$ as claimed. As $f_{rs}(t)$ splits in $L$ and has distinct roots, we see from Proposition~\ref{prop-split-factor} that $f_r(t)$ is also split and has distinct roots. This means that the number of roots of $f_r(t)$ is precisely equal to its degree, so $|K|=p^r$. Now if $K'$ is any other subfield of order $p^r$ we can apply Proposition~\ref{prop-factor} to $K'$ to see that $K'$ is the set of roots of $f_r(t)$, so $K'=K$. Finally, put $t=[L:K]=\dim_K(L)$, which means that $L$ is isomorphic to $K^t$ as vector spaces over $K$. We have $|K^t|=|K|^t=p^{rt}$ whereas $|L|=p^{rs}$; it follows that $t=s$ as claimed. \end{proof} \begin{corollary}\lbl{cor-finite-galois} If $K$ and $L$ are as in Proposition~\ref{prop-finite-subfield} then $L$ is normal over $K$ and $G(L/K)$ is cyclic of order $s$, generated by $\sg^r$. \end{corollary} \begin{proof} First, we have seen that $L$ is the set of roots of the polynomial $f_{rs}(t)=t^{p^{rs}}-t\in\F_p[t]\sse K[t]$, so it is the splitting field over $K$ of that polynomial, so it is normal over $K$. It follows as in Lemma~\ref{lem-splitting-ext} that $|G(L/K)|=[L:K]=s$. Next, we have seen that $L^\tm$ is cyclic, of order $p^{rs}-1$. Choose an element $\bt$ that generates $L^\tm$, so $L^\tm$ consists of the powers $\bt^i$ for $0\leq i 0$. We have seen that $a^{p^d}=a$ for all $a\in K$. Put $f(x)=x^{p^d}-x+1\in K[x]$, so $f(a)=1$ for all $a\in K$. It follows that $f(x)$ has no roots in $K$, so $K$ is not algebraically closed. \end{solution} \section{Multiquadratic extensions} \label{sec-mquad} We will next discuss another extended example, extending Proposition~\ref{prop-biquadratic} and related to Corollary~\ref{cor-mquad-cyclotomic}. Let $p_1,\dotsc,p_n$ be distinct prime numbers. For any subset $T\sse\{1,\dotsc,n\}$, put $r_T=\prod_{i\in T}\sqrt{p_i}$. For the case where $T$ is the empty set, this should be interpreted as $r_\emptyset=1$. We will allow ourselves to write $r_{245}$ rather than $r_{\{2,4,5\}}$ and so on. Let $K(m)$ be the $\Q$-linear span of all the numbers $r_T$ for $T\sse\{1,\dotsc,m-1\}$. For example: \begin{itemize} \item $K(0)$ should be interpreted as $\Q$. \item $K(1)$ is the set of all real numbers that can be written as $a_\emptyset+a_1\sqrt{p_1}$ for some rational numbers $a_\emptyset,a_1\in\Q$. \item $K(2)$ is the set of all real numbers that can be written as \[ a_\emptyset+a_1\sqrt{p_1}+a_2\sqrt{p_2}+a_{12}\sqrt{p_1p_2} \] for some rational numbers $a_\emptyset,a_1,a_2,a_{12}\in\Q$. \item In general, $K(m)$ could also be described as $\Q(\sqrt{p_1},\dotsc,\sqrt{p_m})$. \end{itemize} We will call fields of this type \emph{multiquadratic} extensions of $\Q$. Our main result in this section is as follows: \begin{theorem}\lbl{thm-mquad} \begin{itemize} \item[(a)] $K(m)$ is a subfield of $\R$. \item[(b)] The elements $\{r_T\st T\sse\{1,\dotsc,m\}\}$ form a basis for $K(m)$ over $\Q$ (so $K(m)$ has degree $2^m$ over $\Q$). \item[(c)] If $u\in K(m)$ and $u^2\in\Q$ then $u=u_Tr_T$ for some $T\sse\{1,\dotsc,m\}$ and $u_T\in\Q$. \end{itemize} \end{theorem} We pause to explore the meaning of this a little. Firstly, you can check that \[ 2 - 3 \sqrt{2} + 4\sqrt{3} -\sqrt{2}\sqrt{3} - \sqrt{5} = 0.000004822873256233\dotsc \simeq 0. \] Could there be any ``coincidental'' relationship between square roots that holds \emph{exactly}? Part~(b) of the theorem says that this is impossible. Next, suppose we have a nonzero real number $a$ that can be expressed in terms of the square roots of certain primes. The hardest part of part~(a) of the Theorem tells us that $1/a$ can also be expressed in terms of the square roots of the same primes. For example, if $a$ is the small number mentioned above, \[ a = 2 - 3 \sqrt{2} + 4\sqrt{3} -\sqrt{2}\sqrt{3} - \sqrt{5} \] it works out that \[ a^{-1} = 25918+18327\,\sqrt{2}+14964\,\sqrt{3}+10581\,\sqrt{2}\sqrt{3}+ 11591\,\sqrt{5}+8196\,\sqrt{2}\sqrt{5}+6692\,\sqrt{3}\sqrt{5}+ 4732\,\sqrt{2}\sqrt{3}\sqrt{5}. \] We now prove some preliminary results, which will lead in to the proof of Theorem~\ref{thm-mquad}. \begin{lemma}\lbl{lem-prod} If $T,U\sse\{1,\dotsc,m\}$ then $r_Tr_U\in K(m)$. More precisely, if $W=T\cap U$ and $V=(T\cup U)\sm W$ and $w=\prod_{i\in W}p_i\in\Q$ then $r_Tr_U=wr_V$. \end{lemma} \begin{proof} If $i\in W$ then $\sqrt{p_i}$ occurs in both $r_T$ and $r_U$ giving a factor of $p_i$ in $r_Tr_U$. If $i\in V$ then $\sqrt{p_i}$ occurs either in $r_T$ or in $r_U$ but not both, giving a factor of $\sqrt{p_i}$ in $r_Tr_U$. Thus, we have $r_Tr_U=wr_V$. As $K(m)$ was defined to be the $\Q$-linear span of a set of elements including the element $r_V$, it follows that $r_Tr_U\in K(m)$. \end{proof} \begin{lemma}\lbl{lem-mquad-subring} $K(m)$ is a subring of $\R$. \end{lemma} \begin{proof} We must show that $K(m)$ contains $0$ and $1$ and that it is closed under addition, subtraction and multiplication. Note that $K(m)$ was defined as a span, so it is certainly a $\Q$-linear subspace of $\R$, so it contains $0$ and is closed under addition and subtraction and under multiplication by elements of $\Q$. We can think of $1$ as $r_\emptyset$, so we also have $1\in K(m)$. Now consider elements $a,b\in K(m)$. From the definition of $K(m)$, we can write $a=\sum_Ta_Tr_T$ and $b=\sum_Ub_Ur_U$ for some numbers $a_T,b_U\in\Q$. It follows that \[ ab = \sum_{T,U} a_Tb_U r_Tr_U. \] Here $r_Tr_U\in K(m)$ by Lemma~\ref{lem-prod}, and $a_Tb_U$ is just a rational number, so $a_Tb_Ur_Tr_U\in K(m)$. Moreover, $K(m)$ is closed under addition, so $\sum_{T,U}a_Tb_Ur_Tr_U\in K(m)$, or in other words $ab\in K(m)$. Thus $K(m)$ is closed under multiplication, as required. \end{proof} We could now use Proposition~\ref{prop-subring-subfield} to show that $K(m)$ is a subfield of $\R$. However, we will instead give a more direct and elementary argument, which might be considered more illuminating. \begin{lemma}\lbl{lem-not-square} Suppose that $T\sse\{1,\dotsc,m-1\}$ and $u_T\in\Q$. Then $(u_Tr_T)^2\neq p_m$. \end{lemma} \begin{proof} Suppose that $(u_Tr_T)^2=p_m$; we will derive a contradiction. Clearly we must have $u_T\neq 0$, so we can write $u_T=\pm u/v$, where $u$ and $v$ are positive integers with no common factors. We then have \[ p_m=(u_Tr_T)^2 = u^2r_T^2/v^2=u^2v^{-2}\prod_{i\in T}p_i, \] so \[ p_mv^2 = u^2\prod_{i\in T}p_i. \] This is now an equation in $\Z$; it implies that $p_m$ divides $u^2\prod_Tp_i$. By assumption the primes $p_i$ on the right hand side are all different from $p_m$, so $p_m$ must divide $u$ instead. We can write $u=p_mw$ and rearrange to get \[ v^2 = p_mw^2\prod_{i\in T}p_i. \] Here the right hand side is divisible by $p_m$, so the left hand side must be divisible by $p_m$, so $v$ must be divisible by $p_m$. This contradicts the fact that $u$ and $v$ have no common factors. \end{proof} \begin{lemma}\lbl{lem-step} Suppose that Theorem~\ref{thm-mquad} holds for $K(m-1)$, and that $b,c\in K(m-1)$. Put $a=b+c\sqrt{p_m}$ and $a'=b-c\sqrt{p_m}$, so $a,a'\in K(m)$. Then \begin{itemize} \item $aa'=b^2-c^2p_m\in K(m-1)$ \item If $aa'=0$ then $b=c=0$ and so $a=0$. \item If $aa'\neq 0$ then $1/a\in K(m)$. \end{itemize} \end{lemma} \begin{proof} It is simple algebra to check that $aa'=b^2-c^2p_m$. As $b,c\in K(m-1)$ and $p_m\in\Z$ it follows that $aa'\in K(m-1)$. Now suppose that $aa'=0$, so $b^2=c^2p_m$. Suppose that $c$ is nonzero, so $p_m=(b/c)^2$. By assumption $K(m-1)$ is a field, so the element $u=b/c$ lies in $K(m-1)$, and $u^2=p_m\in\Q$. Part~(c) of Theorem~\ref{thm-mquad} tells us that $u=u_Tr_T$ for some $T\sse\{1,\dotsc,m-1\}$ and $u_T\in\Q$, so $(u_Tr_T)^2=p_m$. Lemma~\ref{lem-not-square} tells us that this is impossible. This contradiction means that we must in fact have $c=0$. We also have $b^2=c^2p_m$, so it follows that $b=0$ as well. Now suppose instead that the element $aa'$ is nonzero. As $K(m-1)$ is a field and $aa'\in K(m-1)$ it follows that $(aa')^{-1}\in K(m-1)\sse K(m)$. As $K(m)$ is a subring of $\R$ and $a', (aa')^{-1}\in K(m)$ it follows that $a'.(aa')^{-1}\in K(m)$; but $a'.(aa')^{-1}=a^{-1}$, so $a^{-1}\in K(m)$ as claimed. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm-mquad}] We can assume by induction that the theorem holds for $K(m-1)$ (as the initial case of $K(0)$ is trivial). \begin{itemize} \item[(b)] The elements $r_T$ span $K(m)$ by definition, so we need only show that they are linearly independent. Suppose we have rational numbers $a_T$ for all $T\sse\{1,\dotsc,m\}$, giving an element $a=\sum_Ta_Tr_T\in K(m)$. We must show that if $a=0$, then the individual coefficients $a_T$ are all zero. We put \begin{align*} b &= \sum_{U\sse\{1,\dotsc,m-1\}} a_Ur_U \in K(m-1) \\ c &= \sum_{U\sse\{1,\dotsc,m-1\}} a_{U\cup\{m\}}r_U \in K(m-1) \end{align*} so that $a=b+c\sqrt{p_m}$. We then put $a'=b-c\sqrt{p_m}$ as in Lemma~\ref{lem-step}. If $a=0$ then certainly $aa'=0$ so the Lemma tells us that $b=c=0$. As $b=0$ we have $\sum_{U\sse\{1,\dotsc,m-1\}}a_Ur_U=0$ but the set $\{r_U\st U\sse\{1,\dotsc,m-1\}\}$ is linearly independent by our inductive assumption, so we must have $a_U=0$ for all $U\sse\{1,\dotsc,m-1\}$. By applying the same logic to $c$, we see that $a_{U\cup\{m\}}$ is also zero for all $U\sse\{1,\dotsc,m-1\}$. These two cases cover all the coefficients $a_T$, so $a_T=0$ for all $T\sse\{1,\dotsc,m\}$, as required. \item[(a)] We showed in Lemma~\ref{lem-mquad-subring} that $K(m)$ is a subring of $\R$, so all that is left is to show that if $a\in K(m)$ is nonzero then $a^{-1}$ is also in $K(m)$. We can write it as $a=b+c\sqrt{p_m}$ and put $a'=b-c\sqrt{p_m}$, just as before. If $aa'=0$ then Lemma~\ref{lem-step} tells us that $a=0$, contrary to assumption. Thus $aa'\neq 0$ and the other part of Lemma~\ref{lem-step} tells us that $a^{-1}\in K(m)$, as required. \item[(c)] Suppose that $u\in K(m)$ and $u^2=q\in\Q$. Just as above, we can write $u=x+y\sqrt{p_m}$ with $x,y\in K(m-1)$. It follows that $(x^2+p_my^2-q)+2xy\sqrt{p_m}=u^2-q=0$. Here $x^2+p_my^2-q$ and $2xy$ are in $K(m-1)$, and it follows easily from part~(a) that $\{1,\sqrt{p_m}\}$ is a basis for $K(m)$ over $K(m-1)$. We must therefore have $x^2+p_my^2-q=0$ and $2xy=0$, so either $x=0$ or $y=0$. Suppose that $y=0$, so the equation $x^2+p_my^2-q=0$ reduces to $u^2=x^2=q$. This means that $u\in K(m-1)$ and $u^2\in\Q$, so part~(c) of the theorem for $K(m-1)$ tells us that $u=u_Tr_T$ for some $T\sse\{1,\dotsc,m-1\}$ and $u_T\in\Q$, as required. Suppose instead that $x=0$, so $y^2=q/p_m$ with $y\in K(m-1)$ and $y^2\in\Q$. It follows that $y=y_Tr_T$ for some $T\sse\{1,\dotsc,m-1\}$ and $y_T\in\Q$, so $u=y\sqrt{p_m}=y_Tr_{T\cup\{m\}}$, which again has the required form. \end{itemize} \end{proof} We next examine the Galois groups of multiquadratic extensions. \begin{proposition}\lbl{prop-mquad-galois} For $i=1,\dotsc,m$ there is an automorphism $\tau_i$ of $K(m)$ with $\tau_i(\sqrt{p_1})=-\sqrt{p_i}$ and $\tau_i(\sqrt{p_j})=\sqrt{p_j}$ for all $j\neq i$. Moreover, the full Galois group $G(K(m)/\Q)$ is the product of all the groups $\{1,\tau_i\}\simeq C_2$, so $G(K(m)/\Q)\simeq C_2^m$. \end{proposition} \begin{proof} As the elements $r_T$ form a basis for $K(m)$, we can certainly define a $\Q$-linear map $\tau_i\:K(m)\to K(m)$ by \[ \tau_i(r_T) = \begin{cases} -r_T & \text{ if } i\in T \\ +r_T & \text{ if } i\not\in T. \end{cases} \] Note that $\tau_i(0)=0$ and $\tau_i(1)=\tau_i(r_\emptyset)=1$. Now consider a pair of basis elements $r_T,r_U$ with $r_Tr_U=wr_V$ as in Lemma~\ref{lem-prod}. We claim that $\tau_i(r_Tr_U)=\tau_i(r_T)\tau_i(r_U)$. There are four cases to consider, depending on whether $i\in T$ or not, and whether $i\in U$ or not; we leave details to the reader. Now consider arbitrary elements $a,b\in K(m)$, say $a=\sum_Ta_Tr_T$ and $b=\sum_Ub_Ur_U$ with $a_T,b_U\in\Q$. We then have \begin{align*} \tau_i(ab) &= \tau_i\left(\sum_{T,U}a_Tb_Ur_Tr_U\right) = \sum_{T,U}a_Tb_U\tau_i(r_Tr_U) = \sum_{T,U}a_Tb_U\tau_i(r_T)\tau_i(r_U) \\ &= \left(\sum_Ta_T\tau_i(r_T)\right)\left(\sum_Ub_U\tau_i(r_U)\right) = \tau_i(a)\tau_i(b). \end{align*} This proves that $\tau_i$ is a homomorphism from $K(m)$ to itself. It is clear that $\tau_i^2(r_T)=r_T$ for all $T$, so $\tau_i^2=1$. Now suppose that $i\neq j$. We find that $\tau_i\tau_j(r_T)$ is either $+r_T$ (if $\{i,j\}\sse T$ or $\{i,j\}\cap T=\emptyset$) or $-r_T$ (if $|\{i,j\}\cap T|=1$). From this it is clear that $\tau_i\tau_j=\tau_j\tau_i$, so the elements $\tau_i$ generate a commutative subgroup $T\leq G(K(m)/\Q)$. For any sequence $\ep_1,\dotsc,\ep_m$ in $\{0,1\}$ we have an element $\sg_{\ep}=\tau_1^{\ep_1}\dotsb\tau_m^{\ep_m}\in T$. Note that $\sg_\ep(\sqrt{p_i})$ is $+\sqrt{p_i}$ if $\ep_i=0$, and $-\sqrt{p_i}$ if $\ep_i=1$. Using this we see that if $\sg_\ep=\sg_\dl$ then $\ep=\dl$. We thus have $2^m$ different elements of $T\sse G(K(m)/\Q)$. It follows using Proposition~\ref{prop-normal} that $K(m)$ is normal over $\Q$ and that $T$ is the full Galois group. \end{proof} It will be proved as Theorem~\ref{thm-primitive} that every field extension of finite degree has a primitive element. It turns out that there is a nice explicit example of this for multiquadratic fields. \begin{proposition}\lbl{prop-mquad-primitive} If $\tht_n=\sum_{i=1}^n\sqrt{p_i}$ then $\Q(\tht_n)=K(n)$. \end{proposition} (I thank Jayanta Manoharmayum for this fact and its proof.) \begin{proof} This is clear for $n=1$, so we may assume inductively that $K(n-1)=\Q(\tht_{n-1})$. We have seen that $K(n)$ has degree $2^n$ over $\Q$, and $\Q\leq\Q(\tht_n)\leq K(n)$ so the degree of $\Q(\tht_n)$ over $\Q$ must have the form $2^m$ for some $m$ with $0\leq m\leq n$; we must show that $m=n$. Let the minimal polynomial of $\tht_n$ over $\Q$ be \[ f(t) = \sum_{i=0}^{2^m} a_i t^i, \] and put \begin{align*} g(t) &= \sum_{i=0}^{2^m} \left(\sum_{2j\leq 2^m-i} \bcf{i+2j}{2j}p_n^ja_{i+2j}\right) t^i \\ h(t) &= \sum_{i=0}^{2^m-1} \left(\sum_{2j<2^m-i} \bcf{i+2j+1}{2j+1}p_n^ja_{i+2j+1}\right) t^i. \end{align*} By expanding out the relation $f(\tht_{n-1}+\sqrt{p_n})=f(\tht_n)=0$ we obtain $g(\tht_{n-1})+h(\tht_{n-1})\sqrt{p_n}=0$, with $g(\tht_{n-1}), h(\tht_{n-1})\in K(n-1)$. We have seen that $\{1,\sqrt{p_n}\}$ is a basis for $K(n)$ over $K(n-1)$, so $g(\tht_{n-1})=h(\tht_{n-1})=0$. The coefficient of $t^{2^m-1}$ in $h(t)$ is $2^m$, so $h$ is nonzero and has degree precisely $2^m-1$. It follows that $2^m-1$ must be at least as large as the degree of $\tht_{n-1}$ over $\Q$, which is $2^{n-1}$ by inductive assumption. This gives $m>n-1$ but we also had $0\leq m\leq n$, so $m=n$ as required. \end{proof} \section{The Galois correspondence} \label{sec-correspondence} The following theorem is the main result of Galois theory. \begin{theorem}\lbl{thm-correspondence} Let $M$ be a normal extension of $K$, with Galois group $G=G(M/K)$. \begin{itemize} \item[(a)] For any subgroup $H\leq G$, the set \[ L=M^H=\{a\in M\st \sg(a)=a\text{ for all } \sg\in H\} \] is a subfield of $M$ containing $K$, and $M$ is normal over $L$ with $G(M/L)=H$. \item[(b)] For any subfield $L\sse M$ containing $K$, the Galois group $H=G(M/L)$ is a subgroup of $G$ and we have $M^H=L$. \item[(c)] If $L$ and $H$ are as above, then $L$ is a normal extension of $K$ if and only if $H$ is a normal subgroup of $G$, and if so, then $G(L/K)=G/H$. \end{itemize} \end{theorem} This will be proved in three parts, as Corollary~\ref{cor-correspondence-a}, Proposition~\ref{prop-correspondence-b} and Proposition~\ref{prop-correspondence-c} below. \begin{remark}\lbl{rem-correspondence} Let $\CL$ be the set of all subfields $L$ with $K\sse L\sse M$. Let $\CH$ be the set of all subgroups of $G$. We can define $\Phi\:\CL\to\CH$ by $\Phi(L)=G(M/L)$, and we can define $\Psi\:\CH\to\CL$ by $\Psi(H)=M^H$. Parts~(a) and~(b) of the theorem can be rephrased as saying that $\Phi$ and $\Psi$ are inverse to each other, so both are bijections. \end{remark} \begin{remark}\lbl{rem-finite-galois} Suppose that $K=\F_p$, so that $M$ is also finite, of order $p^n$ say. Then $G=G(M/K)$ is cyclic of order $n$, generated by the Frobenius automorphism $\sg\:a\mapsto a^p$. For each divisor $d$ of $n$ we have a cyclic subgroup of $G$ of order $d$ generated by $\sg^{n/d}$, and these are all the subgroups of $G$. Given this, all the claims in Theorem~\ref{thm-correspondence} follow easily from Theorem~\ref{thm-finite-fields}. The same is true with just a little more work if $M$ is finite and $K$ is any subfield of $M$. \end{remark} \begin{example}\lbl{eg-even-quartic-galois} Consider again the field $K=\Q(\al,\bt)$, where $\al=\sqrt{3+\sqrt{7}}$ and $\bt=\sqrt{3-\sqrt{7}}$, as in Example~\ref{eg-even-quartic}. We will make the Galois correspondence explicit in this case. First note that $\al^2-3=3-\bt^2=\sqrt{7}$ and $\al\bt=\sqrt{2}$. It follows using this that $(\al+\bt)^2=\al^2+\bt^2+2\al\bt=6+2\sqrt{2}$, and $\al+\bt>0$ so $\al+\bt=\sqrt{6+2\sqrt{2}}$. In the same way we also see that $\al-\bt=\sqrt{6-2\sqrt{2}}$. We also note that $\al^2-\bt^2=2\sqrt{7}$, and we can divide this by the equation $\al\bt=\sqrt{2}$ to get $\al/\bt-\bt/\al=\sqrt{14}$. The subgroups of $D_8$ (other than $\{1\}$ and $D_8$ itself) can be listed as follows: \begin{align*} A_0 &= \{1,\quad (\al\; -\bt)(-\al\;\bt)\} \\ A_1 &= \{1,\quad (\al\; -\al)\} \\ A_2 &= \{1,\quad (\al\; \bt)(-\al\;-\bt)\} \\ A_3 &= \{1,\quad (\bt\; -\bt)\} \\ Z &= \{1,\quad (\al\; -\al)(\bt\; -\bt)\} \\ B_0 &= A_0A_2 \simeq C_2^2 \\ B_1 &= A_1A_3 \simeq C_2^2 \\ C_4 &= \text{ subgroup generated by } (\al\; -\bt\; -\al\; \bt). \end{align*} We can display the subgroups and subfields in the following diagram: \begin{center} \begin{tikzpicture}[scale=2.7] \def\ya{0.7} \def\yb{1.4} \def\yc{2.1} \def\xa{0.6} \def\xb{1.2} \def\Da{( 0.0, 0.0)} \def\Ba{(-\xa, \ya)} \def\Bb{( \xa, \ya)} \def\Ca{( 0, \ya)} \def\Aa{(-\xb, \yb)} \def\Ab{( \xa, \yb)} \def\Ac{(-\xa, \yb)} \def\Ad{( \xb, \yb)} \def\Za{( 0, \yb)} \def\Ta{( 0, \yc)} \begin{scope}[xshift=-0.5cm] \draw(0, 0.0) node{$8$}; \draw(0, \ya) node{$4$}; \draw(0, \yb) node{$2$}; \draw(0, \yc) node{$1$}; \end{scope} \begin{scope}[xshift=1cm] \draw \Da node{$D_8$}; \draw \Ca node{$C_4$}; \draw \Ba node{$B_0$}; \draw \Bb node{$B_1$}; \draw \Aa node{$A_0$}; \draw \Ab node{$A_1$}; \draw \Ac node{$A_2$}; \draw \Ad node{$A_3$}; \draw \Za node{$Z$}; \draw \Ta node{$\{1\}$}; \draw[<-,shorten <=11pt,shorten >=11pt] \Da -- \Ba; \draw[<-,shorten <=11pt,shorten >=11pt] \Da -- \Bb; \draw[<-,shorten <=11pt,shorten >=11pt] \Da -- \Ca; \draw[<-,shorten <=11pt,shorten >=11pt] \Ca -- \Za; \draw[<-,shorten <=11pt,shorten >=11pt] \Ba -- \Aa; \draw[<-,shorten <=11pt,shorten >=11pt] \Ba -- \Ac; \draw[<-,shorten <=11pt,shorten >=11pt] \Ba -- \Za; \draw[<-,shorten <=11pt,shorten >=11pt] \Bb -- \Ab; \draw[<-,shorten <=11pt,shorten >=11pt] \Bb -- \Ad; \draw[<-,shorten <=11pt,shorten >=11pt] \Bb -- \Za; \draw[<-,shorten <=11pt,shorten >=11pt] \Aa -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Ab -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Ac -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Ad -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Za -- \Ta; \end{scope} \begin{scope}[xshift=4cm] \draw \Da node{$\Q$}; \draw \Ca node{$\Q(\sqrt{14})$}; \draw \Ba node{$\Q(\sqrt{2})$}; \draw \Bb node{$\Q(\sqrt{7})$}; \draw \Aa node{$\Q(\al-\bt)$}; \draw \Ab node{$\Q(\bt)$}; \draw \Ac node{$\Q(\al+\bt)$}; \draw \Ad node{$\Q(\al)$}; \draw \Za node{$\Q(\sqrt{2},\sqrt{7})$}; \draw \Ta node{$K$}; \draw[->,shorten <=11pt,shorten >=11pt] \Da -- \Ba; \draw[->,shorten <=11pt,shorten >=11pt] \Da -- \Bb; \draw[->,shorten <=11pt,shorten >=11pt] \Da -- \Ca; \draw[->,shorten <=11pt,shorten >=11pt] \Ca -- \Za; \draw[->,shorten <=11pt,shorten >=11pt] \Ba -- \Aa; \draw[->,shorten <=11pt,shorten >=11pt] \Ba -- \Ac; \draw[->,shorten <=11pt,shorten >=11pt] \Ba -- \Za; \draw[->,shorten <=11pt,shorten >=11pt] \Bb -- \Ab; \draw[->,shorten <=11pt,shorten >=11pt] \Bb -- \Ad; \draw[->,shorten <=11pt,shorten >=11pt] \Bb -- \Za; \draw[->,shorten <=11pt,shorten >=11pt] \Aa -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Ab -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Ac -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Ad -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Za -- \Ta; \end{scope} \end{tikzpicture} \end{center} The first lattice shows all the subgroups, with the smaller groups towards the top. The orders of the groups are shown at the left. Arrows indicate inclusions between subgroups, so they point downwards. For each subgroup $H$ shown on the left, we display the field $K^H$ in the corresponding place on the right. As the Galois correspondence is order-reversing, the largest fields appear at the top and the inclusion arrows point upwards. For example, consider the group $C_4$. We observed above that $\sqrt{14}=\al/\bt-\bt/\al$. Let $\rho$ be the generator of $C_4$. This sends $\al$ to $-\bt$ and $\bt$ to $\al$. It therefore sends $\al/\bt-\bt/\al$ to $(-\bt)/\al-\al/(-\bt)$, which is the same as $\al/\bt-\bt/\al$. In other words, we have $\rho(\sqrt{14})=\sqrt{14}$, so $\sqrt{14}\in K^{C_4}$. On the other hand, we always have $[K:K^H]=|H|$ and so $[K^H:\Q]=8/|H|$, so in particular $[K^{C_4}:\Q]=2$. Similarly, it is clear that $\al$ is fixed by $A_3$, so $\Q(\al)\sse K^{A_3}$, and $[\Q(\al):\Q]=4=[K^{A_3}:\Q]$, so we must have $K^{A_3}=\Q(\al)$. All the other subgroups can be handled in the same way. \end{example} \begin{lemma}\lbl{lem-correspondence-a} Let $H$ be any subgroup of $G$; then $H\leq G(M/M^H)$ and $|G(M/M^H)|=[M:M^H]\geq |H|$. \end{lemma} \begin{proof} By definition we have \begin{align*} M^H &= \{a\in M\st\sg(a)=a\text{ for all } \sg\in H\} \\ G(M/M^H) &= \{\sg\:M\to M\st \sg(a)=a\text{ for all } a\in M^H\}. \end{align*} If $\sg\in H$ then $\sg(a)=a$ for all $a\in M^H$ by the very definition of $M^H$, so $\sg\in G(M/M^H)$. This shows that $H\leq G(M/M^H)$, and so $|H|\leq|G(M/M^H)|$. Next, as $M$ is normal over $K$ we see from Proposition~\ref{prop-top-normal} that it is normal over any intermediate field, such as $M^H$. We therefore see from Proposition~\ref{prop-normal} that $[M:M^H]=|G(M/M^H)|\geq|H|$ as claimed. \end{proof} \begin{lemma}\lbl{lem-V-zero} Let $H$ be any subgroup of $G$, and let $e_1,\dotsc,e_n$ be a basis for $M$ over $M^H$ (so $[M:M^H]=n$). Put \[ V = \{b=(b_1,\dotsc,b_n)\in M^n\st \sum_{i=1}^n b_i\sg(e_i)=0 \text{ for all } \sg\in H\}. \] Then $V=0$. \end{lemma} \begin{proof} We first note some properties of $V$. \begin{itemize} \item[(a)] $V$ is clearly a vector subspace of $M^n$. \item[(b)] If $b\in V$ then we can take $\sg=1$ in the definition to see that $\sum_ib_ie_i=0$. \item[(c)] We next claim that $V\cap(M^H)^n=\{0\}$. Indeed, if $b\in V\cap(M^H)^n$ then the relation $\sum_ib_ie_i=0$ above is an $M^H$-linear relation between the elements $e_i$, which by assumption are linearly independent over $M^H$; so we must have $b_1=\dotsb= b_n=0$. \item[(d)] Suppose that $(b_1,\dotsc,b_n)\in V$ and $\tau\in H$; we claim that $(\tau(b_1),\dotsc,\tau(b_n))\in V$ also. Indeed, we have $\sum_ib_i\sg(e_i)=0$ for all $\sg$, and as $\sg$ is arbitrary we can replace it by $\tau^{-1}\sg$ to see that $\sum_ib_i\tau^{-1}\sg(e_i)=0$. We then apply $\tau$ to this equation to obtain $\sum_i\tau(b_i)\sg(e_i)=0$, which proves the claim. \end{itemize} Next, for any vector $b\in V$, we define the \emph{size} of $b$ to be the number of nonzero entries. We must show that for $r>0$ there are no elements of size $r$, which we do by induction on $r$. Consider an element $b\in V$ of size one, so there is an index $i$ with $b_i\neq 0$, and all other entries are zero. Fact~(b) above therefore reduces to $b_ie_i=0$. As $e_i$ is a basis element it is nonzero, and $b_i\neq 0$ by assumption, so we have a contradiction. Thus, there are no elements in $V$ of size one, which starts the induction. Now suppose that $r>0$, and we have shown already that there are no elements in $V$ of size $s$ for all $0

