%%This is a standard LaTeX2e article document template. personal version 12/5/200%% \documentclass[11pt,twoside]{article} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%packages%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \pagestyle{empty} \usepackage{latexsym} \usepackage{amssymb} \usepackage{amsfonts} \usepackage{amsthm} \usepackage{amstext} \usepackage{multicol} \usepackage[all]{xy} \usepackage{amsxtra} \usepackage[pdftex, breaklinks, linktocpage=true, bookmarksopen=true,bookmarksopenlevel=0,bookmarksnumbered=true]{hyperref} \hypersetup{ pdftitle = {Math 371 Assignment 7}, pdfauthor = {\textcopyright\ Bryden Cais}, pdfcreator = {\LaTeX\ with package \flqq hyperref\frqq}, colorlinks = {true} } %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%formatting%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \setlength{\topmargin}{0in} %%% This sets all the spacing stuff to use the page more \setlength{\oddsidemargin}{0in} %%% efficiently than the normal "article" setup would. \setlength{\evensidemargin}{0in} %%% It's OK to play with these some. \setlength{\textheight}{8.5in} %%% \setlength{\textwidth}{6.5in} %%% \setlength{\headsep}{0in} %%% \setlength{\headheight}{0in} %%% %\setlength{\footskip}{0in} %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Tor}{Tor} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\End}{End} \newcommand{\Q}{\mathbf{Q}} \newcommand{\R}{\mathbf{R}} \newcommand{\Z}{\mathbf{Z}} \newcommand{\C}{\mathbf{C}} \newcommand{\F}{\mathbf{F}} \newcommand{\p}{\mathfrak{p}} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}{Corollary} \begin{document} \begin{center} {\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ {\large Solutions 7}\\Due Friday, April 9 at 08:35 \end{center} \begin{enumerate} \item Let $p$ be a prime and let $K$ be a splitting field of $X^p-2\in \Q[X]$, so $K/\Q$ is a Galois extension. Show that $K=\Q(a,\zeta)$ for $a\in K$ satisfying $a^p=2$ and $\zeta\in K$ a primitive $p$ th root of unity. Describe generators of $G:=\Gal(K/\Q)$ in terms of their actions on $a$ and $\zeta$, and describe $G$ as an abstract group (in terms of generators and relations, say). Write out the diagrams of intermediate fields and groups, indicating clearly the various containments. Also indicate which subfields of $K$ are Galois over $\Q$. \noindent{\bf Solution:} We treat the case that $p>2$ as $p=2$ is a bit different and should be analyzed separately (it's not at all hard). Let $a=a_1,\ldots,a_p\in K$ be the $p$ distinct roots of the separable polynomial $X^p-2$ in $K$. Since $(a/a_i)^p=1$ we see that $\zeta_i:=a/a_i$ is a $p$-th root of unity in $K$ for all $i$. Moreover, for $i\neq j$ we have $\zeta_i\neq \zeta_j$ as the $a_i$ are distinct. This gives $p$ distinct roots of unity in $K$ so since $\mu_p(\overline{K})$ has size $p$, we conclude that $K$ contains all the $p$-th roots of unity. Let $\zeta\in K$ be any primitive $p$-th root of unity. Then up to relabeling, we have $a_i = \zeta^i a$ and it follows that $K\subseteq \Q(a,\zeta)$. The reverse containment has already been noted, we have equality. Any automorphism of $K$ over $\Q$ is therefore uniquely determined by its action on $a$ and $\zeta$. For $u\in (\Z/p\Z)^{\times}$ and $v\in \Z/p\Z$, it is not hard to check that the map $$ \sigma_{u,v}:\begin{cases} a\mapsto \zeta^v a \\ \zeta \mapsto \zeta^u\end{cases} $$ is a well-defined automorphism of $K$ that fixes $\Q$. Furthermore, every automorphism of $K$ fixing $\Q$ must have this form, as the roots of the minimal polynomials of $\zeta$ and of $a$ are permuted among themselves. %We easily determine the relations %$$ % \sigma^p = \tau^{p-1} = 1,\quad \tau\sigma = \sigma^u\tau %$$ %and that $\{\sigma^i\tau^j\}$ for $i=1,\ldots, p$ and $j=1,\ldots,p-1$ are all distinct. Via the canonical isomorphism of groups $(\Z/p\Z)^{\times}\simeq \Aut(\Z/p\Z)$, we have a natural action of $(\Z/p\Z)^{\times}$ on $\Z/p\Z$, and one checks that the map $$(\Z/p\Z)^{\times} \ltimes \Z/p\Z \rightarrow G\qquad (u,v)\mapsto \sigma_{u,v}$$ is an isomorphism of groups. %This gives $p(p-1)$ automorphisms. On the other hand, since $p\neq 2$, using arguments %from algebraic number theory (specifically that the prime (2) does not ramify in $\Q(\zeta)$ %since the discriminant of the latter is a power of $p$) one shows by a variant of Eisenstein's %criterion discusses in class that $X^p-2$ is irreducible over $\Q(\zeta)$. It follows that $K$ has %degree $p(p-1)$ over $\Q$ and hence we have found all of the automorphisms. Thus, $G:=\Gal(K/\Q)$ %is the group generated by $\sigma,\tau$ with the relations as above. It's not hard to see that %$$G\simeq \Z/p\Z \rtimes (\Z/p\Z)^{\times}$$ %with $