~~=11pt] \Ga -- \Aa; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Bb; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Cc; \draw[<-,shorten <=11pt,shorten >=11pt] \Aa -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Bb -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Cc -- \Ta; \end{scope} \begin{scope}[xshift=4cm] \draw \Ga node{$\Q$}; \draw \Aa node{$K(\al)$}; \draw \Bb node{$K(\bt)$}; \draw \Cc node{$K(\al\bt)$}; \draw \Ta node{$L$}; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Aa; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Bb; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Cc; \draw[->,shorten <=11pt,shorten >=11pt] \Aa -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Bb -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Cc -- \Ta; \end{scope} \end{tikzpicture} \end{center} \end{solution} \begin{exercise}\exlabel{ex-golden} Put \begin{align*} \zt &= e^{{2\pi i}/{5}} \\ \al &= \zt + \zt^{-1} = 2\cos(2\pi/5) \\ \bt &= \zt - \zt^{-1} = 2i\sin(2\pi/5). \end{align*} Given that $\zt^4+\zt^3+\zt^2+\zt+1=0$, show that $\al=(-1+\sqrt{5})/2$, and deduce that $\sqrt{5}\in\Q(\zt)$. Then check that $\bt^2=\al^2-4$, and thus that $\bt=\sqrt{-(1+\sqrt{5})/2}$. Draw the subfield and subgroup lattices for the field extension $\Q(\zt)/\Q$. \end{exercise} \begin{solution} Since $\zt^4+\zt^3+\zt^2+\zt+1=0$, we have $\zt^2+\zt+1+\zt^{-1}+\zt^{-2}=0$. Since $\al^2=\zt^2+2+\zt^{-2}$, we see that $\al^2+\al-1=0$. Thus $\al$ is one of the roots of $x^2+x-1=0$, namely, $\al=(-1\pm\sqrt{5})/2$. However, $\zt+\zt^{-1}=\zt+\ov{\zt}=2\cos(2\pi/5)>0$, so we must have $\al=(-1+\sqrt{5})/2$. It follows that $\sqrt{5}=2\al+1=2\zt+2\zt^{-1}+1$, so $\sqrt{5}=2\al+1\in\Q(\zt)$. Next, we have \[ \bt^2=\zt^2-2+\zt^{-2}=\al^2-4= \left(\frac{-1+\sqrt{5}}{2}\right)^2 - 4 = \frac{6-2\sqrt{5}}{4}-4 = - \frac{1+\sqrt{5}}{2}. \] We also observe that $\sin(2\pi/5)>0$, and recall that when $t<0$ the symbol $\sqrt{t}$ refers to the square root in the upper half plane; we thus have $\bt=\sqrt{-(1+\sqrt{5})/2}$. We now put $G=G(\Q(\mu_5)/\Q)$ and look at the subgroup lattice. We know that \[ G = G(\Q(\mu_5)/\Q)=\{\sg_k\st k\in (\Z/5\Z)^\tm\} = = \{\sg_{-2},\sg_{-1},\sg_{1},\sg_{2}\}, \] and this is cyclic of order $4$, generated by $\sg_{2}$. It follows that the only subgroups are the trivial group, the whole group, and the subgroup $A=\{\ov{1},\ov{-1}\}$. This means that the only subfields are $\Q(\mu_5)$, $\Q$ and the intermediate field $M=\Q(\mu_5)^A$. Now $\sg_{-1}$ exchanges $\zt$ and $\zt^{-1}$ so it fixes $\al$ and sends $\bt$ to $-\bt$. We therefore see that $M=\Q(\al)=\Q(\sqrt{5})$, and that $\Q(\bt)$ cannot be $M$ so it must be all of $\Q(\zt)$. (In fact, one can check that $\zt=(\bt-\bt^2-3)/2$, which shows more explicitly that $\Q(\bt)=\Q(\zt)$.) The lattices can now be displayed as follows: \begin{center} \begin{tikzpicture}[scale=2] \def\Ga{( 0.0, 0.0)} \def\Ha{( 0.0, 1.0)} \def\Ta{( 0.0, 2.0)} \begin{scope} \draw(0, 0.0) node{$4$}; \draw(0, 1.0) node{$2$}; \draw(0, 2.0) node{$1$}; \end{scope} \begin{scope}[xshift=1cm] \draw \Ga node{$G$}; \draw \Ha node{$H$}; \draw \Ta node{$\{1\}$}; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Ha; \draw[<-,shorten <=11pt,shorten >=11pt] \Ha -- \Ta; \end{scope} \begin{scope}[xshift=4cm] \draw \Ga node{$\Q$}; \draw \Ha node{$\Q(\sqrt{5})$}; \draw \Ta node{$\Q(\sqrt{-(1+\sqrt{5})/2})$}; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Ha; \draw[->,shorten <=11pt,shorten >=11pt] \Ha -- \Ta; \end{scope} \end{tikzpicture} \end{center} \end{solution} \begin{exercise}\exlabel{ex-mu-eleven} Put $\zt=e^{{2\pi i}/{11}}$ and $K=\Q(\zt)=\Q(\mu_{11})$. Recall that the corresponding cyclotomic polynomial is \[ \vph_{11}(x) = x^{10}+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1, \] and that the roots of this are $\zt,\zt^2,\dotsc,\zt^{10}=\zt^{-1}$. Define \begin{align*} \bt &= \zt+\zt^{-1} = 2\cos(2\pi/11) \\ \gm &= \zt+\zt^3+\zt^4+\zt^5+\zt^9. \end{align*} \begin{itemize} \item[(a)] Explain why $\bt$ satisfies a quintic equation over $\Q$, and write it down. \item[(b)] Expand $\gm^2$ in powers of $\zt$, and hence deduce that $\gm^2+\gm+3=0$. Show that $\Q(\sqrt{-11})\subseteq\Q(\zt)$. \item[(c)] Use the general theory of cyclotomic extensions to find the structure of $G(K/\Q)$, and draw its lattice of subgroups. \item[(d)] Using the earlier parts of the question, draw the subfield lattice. \end{itemize} \end{exercise} \begin{solution}\ \\ \begin{itemize} \item[(a)] Since $\zt^{10}=\zt^{-1}$ etc., we can rewrite the given equation as \[ \zt^5+\zt^4+\zt^3+\zt^2+\zt+1 + \zt^{-1}+\zt^{-2}+\zt^{-3}+\zt^{-4}+\zt^{-5} = 0. \] Now \[ \begin{array}{rrrrrrrrrrrrr} \bt &= &&&&& \zt & & +\zt^{-1} &&&& \\ \bt^2 &= &&&& \zt^2 & & + 2 & & +\zt^{-2} &&& \\ \bt^3 &= &&& \zt^3 && +3\zt && +3\zt^{-1} && + \zt^{-3} && \\ \bt^4 &= && \zt^4 && +4\zt^2 && +6 && +4\zt^{-2} && +\zt^{-4} &\\ \bt^5 &= & \zt^5 && +5\zt^3 && +10\zt && +10\zt^{-1} && + 5\zt^{-3} && +\zt^{-5}. \end{array} \] By combining these, we find that $\bt^5+\bt^4-4\bt^3-3\bt^2+3\bt+1=0$. \item[(b)] We have \begin{align*} \gm^2 &= \zt^2+\zt^8+\zt^7+\zt^{10}+\zt^6+\\ & \qquad 2(\zt^5+\zt^{10}+\zt^6+\zt^4+ \zt^2+\zt^9+\zt^7+\zt^3+\zt+\zt^8)\\ &= (-1-\zt-\zt^3-\zt^4-\zt^5-\zt^9)+2(-1)\\ &= -3-\gm, \end{align*} so $\gm^2+\gm+3=0$. Since $\gm$ is a root of $x^2+x+3=0$, we see that $\gm=(-1\pm\sqrt{-11})/2$. The terms in $\gm$ are distributed in the complex plane as follows: \begin{center} \begin{tikzpicture}[scale=2] \draw[->] (-1.3,0) -- (1.3,0); \draw[->] (0,-1.3) -- (0,1.3); \fill[black!20] ( 0:1) circle(0.03); \fill ( 33:1) circle(0.03); \draw ( 33:1.2) node {$\zt$}; \fill[black!20] ( 65:1) circle(0.03); \fill ( 98:1) circle(0.03); \draw ( 98:1.2) node {$\zt^3$}; \fill (131:1) circle(0.03); \draw (131:1.2) node {$\zt^4$}; \fill (164:1) circle(0.03); \draw (164:1.2) node {$\zt^5$}; \fill[black!20] (196:1) circle(0.03); \fill[black!20] (229:1) circle(0.03); \fill[black!20] (262:1) circle(0.03); \fill (295:1) circle(0.03); \draw (295:1.2) node {$\zt^9$}; \fill[black!20] (327:1) circle(0.03); \fill[black!20] (360:1) circle(0.03); \end{tikzpicture} \end{center} It is clear from this that the imaginary part of $\gm$ is positive, so $\gm=(-1+\sqrt{-11})/2$, so $\sqrt{-11}=2\gm+1$. It is also clear from the definition that $\gm\in\Q(\zt)$, so $\sqrt{-11}\in\Q(\zt)$. \item[(c),(d)] The general cyclotomic theory says that $G(K/\Q)=\{\sg_k\st k\in(\Z/11)^\tm\}$. We have \[ (\Z/11)^\tm = \{-5,-4,-3,-2,-1,1,2,3,4,5\}. \] The powers of $2$ mod $11$ are as follows: \[ 2^0=1, \;\; 2^1=2, \;\; 2^2=4, \;\; 2^3=-3, \;\; 2^4=5, \;\; 2^5=-1, \;\; 2^6=-2, \;\; 2^7=-4, \;\; 2^8=3, \;\; 2^9=-5, \;\; 2^{10}=1. \] This shows that $(\Z/11)^\tm$ is cyclic of order $10$, generated by $2$, and thus $G(K/\Q)$ is cyclic of order $10$, generated by $\sg_2$. We write \begin{align*} C_{10} &= G(K/\Q) = \ip{\sg_2} \\ C_5 &= \ip{\sg_2^2} = \ip{\sg_4} = \{1,\sg_4,\sg_5,\sg_{-2},\sg_3\} \\ C_2 &= \ip{\sg_2^5} = \ip{\sg_{-1}} = \{1,\sg_{-1}\} \\ C_1 &= \{1\}. \end{align*} These are all the subgroups of the Galois group. It follows that the only subfields of $K$ are $K^{C_{10}}=\Q$, $K^{C_5}$, $K^{C_2}$ and $K^{C_1}=K$. The terms in $\gm$ are precisely the orbit of $\zt$ under $C_5$, so $\gm\in K^{C_5}$, so $\sqrt{-11}\in K^{C_5}$. We also know that $[K^{C_5}:\Q]=|C_{10}|/|C_5|=2$, which is the same as the degree of $\Q(\sqrt{-11})$, so we must have $K^{C_5}=\Q(\sqrt{-11})$. Similarly, we have \[ \sg_{-1}(\bt) = \sg_{-1}(\zt)+\sg_{-1}(\zt)^{-1}=\zt^{-1}+\zt=\bt, \] so $\bt\in K^{C_2}$, and it follows that $K^{C_2}=\Q(\bt)$. The subgroup and subfield lattices can thus be displayed as follows: \begin{center} \begin{tikzpicture}[scale=2] \def\Ga{( 0.8, 0.0)} \def\Gb{( 0.0, 0.6)} \def\Gc{( 2.0, 1.6)} \def\Gd{( 1.2, 2.2)} \begin{scope} \draw(0, 0.0) node{$10$}; \draw(0, 0.6) node{$5$}; \draw(0, 1.6) node{$2$}; \draw(0, 2.2) node{$1$}; \end{scope} \begin{scope}[xshift=1cm] \draw \Ga node{$C_{10}$}; \draw \Gb node{$C_5$}; \draw \Gc node{$C_2$}; \draw \Gd node{$\{1\}$}; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Gb; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Gc; \draw[<-,shorten <=11pt,shorten >=11pt] \Gb -- \Gd; \draw[<-,shorten <=11pt,shorten >=11pt] \Gc -- \Gd; \end{scope} \begin{scope}[xshift=4cm] \draw \Ga node{$\Q$}; \draw \Gb node{$\Q(\sqrt{-11})$}; \draw \Gc node{$\Q(\bt)$}; \draw \Gd node{$\Q(\mu_{11})$}; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Gb; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Gc; \draw[->,shorten <=11pt,shorten >=11pt] \Gb -- \Gd; \draw[->,shorten <=11pt,shorten >=11pt] \Gc -- \Gd; \end{scope} \end{tikzpicture} \end{center} \end{itemize} \end{solution} \begin{exercise}\exlabel{ex-two-group} Let $G$ be a finite group of order $2^r$ for some $r$. It is a standard fact from group theory that one can find subgroups \[ \{1\} = H_0 < H_1 < \dotsb < H_{r-1} < H_r = G \] such that $|H_i|=2^i$ for all $i$, and $H_i$ is normal in $G$. Now suppose that $G$ is the Galois group of some normal extension $L/K$. What can we deduce about the structure of $L$? \end{exercise} \begin{solution} Put $M_i=L^{H_i}$, so $L=M_0\supset M_1\supset\dotsb\supset M_r=K$. The Galois Correspondence tells us that $L$ is normal over $M_i$, with Galois group $H_i$ (so $[L:M_i]=2^i$) and $M_i$ is normal over $K$ (with Galois group $G/H_i$). It follows that $[M_i:M_{i+1}]=2$, so the standard analysis of degree two extensions says that $M_i=M_{i+1}(\al_i)$ for some $\al_i$ with $\al_i^2\in M_{i+1}$. This means that $L=K(\al_0,\dotsc,\al_{r-1})$. More precisely, for any subset $I\sse\{0,1,\dotsc,r-1\}$ we can let $\al_I$ denote the product of the elements $\al_i$ for $i\in I$. We then find that these elements $\al_I$ give a basis for $L$ over $K$. This does not yet capture all the information that one might want, as revealed by the following question. Suppose we have fields $K\subset K(\al_1)\subset K(\al_0,\al_1)$, with $\al_1^2\in K$ and $\al_0^2\in K(\al_1)$. When is it true that $K(\al_0,\al_1)$ is normal over $K$? This is usually false but sometimes true. We do not know a good general criterion even in this case where $r=2$, let alone the case of general $r$. \end{solution} \section{Cubics} \label{sec-cubics} In this section we will work with cubic polynomials over $\Q$, for convenience. Not much would change if we instead considered cubics over an arbitrary field $K$ (although there would be some special features if the characteristic of $K$ was $2$ or $3$). Consider a polynomial $f(x)=x^3+ax^2+bx+c$ with $a,b,c\in\Q$. If $f(x)$ is reducible then it must factor as $g(x)h(x)$ with $\deg(g(x))=1$ and $\deg(h(x))=2$. It is then easy to understand the roots of $g(x)$ and $h(x)$, and this determines the roots of $f(x)$. From now on we will ignore this case and assume instead that $f(x)$ is irreducible over $\Q$. We can factor this over $\C$ as $f(x)=(x-\al)(x-\bt)(x-\gm)$ say. Moreover, Proposition~\ref{prop-distinct-roots} assures us that $\al$, $\bt$ and $\gm$ are all distinct, so the set $R=\{\al,\bt,\gm\}$ has size three. By expanding out the relation \[ x^3+ax^2+bx+c = (x-\al)(x-\bt)(x-\gm) \] we find that \begin{align*} a &= -(\al+\bt+\gm) \\ b &= \al\bt+\bt\gm+\gm\al \\ c &= -\al\bt\gm. \end{align*} Now put $K=\Q(\al,\bt,\gm)$, which is the splitting field of $f(x)$. Put $G=G(K/\Q)$, which can be considered as a subgroup of $\Sg_R\simeq\Sg_3$. The subgroups of $\Sg_R$ can be enumerated as follows. \begin{itemize} \item[(a)] There is the trivial subgroup, of order one. \item[(b)] There are three different transpositions, namely $(\al\;\bt)$, $(\bt\;\gm)$ and $(\gm\;\al)$. For each transposition $\tau$, the set $\{1,\tau\}$ is a subgroup of $\Sg_R$ of order two. \item[(c)] The set $A_R=\{1,(\al\;\bt\;\gm),(\gm\;\bt\;\al)\}$ is a subgroup of order $3$, isomorphic to $C_3$. \item[(d)] The full group $\Sg_R$ has order $6$. \end{itemize} It is straightforward to check that this gives all possible subgroups of $\Sg_R$. We also know from Proposition~\ref{prop-root-perms} that the subgroup $G$ acts transitively: for any pair of elements in $R$, there is an element $\sg\in G$ that sends one to the other. It is easy to check that the subgroups of order $1$ or $2$ do not have this property. We must therefore have $G=A_R$ or $G=\Sg_R$. To distinguish between these cases we introduce the element \[ \dl = (\al-\bt)(\bt-\gm)(\gm-\al) \] and the element $\Dl=\dl^2$, which is known as the \emph{discriminant} of $f(x)$. \begin{proposition}\lbl{prop-cubic} \begin{itemize} \item[(a)] If $\sg\in G$ then $\sg(\dl)=\sgn(\sg)\dl$, where $\sgn(\sg)$ denotes the signature of the corresponding permutation. \item[(b)] We also have $\Dl\in\Q$, so $\sg(\Dl)=\Dl$ for all $\sg\in G$. \item[(c)] If $\dl\in\Q$ (or equivalently, $\Dl$ is a square in $\Q$) then $G=A_R\simeq C_3$, and $K=\Q(\al)$. \item[(d)] Suppose instead that $\dl\not\in\Q$. Then $G=\Sg_R$, and $K=\Q(\dl,\al)$, and $K^{A_R}=\Q(\dl)$. \end{itemize} \end{proposition} \begin{proof} \begin{itemize} \item[(a)] Suppose that $\sg$ acts on $R$ as the transposition $(\al\;\bt)$. We then have \[ \sg(\dl)=\sg((\al-\bt)(\bt-\gm)(\gm-\al)) = (\bt-\al)(\al-\gm)(\gm-\bt) = -(\al-\bt)(\bt-\gm)(\gm-\al) = -\dl. \] Similarly, if $\sg=(\bt\;\gm)$ or $\sg=(\gm\;\al)$ we see that $\sg(\dl)=-\dl$. Now suppose instead that $\sg$ acts as the $3$-cycle $(\al\;\bt\;\gm)$. We then have \[ \sg(\dl)=(\bt-\gm)(\gm-\al)(\al-\bt)=\dl. \] If $\sg=(\gm\;\bt\;\al)$ we also have $\sg(\dl)=\dl$, by a very similar argument. This covers all possible permutations (except for the identity, which is trivial) and so proves claim~(a). \item[(b)] For all $\sg\in G$ we have $\sg(\dl)=\pm\dl$, and so $\sg(\Dl)=\sg(\dl^2)=\sg(\dl)^2=(\pm\dl)^2=\dl^2=\Dl$. This proves that $\Dl\in K^G$, which is just $\Q$ by Theorem~\ref{thm-correspondence}. \item[(c)] Suppose that $\dl\in\Q$. It follows that for all $\sg\in G=G(K/\Q)$ we must have $\sg(\dl)=\dl$, which is only consistent with~(a) if $G\sse A_R$. We also saw previously (using transitivity) that $G$ must either be $A_R$ or $\Sg_R$, so now we see that $G=A_R$. In particular we have $|G|=3$ and so $[K:\Q]=3$, but as $f(x)$ is irreducible we also have $[\Q(\al):\Q]=3$, so it must be that $K=\Q(\al)$. \item[(d)] Suppose instead that $\dl\not\in\Q=K^G$, so there must exist $\sg\in G$ with $\sg(\dl)\neq\dl$. We then see from~(a) that $\sg$ gives an odd permutation of $R$, and that $\sg(\dl)=-\dl$. This means that we cannot have $G=A_R$, so we must have $G=\Sg_R$ instead. This means in particular that $[K:\Q]=|G|=6$. Consider the field $K'=\Q(\dl,\al)\sse K$. We then see that $[K':\Q]$ divides $[K:\Q]=6$, so $[K':\Q]\in\{1,2,3,6\}$. On the other hand, as $\Q\sse\Q(\dl)\sse K'$ and $\Q\sse\Q(\al)\sse K'$ we see that $[K':\Q]$ is divisible by both $[\Q(\dl):\Q]=2$ and $[\Q(\al):\Q]=3$. It follows that $[K':\Q]=6$, and thus that $K'=K$. It is also clear from~(a) that $\Q(\dl)\sse K^{A_R}$ and $[K^{A_R}:\Q]=|G/A_R|=|\Sg_R/A_R|=2=[\Q(\dl):\Q]$ so $K^{A_R}=\Q(\dl)$ as claimed. \end{itemize} \end{proof} We will now explore the Galois correspondence in the case where $G(K/\Q)=\Sg_R$. Put \[ A = \{1,(\bt\;\gm)\} \hspace{4em} B = \{1,(\gm\;\al)\} \hspace{4em} C = \{1,(\al\;\bt)\} \] The lattice of subgroups is then as shown on the left below, and the corresponding lattice of subfields is as shown on the right. \begin{center} \begin{tikzpicture}[scale=2] \def\ya{0.7} \def\yb{1.4} \def\yc{2.1} \def\Ga{( 0.9, 0.0)} \def\Ca{( 0.0, \ya)} \def\Aa{( 0.9, \yb)} \def\Bb{( 1.5, \yb)} \def\Cc{( 2.1, \yb)} \def\Ta{( 0.9, \yc)} \begin{scope} \draw(0, 0.0) node{$6$}; \draw(0, \ya) node{$3$}; \draw(0, \yb) node{$2$}; \draw(0, \yc) node{$1$}; \end{scope} \begin{scope}[xshift=1cm] \draw \Ga node{$\Sg_3$}; \draw \Ca node{$C_3$}; \draw \Aa node{$A$}; \draw \Bb node{$B$}; \draw \Cc node{$C$}; \draw \Ta node{$\{1\}$}; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Ca; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Aa; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Bb; \draw[<-,shorten <=11pt,shorten >=11pt] \Ga -- \Cc; \draw[<-,shorten <=11pt,shorten >=11pt] \Ca -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Aa -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Bb -- \Ta; \draw[<-,shorten <=11pt,shorten >=11pt] \Cc -- \Ta; \end{scope} \begin{scope}[xshift=4cm] \draw \Ga node{$\Q$}; \draw \Ca node{$\Q(\dl)$}; \draw \Aa node{$\Q(\al)$}; \draw \Bb node{$\Q(\bt)$}; \draw \Cc node{$\Q(\gm)$}; \draw \Ta node{$K$}; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Ca; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Aa; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Bb; \draw[->,shorten <=11pt,shorten >=11pt] \Ga -- \Cc; \draw[->,shorten <=11pt,shorten >=11pt] \Ca -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Aa -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Bb -- \Ta; \draw[->,shorten <=11pt,shorten >=11pt] \Cc -- \Ta; \end{scope} \end{tikzpicture} \end{center} \begin{remark}\lbl{rem-disc-formula} One can in fact show that \begin{align*} \Delta(\al,\bt,\gm) &= a^2b^2-4a^3c-4b^3+18abc-27c^2 \\ &= - \det\bbm 1&0&3&0&0 \\ a&1&2a&3&0 \\ b&a&b&2a&3 \\ c&b&0&b&2a \\ 0&c&0&0&b \ebm. \end{align*} It would be long, but essentially straightforward, to check this by hand. Alternatively, one can just enter the following in Maple: \begin{verbatim} a := - alpha - beta - gamma; b := alpha * beta + beta * gamma + gamma * alpha; c := - alpha * beta * gamma; delta := (alpha-beta) * (beta - gamma) * (gamma - alpha); M := <<1|0|3|0|0>,,~~,,<0|c|0|0|b>>; expand(delta^2 - (a^2*b^2-4*a^3*c-4*b^3+18*a*b*c-27*c^2)); expand(LinearAlgebra[Determinant](M) + delta^2); \end{verbatim} There is also a more conceptual argument using the determinant formula, which we will not explain here, except to mention that the first two columns contain the coefficients of $f(t)$ and the last three columns contain the coefficients of $f'(t)$. The determinant formula can be generalised to cover polynomials of any degree, not just cubics. \end{remark} \begin{remark}\label{rem-disc-simple} In the case where $a=0$, the formula reduces to $\Dl=-4b^3-27c^2$. One can always reduce to this case: if $f(x)=x^3+ax^2+bx+c$, then $f(x-a/3)=x^3+Bx+C$, where $B=b-a^2/3$ and $C=2a^3/27-ab/3+c$. \end{remark} We next explain how to find the roots $\al$, $\bt$ and $\gm$ in terms of the coefficients of $f(x)$. Traditionally this is usually done by starting with some preliminary steps that simplify the algebra but obscure some of the symmetry. Here we will assume that the algebra can be handled by a system such as Maple or Mathematica, so we will bypass these preliminary steps. First, we define $\Dl=a^2b^2-4a^3c-4b^3+18abc-27c^2$. If this is zero then $f(x)$ must be reducible, and so must have a root in $\Q$. We will ignore this case from now on, and assume that $\Dl\neq 0$. We will also assume for the moment that $b\neq a^2/3$; the significance of this will appear later. Let $\dl$ be one of the square roots of $\Dl$. For definiteness, we choose $\dl>0$ if $\Dl>0$, and we take $\dl$ to be a positive multiple of $i$ if $\Dl<0$. Then put \begin{align*} m &= (9ab - 2a^3 - 27c + 3\sqrt{-3}\delta)/2 \\ n &= (9ab - 2a^3 - 27c - 3\sqrt{-3}\delta)/2. \end{align*} We find (with computer assistance if necessary) that \begin{align*} m+n &= 9ab-2a^3-27c \\ mn &= ((9ab-2a^3-27c)^2+27\Dl)/4 = (a^2-3b)^3. \end{align*} In particular, as we have assumed that $b\neq a^2/3$ we see that $mn\neq 0$ and so $m,n\neq 0$. Now let $\mu$ be any cube root of $m$. (If $\Dl>0$ then $m$ lies in the upper half plane and we can take $\mu$ to be the unique cube root with $0<\arg(\mu)<\pi/3$; if $\Dl<0$ then $m$ is real and we can take $\mu$ to be the unique real cube root of $m$.) Now put $\nu=(a^2-3b)/\mu$ and observe (using the above formula for $mn$) that $\nu$ is a cube root of $n$. We now have \begin{align*} \mu^3+\nu^3 &= 9ab-2a^3-27c \\ \mu\nu &= a^2-3b. \end{align*} Now consider the number $\om=e^{2\pi i/3}=(\sqrt{3}i-1)/2$, so that $\om^3=1$ and $\om^2=\ov{\om}=\om^{-1}=-1-\om$. It is easy to check that the above equations will still hold if we replace $(\mu,\nu)$ by $(\om\mu,\ob\nu)$ or $(\ob\mu,\om\nu)$. Finally, we put \begin{align*} \al &= (\mu+\nu-a)/3 \\ \bt &= (\om\mu+\ob\nu-a)/3 \\ \gm &= (\ob\mu+\om\nu-a)/3. \end{align*} We claim that these are the roots of $f(x)$. To see this, we note by direct expansion that \[ f(\al) = f((\mu+\nu-a)/3) = (\mu^3+\nu^3+2a^3-9ab+27c)/27 + (\mu\nu+3b-a^2)(\mu+\nu)/9. \] However, we saw above that $\mu^3+\nu^3+2a^3-9ab+27c=0$ and $\mu\nu+3b-a^2=0$, so it follows that $f(\al)=0$. We can now replace $(\mu,\nu)$ by $(\om\mu,\ob\nu)$ and argue in the same way to see that $f(\bt)=0$, and similarly $f(\gm)=0$. If we can show that $\al$, $\bt$ and $\gm$ are distinct, it will follow from Proposition~\ref{prop-several-roots} that $f(x)=(x-\al)(x-\bt)(x-\gm)$ as expected. To check for distinctness, first note that $\mu^3-\nu^3=m-n=3\sqrt{-3}\dl\neq 0$, which implies that $\mu\neq\nu$, so $\mu-\nu\neq 0$. We also have \[ \bt-\gm = (\om-\ob)(\mu-\nu)/3 = \sqrt{-3}(\mu-\nu)/3 \neq 0, \] so $\bt\neq\gm$. One can show that $\al\neq\bt$ and $\al\neq\gm$ in a similar way. This completes our discussion of the general case where $b\neq a^2/3$. We conclude by discussing briefly the special case where $b=a^2/3$. Here we find that $\Dl=-(a^3-27c)^2/27$, so $\dl=\pm(a^3-27c)/(3\sqrt{-3})$. We also have $mn=0$, so either $m$ or $n$ is zero. On the other hand, we have $m-n=3\sqrt{-3}\dl\neq 0$, so $m$ and $n$ are not both zero. If $m\neq 0$ then we proceed exactly as before, noting that $\nu=(a^2-3b)/\mu=0$. If $m=0$ then we instead define $\nu$ to be the standard cube root of $n$ and put $\mu=0$, and then the rest of the argument works as previously. %============================================================ %============================================================ \begin{center} \Large \textbf{Exercises} \end{center} \begin{exercise}\exlabel{ex-classify-cubics} Show that the cubics $g_0(x)=x^3-3x+1$ and $g_1(x)=x^3+3x+1$ are irreducible, and find their Galois groups. \end{exercise} \begin{solution} We first claim that $g_0(x)$ is irreducible over $\Q$. If not, it would have to have a monic linear factor, say $x-a$ with $a\in\Q$. Then Gauss's Lemma (Proposition~\ref{prop-gauss}) would tell us that $a\in\Z$. We would also have $g_0(a)=0$, which rearranges to give $a(3-a^2)=1$, so $a$ divides $1$, so $a=\pm 1$. However $g_0(1)$ and $g_0(-1)$ are nonzero, so this is impossible. By essentially the same argument, $g_1(x)$ is irreducible over $\Q$. This can also be proved by applying Eisenstein's criterion (with $p=3$) to $g_0(x-1)$ and $g_1(x-1)$. We now see from the general theory that the Galois groups are either $A_3=C_3$ (if the discriminant is a square) or $\Sg_3$ (if the discriminant is not a square). Using the formula in Remark~\ref{rem-disc-simple} we see that the discriminant of $g_0(x)$ is $-4\tm(-27)-27=81=9^2$, whereas the discriminant of $g_1(x)$ is $-4\tm 27-27=-135$. Thus, the Galois group for $g_0(x)$ is $A_3$, and the Galois group for $g_1(x)$ is $\Sg_3$. \end{solution} \begin{exercise}\exlabel{ex-cyclic-cubic} Let $q$ be a rational number, and put $r=1+q+q^2$. Consider the polynomials \begin{align*} f(x) &= x^3 - (3x - 2q - 1)r \\ g(x) &= x^3 + 3qx^2 - 3(q+1)x - (4q^3+6q^2+6q+1) \\ s(x) &= x^2 +qx -2r. \end{align*} Check (with assistance from Maple if necessary) that $f(s(x))=f(x)g(x)$. For the rest of the exercise we will assume that $q$ has been chosen so that $f(x)$ is irreducible. Now suppose we have a field $L$ and an element $\al\in L$ with $f(\al)=0$. Show that $s(\al)$ is also a root of $f(x)$ in $\Q(\al)$, and is different from $\al$. Deduce that $\Q(\al)$ is a splitting field for $f(x)$ over $\Q$, and that $G(\Q(\al)/\Q)$ is cyclic of order $3$. \end{exercise} \begin{solution} The first claim can be checked using Maple as follows: \begin{verbatim} r := 1 + q + q^2; f := (x) -> x^3 - (3*x - 2*q - 1)*r; g := (x) -> (x^3+3*q*x^2-3*(q+1)*x-(4*q^3+6*q^2+6*q+1)); s := (x) -> x^2+q*x-2*r; expand(f(s(x)) - f(x)*g(x)); \end{verbatim} It is possible but painful to do this by hand; $f(s(x))$ has 25 terms when fully expanded. Now suppose we have $\al\in L$ with $f(\al)=0$, and we put $\bt=s(\al)\in\Q(\al)$. We can substitute $x=\al$ in the relation $f(s(x))=f(x)g(x)$ to see that $f(\bt)=f(\al)g(\al)=0$, so $\bt$ is another root of $f(x)$. Next, as $f(x)$ is assumed to be irreducible, it must be the minimal polynomial of $\al$, so $\Q(\al)\simeq\Q[x]/f(x)$. This means that homomorphisms from $\Q(\al)$ to any field $M$ biject with roots of $f(x)$ in $M$. In particular, we can take $M=\Q(\al)$ and we find that there is a homomorphism $\sg\:\Q(\al)\to\Q(\al)$ with $\sg(\al)=\bt$. We next claim that $\bt\neq\al$, or equivalently that $\al$ is not a root of the quadratic polynomial $s(x)-x$. This is clear because the minimal polynomial of $\al$ is $f(x)$, which is cubic, so it cannot divide $s(x)-x$. It follows that $f(x)$ is divisible in $\Q(\al)[x]$ by $(x-\al)(x-\bt)$. The remaining factor is a monic polynomial of degree $1$, so it must have the form $x-\gm$ for some $\gm\in\Q(\al)$. We now have a splitting $f(x)=(x-\al)(x-\bt)(x-\gm)$, so $\Q(\al)$ is a splitting field for $f(x)$. This means that it is normal, and the order of the Galois group is $[\Q(\al):\Q]=3$. All groups of order $3$ are cyclic, and $\sg$ is a nontrivial element, so we must have $G(\Q(\al)/\Q)=\{1,\sg,\sg^2\}$. \end{solution} \begin{exercise}\exlabel{ex-inv-sq-sum} Suppose that the polynomial $f(x)=x^3+ux^2+vx+w$ hs three distinct roots, namely $\al$, $\bt$ and $\gm$. Give a formula for \[ p = \frac{1}{\al^2} + \frac{1}{\bt^2} + \frac{1}{\gm^2} \] in terms of $u$, $v$ and $w$. \end{exercise} \begin{solution} First, we have \[ x^3+ux^2+vx+w = f(x) = (x-\al)(x-\bt)(x-\gm) = x^3 - (\al+\bt+\gm) x^2 + (\al\bt+\bt\gm+\gm\al) x - \al\bt\gm, \] so \begin{align*} u &= -\al-\bt-\gm \\ v &= \al\bt+\bt\gm+\gm\al \\ w &= -\al\bt\gm. \end{align*} It follows that \[ w^2p = \al^2\bt^2 + \bt^2\gm^2 + \gm^2\al^2. \] This is similar to $v^2$, but not equal to it. More precisely, we have \[ v^2 = \al^2\bt^2 + \bt^2\gm^2 + \gm^2\al^2 + 2(\al^2\bt\gm + \al\bt^2\gm + \al\bt\gm^2) = w^2p + 2uw. \] Rearranging this gives $p=v^2/w^2-2u/w$. \end{solution} \begin{exercise}\exlabel{ex-vandermonde} Suppose $f(x)=x^3+ax+b$. If $f$ has roots $\al$, $\bt$ and $\gm$, then recall that its discriminant $\Dl(f)$ is $(\al-\bt)^2(\al-\gm)^2(\bt-\gm)^2$. Let $M$ denote the matrix \[ M= \begin{pmatrix} 1&1&1\\ \al&\bt&\gm\\ \al^2&\bt^2&\gm^2 \end{pmatrix}. \] \begin{itemize} \item[(a)] Define $\dl(f)=(\al-\bt)(\bt-\gm)(\gm-\al)$. Show that $\dl(f)=\det(M)$. \item[(b)] Thus $\Dl(f)=\dl(f)^2$. Given that $\det(M)=\det(M^T)$, deduce that $\Dl(f)=\det(MM^T)$. \item[(c)] Write $S_i=\al^i+\bt^i+\gm^i$. Show that \[ MM^T = \begin{pmatrix} S_0&S_1&S_2\\ S_1&S_2&S_3\\ S_2&S_3&S_4 \end{pmatrix}. \] \item[(d)] Clearly $S_0=3$ and $S_1=0$ (as $S_1$ is the sum of the roots, which is zero as the coefficient of $x^2$ in $f$ is zero). Show that $S_2=-2a$ by an explicit computation. \item[(e)] As $\al$, $\bt$ and $\gm$ are roots of $f$, we have \begin{align*} \al^3+a\al+b &= 0\\ \bt^3+a\bt+b &= 0\\ \gm^3+a\gm+b &= 0. \end{align*} By summing these three, find $S_3$ in terms of $S_0$ and $S_1$. Similarly, multiplying these equations by $\al$, $\bt$ and $\gm$ respectively, find $S_4$ in terms of $S_1$ and $S_2$. Compute the values of $S_3$ and $S_4$ in terms of $a$ and $b$. \item[(f)] Combining all the above, show that $\Dl(f)=-(4a^3+27b^2)$. \end{itemize} \end{exercise} \begin{solution} \begin{itemize} \item[(a)] One approach is to simply expand everything out. Alternatively, we can recall the behaviour of determinants under row and column operations, and argue as follows: \[ \det\bsm 1&1&1\\ \al&\bt&\gm\\ \al^2&\bt^2&\gm^2\esm = \det\bsm 1&0&0\\ \al&\bt-\al&\gm-\al\\ \al^2&\bt^2-\al^2&\gm^2-\al^2\esm = (\bt-\al)(\gm-\al) \det\bsm 1&0&0\\ \al&1 &1 \\ \al^2&\bt+\al&\gm+\al\esm = (\bt-\al)(\gm-\al)(\gm-\bt) = \dl(f). \] (At the first stage we subtracted the first column from each of the other two columns, then we extracted factors of $\bt-\al$ and $\gm-\al$ from the second and third columns, then we calculated the final determinant directly.) \item[(b)] We have \[ \det(MM^T) = \det(M)\det(M^T) = \det(M)^2 = \dl(f)^2 = \Dl(f). \] \item[(c)] This is just a direct calculation: \[ \begin{pmatrix} 1&1&1\\ \al&\bt&\gm\\ \al^2&\bt^2&\gm^2 \end{pmatrix} \begin{pmatrix} 1&\al&\al^2\\ 1&\bt&\bt^2\\ 1&\gm&\gm^2 \end{pmatrix} = \begin{pmatrix} 1+1+1&\al+\bt+\gm&\al^2+\bt^2+\gm^2\\ \al+\bt+\gm&\al^2+\bt^2+\gm^2&\al^3+\bt^3+\gm^3\\ \al^2+\bt^2+\gm^2&\al^3+\bt^3+\gm^3&\al^4+\bt^4+\gm^4. \end{pmatrix} \] \item[(d)] We have \[ S_2=\al^2+\bt^2+\gm^2 = (\al+\bt+\gm)^2-2(\al\bt+\bt\gm+\gm\al)=-2a, \] as $\al+\bt+\gm=S_1=0$ and $\al\bt+\bt\gm+\gm\al=a$. \item[(e)] Add the three equations to get \[ (\al^3+\bt^3+\gm^3)+a(\al+\bt+\gm)+b(1+1+1)=0, \] or $S_3+aS_1+bS_0=0$. Thus $S_3=-aS_1-bS_0$. Also, add \begin{align*} \al^4+a\al^2+b\al &= 0 \\ \bt^4+a\bt^2+b\bt &= 0 \\ \gm^4+a\gm^2+b\gm &= 0 \end{align*} to get $S_4=-aS_2-bS_1$. Thus we conclude that \begin{align*} S_3 &= -3b\\ S_4 &= 2a^2. \end{align*} \item[(f)] Substituting the values of $S_0,\ldots,S_4$ into the matrix in (c), we get: \[ MM^T = \begin{pmatrix} 3 & 0 & -2a \\ 0 & -2a & -3b \\ -2a & -3b & 2a^2 \end{pmatrix}. \] By part~(b), $\Dl(f)$ is the determinant of this matrix, which can be evaluated directly to give $\Dl(f)=-(4a^3+27b^2)$. \end{itemize} \end{solution} \section{Quartics} \label{sec-quartics} Let $f(x)$ be an irreducible quartic over $\Q$, with roots $R=\{\al,\bt,\gm,\dl\}$ say. Let $K=\Q(\al,\bt,\gm,\dl)$ be the splitting field, and let $G=G(K/\Q)$ be the Galois group. This is then a transitive subgroup of $\Sg_R$. Our first task will be to classify such subgroups. First note that $|\Sg_R|=4!=24$. The elements can be listed as follows. \begin{itemize} \item The identity element has order $1$. \item There are six transpositions ($(\al;\bt),\;(\al\;\gm),\;(\al\;\dl),\;(\bt\;\gm),\;(\bt\;\dl)$ and $(\gm\;\dl)$), each of order $2$. \item There are three transposition pairs, which again have order $2$: \begin{align*} \tau_1 &= (\al\;\bt)(\gm\;\dl) \\ \tau_2 &= (\al\;\gm)(\bt\;\dl) \\ \tau_3 &= (\al\;\dl)(\bt\;\gm). \end{align*} \item There are eight three-cycles, each of order three. \item There are six four-cycles, each of order $4$. \end{itemize} One crucial fact is as follows: \begin{proposition}\lbl{prop-vier} The set $V=\{1,\tau_1,\tau_2,\tau_3\}$ is a normal subgroup of $\Sg_R$, isomorphic to $C_2\tm C_2$. It is also transitive. For each $\sg\in\Sg_R$ there is a unique permutation $\ov{\sg}\in\Sg_3$ such that $\sg\tau_i\sg^{-1}=\tau_{\ov{\sg}(i)}$ for all $i$. Moreover, the rule $\pi(\sg)=\ov{\sg}$ defines a surjective homomorphism of groups $\pi\:\Sg_R\to\Sg_3$, with kernel $V$. \end{proposition} \begin{remark}\lbl{rem-resolvent} This connection between $\Sg_R$ and $\Sg_3$ allows us to relate cubics to quartics. More precisely, we will later write down a cubic polynomial $h(x)\in\Q[x]$ (called the \emph{resolvent cubic} of $f(x)$) such that $K^{G\cap V}$ is a splitting field for $h(x)$. The full field $K$ can then be obtained by adjoining at most two square roots to $K^{G\cap V}$. \end{remark} Before the proof, we will give a sample calculation with $\pi$. Consider the three-cycle $\sg=(\al\;\bt\;\gm)$, so $\sg^{-1}=(\gm\;\bt\;\al)$. We have \begin{align*} \sg\tau_1\sg^{-1} &= (\al\;\bt\;\gm)(\al\;\bt)(\gm\;\dl)(\gm\;\bt\;\al) = (\al\;\dl)(\bt\;\gm) = \tau_3 \\ \sg\tau_2\sg^{-1} &= (\al\;\bt\;\gm)(\al\;\gm)(\bt\;\dl)(\gm\;\bt\;\al) = (\al\;\bt)(\gm\;\dl) = \tau_1 \\ \sg\tau_3\sg^{-1} &= (\al\;\bt\;\gm)(\al\;\dl)(\bt\;\gm)(\gm\;\bt\;\al) = (\al\;\gm)(\bt\;\dl) = \tau_2. \end{align*} The first line shows that $\ov{\sg}(1)=3$, the second that $\ov{\sg}(2)=1$, and the third that $\ov{\sg}(3)=2$. It follows that $\ov{\sg}=(1\;3\;2)\in\Sg_3$. \begin{proof}[Proof of Proposition~\ref{prop-vier}] One can check directly that $\tau_i^2=1$ for all $i$ and \begin{align*} \tau_1\tau_2 &= \tau_2\tau_1 = \tau_3 \\ \tau_2\tau_3 &= \tau_3\tau_2 = \tau_1 \\ \tau_3\tau_1 &= \tau_1\tau_3 = \tau_2. \end{align*} (More succinctly, the product of any two $\tau$'s is the third one.) This shows that $V$ is a subgroup of $\Sg_R$. The subgroups generated by $\tau_1$ and $\tau_2$ are cyclic of order $2$, and $V$ is the direct product of these subgroups, so $V\simeq C_2\tm C_2$. Next, recall that any conjugate of a transposition pair is another transposition pair. More precisely, for any $\sg\in\Sg_R$ and any transposition pair $(\kp\;\lm)(\mu\;\nu)$ we have \[ \sg(\kp\;\lm)(\mu\;\nu)\sg^{-1} = (\sg(\kp)\;\sg(\lm))(\sg(\mu)\;\sg(\nu)). \] As $\tau_1$, $\tau_2$ and $\tau_3$ are the only transposition pairs, we must have $\sg\tau_i\sg^{-1}=\tau_j$ for some $j$. We define $\ov{\sg}(i)$ to be this $j$, so $\sg\tau_i\sg^{-1}=\tau_{\ov{\sg}(j)}$. Now if we have another permutation $\rho$ we find that \[ \tau_{\ov{\rho\sg}(i)} = \rho\sg\tau_i(\rho\sg)^{-1} = \rho\sg\tau_i\sg^{-1}\rho^{-1} = \rho\tau_{\ov{\sg}(i)}\rho^{-1} = \tau_{\ov{\rho}(\ov{\sg}(i))}, \] so $\ov{\rho\sg}=\ov{\rho}\circ\ov{\sg}$. In particular, we can take $\rho=\sg^{-1}$ and we find that $\ov{\rho}$ is an inverse for $\ov{\sg}$, so $\ov{\sg}$ is a permutation of $\{1,2,3\}$. In particular, we see from this that $\sg V\sg^{-1}=V$, so $V$ is a normal subgroup of $\Sg_R$. We can now define $\pi\:\Sg_R\to\Sg_3$ by $\pi(\sg)=\ov{\sg}$, and the relation $\ov{\rho\sg}=\ov{\rho}\circ\ov{\sg}$ tells us that this is a homomorphism. Note that $V$ is commutative, so if $\sg\in V$ then $\sg\tau_i\sg^{-1}=\tau_i\sg\sg^{-1}=\tau_i$, so $\ov{\sg}$ is the identity. We therefore have $V\leq\ker(\pi)$. Next, using the formula above for conjugating transposition pairs, we find that \begin{align*} (\bt\;\gm) \tau_1 (\bt\;\gm)^{-1} &= \tau_2 & (\gm\;\dl) \tau_1 (\gm\;\dl)^{-1} &= \tau_1 \\ (\bt\;\gm) \tau_2 (\bt\;\gm)^{-1} &= \tau_1 & (\gm\;\dl) \tau_2 (\gm\;\dl)^{-1} &= \tau_3 \\ (\bt\;\gm) \tau_3 (\bt\;\gm)^{-1} &= \tau_3 & (\gm\;\dl) \tau_3 (\gm\;\dl)^{-1} &= \tau_2, \end{align*} so $\pi((\bt\;\gm))=(1\;2)$ and $\pi((\gm\;\dl))=(2\;3)$. Thus, the image of $\pi$ is a subgroup of $\Sg_3$ containing $(1\;2)$ and $(2\;3)$, but it is straightforward to check that the only such subgroup is $\Sg_3$ itself, so $\pi$ is surjective. The First Isomorphism Theorem for groups then gives $\Sg_R/\ker(\pi)\simeq\Sg_3$, so $|\ker(\pi)|=|\Sg_R|/|\Sg_3|=24/6=4$. On the other hand, we also have $V\leq\ker(\pi)$ and $|V|=4$. We must therefore have $\ker(\pi)=V$ as claimed. \end{proof} We next explain in more detail the Galois-theoretic significance of $V$ and $\pi$. We put \begin{align*} \mu_1 &= \tfrac{1}{2}((\al+\bt)-(\gm+\dl)) & \lm_1 &= \mu_1^2 \\ \mu_2 &= \tfrac{1}{2}((\al+\gm)-(\bt+\dl)) & \lm_2 &= \mu_2^2 \\ \mu_3 &= \tfrac{1}{2}((\al+\dl)-(\bt+\gm)) & \lm_3 &= \mu_3^2 \\ K_0 &= \Q(\lm_1,\lm_2,\lm_3) \sse K. \end{align*} The factor of $1/2$ is included for later convenience. Note that $\mu_1+\mu_2=\al-\dl$ and $\mu_1-\mu_2=\bt-\gm$. These are nonzero so $\mu_1\neq\pm\mu_2$, so $\lm_1\neq\lm_2$. We can do the same for $\mu_1\pm\mu_3$ and $\mu_2\pm\mu_3$ so we find that all the numbers $\pm\mu_i$ are distinct, and all the numbers $\lm_i$ are distinct. Because the roots are grouped in $\mu_i$ the same way that they are in $\tau_i$, we find that \[ \sg(\mu_i) = \pm\mu_{\ov{\sg}(i)} \hspace{5em} \sg(\lm_i) = \lm_{\ov{\sg}(i)} \] for all $\sg\in G$ and $i\in\{1,2,3\}$. It follows that $\sg|_{K_0}=1_{K_0}$ iff $\ov{\sg}=1$ iff $\sg\in V\cap G$. This means that $V\cap G=G(K/K_0)$ and so (by the Galois Correspondence) $K_0=K^{V\cap G}$. As $V\cap G$ is normal in $G$ we deduce that $K_0$ is a Galois extension of $\Q$ with Galois group $G/(V\cap G)\simeq\pi(G)\leq\Sg_3$, and also $K$ is Galois over $K_0$ with Galois group $V\cap G$. To understand the extension $K_0/\Q$ in more detail, consider the polynomial \begin{align*} g(x) &= (x-\lm_1)(x-\lm_2)(x-\lm_3) \\ &= x^3 - (\lm_1+\lm_2+\lm_3) x^2 + (\lm_1\lm_2+\lm_2\lm_3+\lm_3\lm_1) x - \lm_1\lm_2\lm_3. \end{align*} As $G$ permutes the elements $\lm_i$ and the coefficients of $g$ are symmetric in these elements, we see that these coefficients lie in $K^G=\Q$, so $g(x)\in\Q[x]$. Thus $g(x)$ is a cubic over $\Q$ (called the \emph{resolvent cubic} for $f(x)$) and $K_0$ is a splitting field for $g(x)$. Later we will give formulae for the coefficients of $g(x)$ in terms of the coefficients of $f(x)$. Once we know $g(x)$ we can find the roots $\lm_i$ by the methods of Section~\ref{sec-cubics}. We can then find $\mu_i=\pm\sqrt{\lm_i}$. We also note that the element $a=-(\al+\bt+\gm+\dl)$ is just the coefficient of $x^3$ in $f(x)$, so we can find the roots of $f(x)$ by the formulae \begin{align*} \al &= (+\mu_1+\mu_2+\mu_3)/2-a/4 \\ \bt &= (+\mu_1-\mu_2-\mu_3)/2-a/4 \\ \gm &= (-\mu_1+\mu_2-\mu_3)/2-a/4 \\ \dl &= (-\mu_1-\mu_2+\mu_3)/2-a/4. \end{align*} The only issue here is to control the signs of the elements $\mu_i=\pm\sqrt{\lm_i}$. Suppose that \[ f(x) = (x-\al)(x-\bt)(x-\gm)(x-\dl) = x^4 + ax^3 + bx^2 + cx + d, \] so that \begin{align*} a &= -(\al+\bt+\gm+\dl) \\ b &= \al\bt+\al\gm+\al\dl+\bt\gm+\bt\dl+\gm\dl \\ c &= -(\al\bt\gm + \al\bt\dl + \al\gm\dl + \bt\gm\dl) \\ d &= \al\bt\gm\dl. \end{align*} One can check directly (perhaps with assistance from Maple) that \[ \mu_1\mu_2\mu_3 = (4ab - a^3 - 8c)/8. \] When solving the quartic, one can choose the signs of $\mu_1$ and $\mu_2$ arbitrarily, but one should then define $\mu_3$ to be $(4ab-a^3-8c)/(8\mu_1\mu_2)$ so that the above identity holds. It then works out that $\mu_3$ is a square root of $\lm_3$, and the roots of $f(x)$ can be found by the formulae displayed above. The formulae simplify considerably if we assume that $f(x)$ has no term in $x^3$, so $\al+\bt+\gm+\dl=0$. This does not really lose any generality: if $f(x)=x^4+ax^3+bx^2+cx+d$ then one can check that the polynomial $f(x-a/4)$ has no term in $x^3$, and if we know the roots of $f(x-a/4)$ we can just subtract $a/3$ to get the roots of $f(x)$. If $\al+\bt+\gm+\dl=0$ then we find that \begin{align*} \mu_1 &= \al+\bt & -\mu_1 &= \gm+\dl & \lm_1 &= (\al+\bt)^2=(\gm+\dl)^2 \\ \mu_2 &= \al+\gm & -\mu_2 &= \bt+\dl & \lm_2 &= (\al+\gm)^2=(\bt+\dl)^2 \\ \mu_3 &= \al+\dl & -\mu_3 &= \bt+\gm & \lm_3 &= (\al+\dl)^2=(\bt+\gm)^2 \end{align*} It follows that $\mu_1\mu_2\mu_3=-c$. We can now expand out the definition of $g(x)$ to obtain the following result: \begin{proposition}\lbl{prop-resolvent} For a quartic polynomial of the form $f(x)=x^4+bx^2+cx+d$, the resolvent cubic is given by \[ g(x) = x^3+2bx^2+b^2x-4dx-c^2. \qed \] \end{proposition} We now continue our investigation of which subgroups of $\Sg_R$ can appear as Galois groups. \begin{proposition}\lbl{prop-vier-converse} Suppose that $H$ is a transitive subgroup of $\Sg_R$ such that $|H|=4$, and that $H$ contains no elements of order $4$. Then $H=V$. \end{proposition} \begin{proof} Suppose that $\sg\in H$ with $\sg\neq 1$. By Lagrange's Theorem the order of $\sg$ must divide $|H|=4$, but by assumption the order is not equal to $4$, so the order must be two. This means that $\sg$ is either a transposition or a transposition pair. Suppose that $\sg$ is a transposition; then there exists a root $\lm$ with $\sg(\lm)=\lm$. Put $K=\stab_H(\lm)=\{\rho\in H\st\rho(\lm)=\lm\}$, so $\{1,\sg\}\sse K$, so $|K|>1$, so $|H|/|K|<4$. However, the Orbit-Stabiliser Theorem tells us that $|H\lm|=|H|/|K|$, so $|H\lm|<4$, so $H\lm\neq R$. This contradicts the assumption that $H$ is transitive. It follows that all nontrivial elements of $H$ must actually be transposition pairs, but there are only three transposition pairs in $\Sg_R$, so all of them must be in $H$, so $H=V$. \end{proof} \begin{definition}\lbl{defn-Qi} For $i\in\{1,2,3\}$ we put $Q_i=\{\sg\in\Sg_R\st\ov{\sg}(i)=i\}$. \end{definition} \begin{proposition}\lbl{prop-dihedral} $Q_i$ is a dihedral group of order $8$, and is transitive. Moreover, these are the only subgroups of order $8$ in $\Sg_R$. \end{proposition} \begin{proof} We first consider $Q_2$. Let $\rho$ be the four-cycle $(\al\;\bt\;\gm\;\dl)$. Note that $\rho^2=(\al\;\gm)(\bt\;\dl)=\tau_2$, so $\rho\tau_2\rho^{-1}=\tau_2$, so $\ov{\rho}(2)=2$, so $\rho\in Q_2$. On the other hand, we have \begin{align*} \rho\tau_1\rho^{-1} &= (\rho(\al)\;\rho(\bt))(\rho(\gm)\;\rho(\dl)) = (\bt\;\gm)(\dl\;\al) = \tau_3 \\ \rho\tau_3\rho^{-1} &= (\rho(\al)\;\rho(\dl))(\rho(\bt)\;\rho(\gm)) = (\bt\;\al)(\gm\;\dl) = \tau_1, \end{align*} so $\ov{\rho}=(1\;3)$. If $\sg\in Q_2$ then $\ov{\sg}$ must either be the identity or $(1\;3)$. If $\ov{\sg}=1$ then $\sg\in\ker(\pi)=V$. If $\ov{\sg}=(1\;3)=\ov{\rho}$ then we find that $\sg\rho^{-1}\in\ker(\pi)=V$, so $\sg\in V\rho$. It follows that $Q_2=V\amalg V\rho$, which has order $8$. One can also check that $\tau_1\rho\tau_1^{-1}=\rho^{-1}$, which mean that $\tau_1$ and $\rho$ generate a group isomorphic to $D_8$, which must be all of $Q_2$. As $V$ is transitive and $V\leq Q_2$ we also see that $Q_2$ is transitive. One can show in the same way that $Q_1$ and $Q_3$ are also transitive and isomorphic to $D_8$. Now let $H$ be an arbitrary subgroup of $\Sg_R$ with $|H|=8$. We then have subgroups $\pi(H)\leq\Sg_3$ and $H\cap V=\ker(\pi\:H\to\pi(H))\leq V$, and the First Isomorphism Theorem tells us that $|H\cap V||\pi(H)|=|H|=8$. Here $|H\cap V|$ must divide $|V|=4$ and $|\pi(H)|$ must divide $|\Sg_3|=6$. The only possibility is $|\pi(H)|=2$ and $|H\cap V|=4=|V|$. This means that $H\cap V=V$ (or in other words, that $V\leq H$) and that $\pi(H)=\{1,\sg\}$ for some transposition $\sg\in\Sg_3$. If $\sg=(1\;2)$ we see that $H\leq Q_3$, but $|H|=8=|Q_3|$ so $H=Q_3$. Similarly, if $\sg=(1\;3)$ then $H=Q_2$, and if $\sg=(2\;3)$ then $H=Q_1$. \end{proof} One can check directly that in any group isomorphic to $D_8$ there is a unique cyclic subgroup of order $4$. We can thus do the following: \begin{definition}\lbl{defn-Pi} We write $P_i$ for the unique cyclic subgroup of order $4$ in $Q_i$. \end{definition} \begin{proposition}\lbl{prop-C-four} The groups $P_i$ are all different, and they are the only cyclic subgroups of order $4$ in $\Sg_R$. \end{proposition} \begin{proof} First, we have $Q_i=P_iV$, and the subgroups $Q_i$ are all different, so the subgroups $P_i$ are all different. Each $P_i$ contains precisely two elements of order $4$ (each inverse to the other). The elements of order $4$ are the four-cycles, and there are only six of them in $\Sg_R$. Thus, there cannot be any further cyclic subgroups of order $4$. \end{proof} \begin{lemma}\lbl{lem-half-normal} Let $G$ be a finite group, and let $H$ be a subgroup such that $|G|=2|H|$. Then $H$ is normal in $G$. \end{lemma} \begin{proof} Put $C=G\sm H$, so $|C|=|G|-|H|=|H|$. Suppose that $g\in G$; we must show that $gHg^{-1}=H$. If $g\in H$ then this is clear. If $g\not\in H$, then the left coset $gH$ is disjoint from $H$ and so is contained in $C$, but $|gH|=|H|=|C|$ so $gH=C$. Similarly, the right coset $Hg$ is disjoint from $H$ and has the same size as $C$ so it is equal to $C$. We now have $gH=Hg$ and we can multiply on the right by $g^{-1}$ to get $gHg^{-1}=H$ as required. \end{proof} \begin{proposition}\lbl{prop-alternating} The only subgroup of $\Sg_R$ of order $12$ is the group $A_R$ of even permutations of $R$. \end{proposition} \begin{proof} Suppose that $|H|=12$. By the lemma, we see that $H$ is normal so we have a quotient group $G/H\simeq C_2$ and a quotient homomorphism $q\:G\to G/H$ with kernel $H$. Let $x$ denote the nontrivial element of $G/H$. Recall that all the transpositions in $\Sg_R$ are conjugate to each other. Thus, if $H$ contains any transposition then it must contain all of them, but the transpositions generate $\Sg_R$, so $H=\Sg_R$, contradicting the fact that $|H|=12$. It follows that for all transpositions $\sg$ we have $q(\sg)=x$. Now if $\rho$ is an even permutation then it can bew written as a product of $2m$ transpositions, say, which gives $q(\rho)=x^{2m}=1$, so $\rho\in H$. This shows that $A_R\leq H$ but $|A_R|=12=|H|$ so $H=A_R$. \end{proof} \begin{proposition}\lbl{prop-transitive} The transitive subgroups of $\Sg_R$ are as follows: $V,P_1,P_2,P_3,Q_1,Q_2,Q_3,A_R$ and $\Sg_R$. Thus, the Galois group $G$ must be one of these groups. \end{proposition} \begin{proof} Let $H$ be a transitive subgroup of $\Sg_R$. As $H$ is transitive, the orbit $H\al$ is all of $R$, so $|H\al|=4$. Put $K=\stab_H(\al)=\{\sg\in H\st\sg(\al)=\al\}$. The Orbit-Stabiliser Theorem tells us that $|H\al|=|H|/|K|$, so $|H|=4|K|$, which is divisible by $4$. On the other hand, Lagrange's Theorem tells us that $|H|$ divides $|\Sg_R|=24$. It follows that $|H|\in\{4,8,12,24\}$. If $|H|=24$ then clearly $H=\Sg_R$. If $|H|=12$ then Proposition~\ref{prop-alternating} tells us that $H=A_R$. If $|H|=8$ then Proposition~\ref{prop-dihedral} tells us that $H=Q_i$ for some $i$. If $|H|=4$ and $H$ contains an element of order $4$ then $H$ must be cyclic and Proposition~\ref{prop-C-four} tells us that $H=P_i$ for some $i$. This just leaves the case where $|H|=4$ but $H$ has no element of order $4$, in which case Proposition~\ref{prop-vier-converse} tells us that $H=V$. \end{proof} \begin{remark}\lbl{rem-transitive} The subgroups $P_i$ are all conjugate to each other, so we can convert between them by just renaming the roots. As the naming of the roots is arbitrary, it is not very meaningful to distingush between these subgroups. The same applies to the subgroups $Q_i$. Thus, we can say that the Galois group is always $V$, $C_4$, $D_8$, $A_4$ or $\Sg_4$. The inclusions between these subgroups can be displayed as follows: \[ \xymatrix{ & & \Sg_4 \\ & D_8 \urto & & A_4 \ulto \\ C_4 \urto & & V \urto \ulto } \] \end{remark} \begin{remark}\lbl{rem-irr-resolvent} Consider a quartic $f(x)$ with resolvent $g(x)$. If the Galois group of $f(x)$ is $H\leq\Sg_4$, then the Galois group of $g(x)$ is the image of $H$ in $\Sg_4/V\simeq\Sg_3$, which we will call $\ov{H}$. If $g(x)$ is irreducible then $\ov{H}$ must be transitive, and so must have order divisible by $3$. It follows that $|H|$ must be divisible by $3$, and by inspecting the above list of possibilities we see that either $H=A_4$ and $\ov{H}=A_3$, or $H=\Sg_4$ and $\ov{H}=\Sg_3$. \end{remark} %============================================================ %============================================================ \begin{center} \Large \textbf{Exercises} \end{center} \begin{exercise}\exlabel{ex-classify-quartics} What are the Galois groups of the quartics $f_0(x)=x^4+8x+12$ and $f_1(x)=x^4+8x-12$? [{\sl Hint: You may assume that these are irreducible. Exercise~\ref{ex-classify-cubics} is relevant.}] \end{exercise} \begin{solution} Using the formula in Proposition~\ref{prop-resolvent}, we see that the resolvent cubic for $f_0(x)$ is $x^3-32x-64=64((x/4)^3-2(x/4)-1)$. In the notation of Exercise~\ref{ex-classify-cubics}, this is $64 g_0(x/4)$, so the Galois group is the same as for $g_0(x)$, namely $A_3$. Using Remark~\ref{rem-irr-resolvent} we deduce that the Galois group for $f_0(x)$ is $A_4$. Similarly, the resolvent cubic for $f_1(x)$ is $64 g_1(x/4)$, and the Galois group for $g_1(x)$ is $\Sg_3$, so the Galois group for $f_1(x)$ is $\Sg_4$. \end{solution} \begin{exercise}\exlabel{ex-biquad-quartic} You are given that a quartic polynomial $f(x)$ has roots as follows: \[ \al_0 = \sqrt{2}+\sqrt{5} \hspace{3em} \al_1 = \sqrt{2}-\sqrt{5} \hspace{3em} \al_2 = -\sqrt{2}+\sqrt{5} \hspace{3em} \al_3 = -\sqrt{2}-\sqrt{5}. \] What is its discriminant? What is the Galois group? \end{exercise} \begin{solution} The discriminant is \begin{align*} \prod_{i 0$ such that $L=K(\al_1,\dotsc,\al_r)$ and $\al_i^{n_i}\in K(\al_1,\dotsc,\al_{i-1})$ for all $i$. \end{definition} \begin{definition}\lbl{defn-solvable-poly} Let $K$ be a field, and let $f(x)$ be a monic polynomial in $K[x]$. We say that $f(x)$ is \emph{solvable by radicals} if there exists a radical extension $L/K$ such that $f(x)$ splits in $K[x]$. \end{definition} \begin{theorem}\lbl{thm-solvable-poly} Suppose that $K$ has characteristic zero. Let $f(x)$ be a monic polynomial in $f(x)$, and let $N$ be a splitting field for $f(x)$. Then $f(x)$ is solvable by radicals if and only if the Galois group $G(N/K)$ is solvable. \end{theorem} One half of this will be proved as Proposition~\ref{prop-radicals-a} below, and the converse half as Corollary~\ref{cor-radicals-b}. First, however, we will give some examples and preliminary results about solvable groups. \begin{example}\lbl{eg-solvable-three} Consider the group $\Sg_3$. The alternating subgroup $A_3$ is cyclic of order $3$, and the quotient $\Sg_3/A_3$ is cyclic of order $2$. We thus have a series $\{1\} 4$. \end{example} \begin{example}\lbl{eg-order-solvable} Let $G$ be a group of order $n$. If $n$ is prime then $G$ is cyclic and therefore solvable. If $n$ is a power of a prime, then $G$ is still solvable. We will not give the proof here but it is a standard exercise in the theory of groups of prime power order. If $n$ involves only two primes, then $G$ is still solvable by a theorem of Burnside which is often covered in advanced undergraduate courses on Representation Theory. More strikingly, if $n$ is odd then $G$ is automatically solvable. This is a famous theorem of Feit and Thompson; the proof takes hundreds of pages and is only accessible to specialists in finite group theory. \end{example} \begin{proposition}\lbl{prop-abelian-solvable} Any finite abelian group is solvable. \end{proposition} \begin{proof} Let $G$ be a finite abelian group. Put $G_0=\{1\}\leq G$. If $G\neq G_0$, we choose an element $a_1\in G\sm G_0$, and let $G_1$ be the subgroup generated by $G_0$ together with $a_1$. If $G\neq G_1$, we choose an element $a_2\in G\sm G_1$, and let $G_2$ be the subgroup generated by $G_1$ together with $a_2$. Continuing in this way, we get a chain of subgroups \[ \{1\} = G_0 < G_1 < G_2 < \dotsb \leq G. \] As $G$ is finite and $G_i$ is strictly bigger than $G_{i-1}$, we must eventually reach a stage where $G_r=G$. As everything is abelian, all subgroups are normal, so we can form quotient groups $G_i/G_{i-1}$. As $G_i$ is generated by $G_{i-1}$ together with $a_i$, we see that $G_i/G_{i-1}$ is generated by the coset $a_iG_{i-1}$ and so is cyclic. We therefore have a solvable series for $G$. \end{proof} \begin{proposition}\lbl{prop-subquotient} Let $G$ be a finite group, and let $H$ be a normal subgroup. Put $\ov{G}=G/H$ and let $\pi\:G\to\ov{G}$ be the quotient homomorphism, so $\pi(g)=gH$. \begin{itemize} \item[(a)] If $K$ is any subgroup of $G$ such that $H\sse K$, then the set $\ov{K}=\pi(K)$ is a subgroup of $\ov{G}$ and is the same as $K/H$. Moreover, we have $K=\{x\in G\st\pi(x)\in\ov{K}\}$. \item[(b)] Conversely, if $\ov{K}$ is any subgroup of $\ov{G}$ then the set $K=\{x\in G\st\pi(x)\in\ov{K}\}$ is a subgroup of $G$ containing $H$, and we have $\ov{K}=\pi(K)=K/H$. \item[(c)] If $K$ and $\ov{K}$ are related as above, then $K$ is normal in $G$ if and only if $\ov{K}$ is normal in $\ov{G}$. If so, then there is an isomorphism $G/K\to\ov{G}/\ov{K}$ given by $gK\mapsto\pi(g)\ov{K}$. \end{itemize} \end{proposition} \begin{proof} \begin{itemize} \item[(a)] The identity element $1_G$ lies in $K$, so the identity element $1_{\ov{G}}=\pi(1_G)$ lies in $\ov{K}$. Suppose we have elements $\ov{a},\ov{b}\in\ov{K}$. By the definition of $\ov{K}$, we can choose $a,b\in K$ with $\ov{a}=\pi(a)$ and $\ov{b}=\pi(b)$. As $K$ is a subgroup, we have $ab\in K$ and $a^{-1}\in K$. It follows that $\pi(ab),\pi(a^{-1})\in\ov{K}$ but $\pi(ab)=\ov{a}\,\ov{b}$ and $\pi(a^{-1})=\ov{a}^{-1}$, so $\ov{a}\,\ov{b}\in\ov{K}$ and $\ov{a}^{-1}\in\ov{K}$. This proves that $\ov{K}$ is a subgroup of $\ov{G}$. The elements are just the cosets $xH$ for $x\in K$, which are the same as the elements of $K/H$; so $\ov{K}=K/H$. Now consider the set $K'=\{x\in G\st \pi(x)\in\ov{K}\}$; we claim that this is the same as $K$. If $x\in K$ then $\pi(x)\in\ov{K}$ by the definition of $\ov{K}$, so $x\in K'$ by the definition of $K'$. Thus $K\sse K'$. Conversely, suppose that $x\in K'$. Then $\pi(x)\in\ov{K}=\pi(K)$, so $\pi(x)=\pi(y)$ for some $y\in K$. This means that $xH=yH$, so $x=yz$ for some $z\in H$. However, we have $H\sse K$ by assumption, so $y$ and $z$ both lie in $K$, so $x\in K$. This shows that $K'\sse K$, so in fact $K'=K$ as claimed. \item[(b)] Now let $\ov{K}$ be an arbitrary subgroup of $\ov{G}$, and put $K=\{x\in G\st\pi(x)\in\ov{K}\}$. Clearly, if $x\in H$ then $\pi(x)=1_{\ov{G}}\in\ov{K}$, so $x\in K$. This proves that $H\sse K$, so in particular $1\in K$. Now suppose we have alements $a,b\in K$. This means that the elements $\pi(a)$ and $\pi(b)$ lie in $\ov{K}$, but $\ov{K}$ is a subgroup, so we have $\pi(a)\pi(b)\in\ov{K}$ and $\pi(a)^{-1}\in\ov{K}$. As $\pi$ is a homomorphism we have $\pi(ab)=\pi(a)\pi(b)$, which lies in $\ov{K}$, so $ab\in K$. Similarly we have $\pi(a^{-1})=\pi(a)^{-1}$, which lies in $\ov{K}$, so $a^{-1}\in K$. This shows that $K$ is a subgroup of $G$ containing $H$. From the very definition of $K$ we have $\pi(K)\sse\ov{K}$. Conversely, if $u\in\ov{K}\sse\ov{G}=G/H$ then we must have $u=xH=\pi(x)$ for some $x\in G$. Now $\pi(x)\in\ov{K}$ so by the definition of $K$ we have $x\in K$. This means that $u\in\pi(K)$. We thus have $\ov{K}\sse\pi(K)$, and so $\ov{K}=\pi(K)$ as claimed. \item[(c)] Let $K$ and $\ov{K}$ be related as discussed above. Suppose that $K$ is normal in $G$. For any $\ov{a}\in\ov{G}$ we can choose $a\in G$ with $\pi(a)=\ov{a}$, and we note that $aKa^{-1}=K$ because $K$ is normal. We thus have \[ \ov{a}\ov{K}\ov{a}^{-1}=\pi(a)\pi(K)\pi(a)^{-1}=\pi(aKa^{-1})= \pi(K)=\ov{K}, \] which proves that $\ov{K}$ is normal in $\ov{G}$. Conversely, suppose that $\ov{K}$ is normal in $\ov{G}$. Consider an element $a\in G$, and the corresponding subgroup $K'=aKa^{-1}\leq G$. Note that $K'$ contains $aHa^{-1}$, but $aHa^{-1}=H$ as $H$ is normal. We can thus apply part~(a) to $K'$ as well as to $K$. The last claim in~(a) says that $K'=\{x\st\pi(x)\in\pi(K')\}$, whereas $K=\{x\st\pi(x)\in\pi(K)\}$. Now $\pi(K')=\pi(a)\ov{K}\pi(a)^{-1}$, but this is just the same as $\ov{K}$, because $\ov{K}$ is assumed to be normal. We thus have $K=K'$, which means that $K$ is normal. Finally, suppose that $K$ (and thus $\ov{K}$) is normal, and define a homomorphism $\phi\:G\to\ov{G}/\ov{K}$ by $\phi(x)=\pi(x)\ov{K}$. This is clearly surjective, and we have \begin{align*} \ker(\phi) &= \{x\in G\st \pi(x)\ov{K} = \ov{K}\} \\ &= \{x\in G\st \pi(x)\in\ov{K}\} = K \end{align*} (where we have again used the last part of~(a)). The First Isomorphism Theorem therefore gives us an induced isomorphism $\ov{\phi}\:G/K=G/\ker(\phi)\to\ov{G}/\ov{K}$, as claimed. \end{itemize} \end{proof} \begin{proposition}\lbl{prop-solvable-layers} Let $G$ be a finite group. \begin{itemize} \item[(a)] If $G$ is solvable then every subgroup of $G$ is solvable. \item[(b)] If $G$ is solvable, then for every normal subgroup $H\leq G$, the quotient $G/H$ is also solvable. \item[(c)] If $G$ has a normal subgroup $H$ such that both $H$ and $G/H$ are solvable, then $G$ is solvable. \end{itemize} \end{proposition} \begin{proof} \begin{itemize} \item[(a)] Suppose that $G$ is solvable, so we have a solvable series $G_0\leq\dotsb\leq G_r$ as in the definition. Let $H$ be a subgroup of $G$. Put $H_i=H\cap G_i$, which is a subgroup of $H$. Note that $H_0=H\cap\{1\}=\{1\}$ and $H_r=H\cap G=H$. We can define a homomorphism $\pi_i\:H_i\to G_i/G_{i-1}$ by $\pi_i(x)=xG_{i-1}$. The kernel of this is the set of elements in $H_i$ that also lie in $G_{i-1}$, so \[ \ker(\pi_i)=H_i\cap G_{i-1}=H\cap G_i\cap G_{i-1}= H\cap G_{i-1} = H_{i-1}. \] Thus, the First Isomorphism Theorem tells us that $H_{i-1}$ is normal in $H_i$ and that $H_i/H_{i-1}$ is isomorphic to $\pi_i(H_i)$. This is a subgroup of the cyclic group $G_i/G_{i-1}$, so is itself cyclic. Thus, the subgroups $H_i$ form a solvable series for $H$. \item[(b)] Now suppose that $H$ is normal, so we have a quotient group $\ov{G}=G/H$ and a quotient homomorphism $\pi\:G\to\ov{G}$ given by $\pi(g)=gH$. Put $\ov{G}_i=\pi(G_i)$, which is a subgroup of $\ov{G}$. Note that $\ov{G}_0=\pi(\{1\})=\{1\}$ and $\ov{G}_r=\pi(G)=\ov{G}$. As $G_{i-1}\sse G_i$ we have $\ov{G}_{i-1}\sse \ov{G}_i$. We next claim that $\ov{G}_{i-1}$ is normal in $\ov{G}_i$. Indeed, if $a\in\ov{G}_i$ and $b\in\ov{G}_{i-1}$ then we must have $a=\pi(x)$ and $b=\pi(y)$ for some $x\in G_i$ and $y\in G_{i-1}$. This means that $aba^{-1}=\pi(xyx^{-1})$, but $G_{i-1}$ is normal in $G_i$, so $xyx^{-1}\in G_{i-1}$, so $aba^{-1}\in\pi(G_{i-1})=\ov{G}_{i-1}$ as claimed. Finally, we claim that $\ov{G}_i/\ov{G}_{i-1}$ is cyclic. To see this, choose $x\in H_i$ such that $xH_{i-1}$ generates the cyclic group $H_i/H_{i-1}$, and put $a=\pi(x)\ov{G}_{i-1}\in \ov{G}_i/\ov{G}_{i-1}$. Any other element $b\in \ov{G}_i/\ov{G}_{i-1}$ has the form $b=\pi(y)\ov{G}_{i-1}$ for some $y\in H_i$. By our choice of $x$ we have $y=x^iz$ for some $i\in\Z$ and $z\in H_{i-1}$, and it follows that $b=a^i$, as required. We have thus constructed a solvable series for $\ov{G}$. \item[(c)] Now suppose instead that $G$ is a finite group with a normal subgroup $H$, and that both $H$ and the quotient group $\ov{G}=G/H$ are solvable. Let $\pi\:G\to\ov{G}$ be the quotient map. Choose solvable series \[ \{1\} = H_0 \leq H_1 \leq \dotsb \leq H_r = H \] \[ \{1\} = \ov{G}_0 \leq\ov{G}_1 \leq \dotsb \leq \ov{G}_s = \ov{G}. \] For $1\leq j\leq s$ we put $H_{r+j}=\{x\in G\st\pi(x)\in\ov{G}_j\}$. (For $j=0$ the group $H_{r+j}$ is already defined and is equal to $H$, and in this case it is still true that $H_{r+j}=\{x\in G\st\pi(x)\in\ov{G}_j\}$.) This defines a chain \[ \{1\} = H_0 \leq \dotsb \leq H_r = H \leq H_{r+1} \leq \dotsb H_{r+s} = G, \] and with the help of Proposition~\ref{prop-subquotient} we see that this is a solvable series for $G$. \end{itemize} \end{proof} \begin{corollary}\lbl{cor-solvable-defn} Let $G$ be a finite group, and suppose that there is a chain \[ \{1\} = G_0 \leq G_1 \leq \dotsb \leq G_r = G \] such that $G_{i-1}$ is normal in $G_i$ and $G_i/G_{i-1}$ is abelian for all $i$. Then $G$ is solvable. \end{corollary} \begin{proof} Recall from Proposition~\ref{prop-abelian-solvable} that all abelian groups are solvable, so $G_i/G_{i-1}$ is solvable for all $i$. This means that $G_1$ and $G_2/G_1$ are solvable, so $G_2$ is solvable by Proposition~\ref{prop-solvable-layers}(c). Now $G_2$ and $G_3/G_2$ are solvable, so $G_3$ is solvable by Proposition~\ref{prop-solvable-layers}(c). Continuing in this way, we see that $G_i$ is solvable for all $i$. In particular, the group $G=G_r$ is solvable as claimed. \end{proof} In Section~\ref{sec-cyclotomic} we analysed cyclotomic extensions of $\Q$. In fact, most of what we said there can be adapted to cover cyclotomic extensions of any field of characteristic zero. Our next result is one instance of that. \begin{proposition}\lbl{prop-cyclotomic-abelian} Suppose we have a field $K$ of characteristic zero and an extension $L=K(\zt)$, where $\zt^n=1$. Then $L$ is normal over $K$ and $G(L/K)$ is abelian. \end{proposition} \begin{proof} Let $d$ be the smallest positive integer such that $\zt^d=1$. We then find that $1,\zt,\dots,\zt^{d-1}$ are $d$ distinct roots of the polynomial $x^d-1$, so we have $x^d-1=\prod_{i=0}^{d-1}(x-\zt^i)$ in $L[x]$. This proves that $L$ is a splitting field for $x^d-1$ over $K$, so it is a normal extension of $K$. Next, for each $\sg\in G(L/K)$ we see that $\sg(\zt)$ is a root of $x^d-1$, so $\sg(\zt)=\zt^{\lm(\sg)}$ say. Here $\lm(\sg)$ is an integer that is well-defined modulo $d$, so we can regard $\lm$ as a function $G(L/K)\to\Z/d\Z$. Note that \[ \tau(\sg(\zt)) = \tau(\zt^{\lm(\sg)}) = \tau(\zt)^{\lm(\sg)}=\zt^{\lm(\tau)\lm(\sg)}, \] which means that $\lm(\tau\sg)=\lm(\tau)\lm(\sg)$. In particular, we have $\lm(\sg^{-1})\lm(\sg)=\lm(1_L)=1$, so $\lm(\sg)$ is invertible in $\Z/n\Z$, and we can regard $\lm$ as a group homomorphism $G(L/K)\to(\Z/n\Z)^\tm$. We claim that this is injective. Indeed, if $\lm(\sg)=1$ then $\sg(\zt)=\zt$, so $\sg$ acts as the identity on $K(\zt)$, but $K(\zt)=L$, so $\sg=1$ as required. We now see that $G(L/K)$ is isomorphic to a subgroup of the abelian group $(\Z/n\Z)^\tm$, so $G(L/K)$ is abelian. \end{proof} \begin{proposition}\lbl{prop-radicals-a} Let $K$ be a field of characteristic zero. Let $L$ be a splitting field for a polynomial $f(x)\in K[x]$, and suppose that $G(L/K)$ is solvable. Then $f(x)$ is solvable by radicals. \end{proposition} \begin{proof} Put $n=[L:K]$, and let $N$ be a splitting field for $x^n-1$ over $L$. This is also a splitting field for $(x^n-1)f(x)$ over $K$, so it is normal over $K$. Next, consider the composite \[ \phi = (G(N/K(\zt)) \xra{\text{include}} G(N/K) \xra{\text{restrict}} G(L/K)). \] If $\sg$ is in the kernel then it acts as the identity on $K(\zt)$ (because $\sg\in G(L/K)$) and on $L$ (as $\phi(\sg)=1$) so it acts as the identity on $L(\zt)=N$, so $\sg=1$. This means that $\phi$ is injective, so $G(N/K(\zt))$ is isomorphic to a subgroup of $G(L/K)$. This means that $|G(N/K(\zt))|$ divides $n$, and also that $G(N/K(\zt))$ is solvable. We can thus find a solvable series \[ \{1\} = H_0 \leq H_1 \leq \dotsb \leq H_r = G(N/K(\zt)). \] We put $N_i=N^{H_i}$, so that \[ N = N_0 \supseteq N_1 \supseteq \dotsb \supseteq N_r = K(\zt). \] As $H_{i-1}$ is normal in $H_i$ we see that $N_{i-1}$ is normal over $N_i$. The Galois group $G(N_{i-1}/N_i)$ is isomorphic to $H_i/H_{i-1}$, so it is cyclic, of order $n_i$ say. Here $n_i$ divides $|H_r|$ which divides $n$, so $x^{n_i}-1$ splits in $K(\zt)\sse N_i$. We can thus use Proposition~\ref{prop-cyclic-ext} to find $\al_{i-1}\in N_{i-1}$ such that $N_{i-1}=N_i(\al_{i-1})$ and $\al_{i-1}^{n_i}\in N_i$. This proves that $N$ is a radical extension of $K(\zt)$, which is clearly a radical extension of $K$. Thus $L$ is contained in a radical extension of $K$, as required. \end{proof} \begin{lemma}\lbl{lem-normal-radical} Let $N$ be a radical extension of $K$. Then there is another extension $M\supseteq N$, an integer $n>0$, and a chain of subfields $K\sse M_0\sse \dotsb \sse M_t=M$ such that: \begin{itemize} \item[(a)] $M$ is normal over $K$. \item[(b)] $M_0=K(\zt)$ for some $\zt$ such that $x^n-1=\prod_{i=0}^{n-1}(x-\zt^i)$ in $M_0[x]$. \item[(c)] For $0 3$. Put $\tau=(a_1\;a_2\;a_3)$. This commutes with all the other cycles in $\sg$, and it follows that $[\tau,\sg]=[\tau,\rho]$. One can check directly that $[\tau,\rho]=(a_1\;a_2\;a_4)$, so $H$ contains a $3$-cycle, so $H=A_n$ by Lemma~\ref{lem-An-simple}. \item[(b)] Now suppose that~(a) does not hold, so $\sg$ involves only $3$-cycles and transpositions. Suppose that there is at least one transposition. As $\sg$ is even and $3$-cycles are even, there must be an even number of transpositions. We can thus write $\sg=\rho\om$, where $\rho=(a\;b)(c\;d)$ and $\om$ is disjoint from $\rho$. Put $\tau=(a\;b\;c)$; we then find that $[\tau,\sg]=[\tau,\rho]=(a\;b)(c\;d)$, so $H$ contains a transposition pair. It follows by Lemma~\ref{lem-An-simple} that $H=A_n$. \item[(c)] Now suppose that neither~(a) nor~(b) holds, so $\sg$ is a product of $3$-cycles. If $\sg$ is a single $3$-cycle then we can immediately use Lemma~\ref{lem-An-simple} to see that $H=A_n$. If there are at least two $3$-cycles then we can write $\sg=\rho\om$, where $\rho=(a\;b\;c)(d\;e\;f)$ and $\om$ is disjoint from $\rho$. We then put $\tau=(a\;b\;d)$ and check that \[ [\tau,\sg] = [\tau,\rho] = (a\;b\;e\;c\;d). \] We can thus apply case~(a) to this $5$-cycle to see that $H=A_n$ again. \end{itemize} \end{proof} \begin{corollary}\lbl{cor-not-solvable} For $n\geq 5$ the groups $\Sg_n$ and $A_n$ are not solvable. \end{corollary} \begin{proof} Suppose we have a solvable series $1=H_0\leq H_1\leq \dotsb\leq H_{r-1}\leq H_r=A_n$. After eliminating any repetitions, we may assume that these inclusions are strict, so $1 ] (-1.4,0) -- (2.4,0); \draw[->] (0,-3.3) -- (0,2.4); \draw (-1,0) -- (-1,-0.1); \draw ( 0,0) -- ( 0,-0.1); \draw ( 1,0) -- ( 1,-0.1); \draw ( 2,0) -- ( 2,-0.1); \draw (-1,-0.25) node{$-1$}; \draw ( 1,-0.25) node{$ 1$}; \draw[red,domain=-1.35:2.35,smooth,samples=200,variable=\x] plot ({\x},{\ff}); \draw[dotted] (1,0) circle(0.15); \end{scope} \begin{scope}[xshift=3cm,xscale=3,yscale=10] \draw (0.85,0) -- (1.15,0); \draw[red,domain=0.85:1.15,smooth,samples=200,variable=\x] plot ({\x},{\ff}); \draw[dotted] (1,0) circle(0.15); \end{scope} \end{tikzpicture} \end{center} More rigorously, we can check that \[ f'(x) = 210(x^6-2x^5-x^4+2x^3) = 210x^3(x-1)(x+1)(x-2), \] which has four real roots, at $-1,0,1,2$. Rolle's Theorem says that between any two real roots of $f(x)$ there is a real root of $f'(x)$, so there are at most five real roots. We also have \begin{align*} f(x) &\to -\infty\qquad\mbox{as }x\to-\infty\\ f(-1) &= 26 \\ f(0) &= -21 \\ f(1) &= 2 \\ f(2) &= -325 \\ f(x) &\to +\infty\qquad\mbox{as }x\to+\infty \end{align*} so (by the Intermediate Value Theorem) $f(x)$ has exactly five real roots. \end{solution} \begin{exercise}\exlabel{ex-affine-five} In this question it will be convenient to think $\Sg_5$ as the group of permutations of the set $\F_5=\{0,1,2,3,4\}$. For $a\in\F_5^\tm$ and $b\in\F_5$ we define $\rho_{ab}\:\F_5\to\F_5$ by $\rho_{ab}(u)=au+b$. We then put \[ U = \{\rho_{ab}\st a\in\F_5^\tm,b\in\F_5. \] \begin{itemize} \item[(a)] Prove that $U$ is a subgroup of $\Sg_5$, which contains a normal cyclic subgroup of order $5$, whose quotient is cyclic of order $4$. \item[(b)] Suppose that $H$ is some other subgroup of $\Sg_5$, and there is a cyclic subgroup $C$ of order $5$ that is normal in $H$. Prove that $H$ is conjugate to a subgroup of $A$. \item[(c)] Prove that any transitive subgroup of $\Sg_5$ is either equal to $\Sg_5$, or equal to $A_5$, or conjugate to a subgroup of $A$. \end{itemize} \end{exercise} \begin{solution}\ \\ \begin{itemize} \item[(a)] First note that \[ \rho_{ab}(\rho_{cd}(u)) = a(cu+d)+b = (ac)u+(ad+b) = \rho_{ac,ad+b}(u). \] It follows that $U$ is closed under composition. We also see that $\rho_{10}$ is the identity, and that $\rho_{1/a,-b/a}$ is an inverse for $\rho_{ab}$. This means that $U$ is a subgroup of $\Sg_5$. Now define $\pi\:U\to\F_5^\tm$ by $\pi(\rho_{ab})=a$. The above composition formula shows that $\pi(\rho_{ab}\rho_{cd})=ac=\pi(\rho_{ab})\pi(\rho_{cd})$, so $\pi$ is a homomorphism. For each $a\in\F_5^\tm$ we have an element $\rho_{a0}\in U$ with $\pi(\rho_{a0})=a$, so $\pi$ is surjective. The kernel is $V=\{\rho_{1b}\st b\in\F_5\}$, which is therefore a normal subgroup. The First Isomorphism Theorem tells us that $U/V\simeq\F_5^\tm=\{-2,-1,1,2\}$, which is cyclic of order $4$, generated by $2$. We also see from the composition formula that $\rho_{1b}\rho_{1d}=\rho_{1,b+d}$, so $\rho_{1b}=\rho_{11}^b$. It follows that $V$ is cyclic of order $5$, generated by $\rho_{11}$. \item[(b)] Let $H$ be a subgroup of $\Sg_5$, and let $C$ be a normal subgroup of $H$ that is cyclic of order $5$. Choose a generator $\sg$ for $C$. This has order $5$, and by considering the possible cycle types in $\Sg_5$ we see that it must be a $5$-cycle, say $\sg=(p_0\;p_1\;p_2\;p_3\;p_4)$. Let $\tht$ be the permutation that sends $i$ to $p_i$, and note that $\tht^{-1}\sg\tht=\rho_{11}$. Put $H'=\tht^{-1}H\tht$ and $C'=\tht^{-1}C\tht$, so $C'$ is normal in $H'$. As $\tht^{-1}\sg\tht=\rho_{11}$ we see that $C'=V$. Now consider an arbitrary element $\tau\in H'$. Put $b=\tau(0)\in\F_5$. As $V$ is normal in $H'$ we see that $\tau\rho_{11}\tau^{-1}$ must be another generator for $V$, so $\tau\rho_{11}\tau^{-1}=\rho_{1a}$ for some $a\in\F_5^\tm$. We now claim that $\tau=\rho_{ab}$, or equivalently that the permutation $\phi=\rho_{ab}^{-1}\tau$ is the identity. Indeed, we have $\rho_{ab}(0)=b=\tau(0)$, so $\phi(0)=0$. We also have \[ \rho_{ab}\rho_{11}\rho_{ab}^{-1}=\rho_{a,a+b}\rho_{1/a,-b/a}= \rho_{1a} = \tau\rho_{11}\tau^{-1}, \] so $\phi\rho_{11}\phi^{-1}=\rho_{11}$. This means that $\phi$ commutes with $\rho_{11}$, and thus also with $\rho_{1m}=\rho_{11}^m$. It follows that \[ \phi(m) = \phi(\rho_{1m}(0)) = \rho_{1m}(\phi(0)) = \rho_{1m}(0) = m, \] so $\phi$ is the identity as claimed, so $\tau=\rho_{ab}$. As $\tau$ was an arbitrary element of $H'$, we conclude that $H'\sse U$, and so $H=\tht H'\tht^{-1}\sse\tht U\tht^{-1}$. \item[(c)] Now instead let $H$ be an arbitrary transitive subgroup of $\Sg_5$. For any $x\in\F_5$, the orbit $Hx$ is then the whole set $\F_5$. We have the standard orbit-stabiliser identity $|H|=|Hx|.|\stab_H(x)|=5|\stab_H(x)|$, so $|H|$ must be divisible by $5$. Moreover, $|H|$ must divide $|\Sg_5|=120$, so it cannot be divisible by $5^2$. Let $C$ be any Sylow $5$-subgroup of $H$; then $|C|=5$ is prime, so $C$ must be cyclic. If $C$ is normal in $H$ then $H$ is conjugate to a subgroup of $U$ by part~(b). From now on we suppose that $C$ is not normal in $H$. Sylow theory tells us that the Sylow subgroups of $H$ are precisely the conjugates of $C$, and that the number $n$ of such conjugates divides $|H|/|C|$ and is congruent to $1$ modulo $5$. Moreover, as $C$ is not normal we have $n>1$, and $|H|/|C|$ must divide $|\Sg_5|/|C|=24$. It follows that $n=6$, and this must divide $|H|/|C|$, so $|H|\in\{30,60,120\}$. If $|H|=120$ then $H$ is all of $\Sg_5$. If $|H|=60$ then $H$ has index two, so it is normal by a standard lemma. It is not hard to deduce that $H=A_5$. \textbf{This just leaves the case where $|H|=30$. I think that there are no subgroups of order $30$ in $\Sg_5$, but this needs a proof. } \end{itemize} \end{solution} \begin{exercise}\exlabel{ex-special-sextic} Find an irreducible polynomial of degree 6 over $\Q$ with 4 real roots, but whose Galois group over $\Q$ is not $\Sg_6$. \end{exercise} \begin{solution} These are not too difficult to construct. Here is one way to do it: \begin{description} \item[1] Choose a cubic with two positive real roots and one negative real root. For example, $x^3-7x+6=(x+3)(x-1)(x-2)$. \item[2] Move this polynomial up or down the $y$-axis slightly to make it irreducible, but still ensuring that there are two positive and one negative real root. (If you do this cleverly, you will be able to use Eisenstein's criterion to check irreducibility!) For example, $x^3-7x+6-\frac{1}{6}=\frac{1}{6}(6x^3-42x+35)$ is irreducible by Eisenstein's criterion with $p=7$. \item[3] Now replace $x$ by $x^2$ to get a polynomial of degree 6. In our example, we can consider the polynomial $6x^6-42x^2+35$. Now this polynomial is still irreducible by Eisenstein with $p=7$, and its roots are the square roots of the roots of the cubic in step 2, two of which were positive, giving 4 real roots, and one negative, giving 2 imaginary roots. Finally, the Galois group cannot be $\Sg_6$, since the polynomial is solvable by radicals (the roots are just the square roots of the roots of the cubic, so are certainly expressible as radicals). \end{description} \end{solution} % \section{Things still to do} % \begin{itemize} % \item Ruler and compass constructions % \item Norms, traces and characteristic polynomials % \item Hilbert's Theorem 90 % \item The Normal Basis Theorem. % \end{itemize} \newpage \section*{Solutions} \label{apx-solutions} \includesolutions % \begin{bibdiv} % \begin{biblist} % \bibselect{% % ../../../BiBTeX/refs,% % ../../../BiBTeX/myrefs% % } % \end{biblist} % \end{bibdiv} \end{document}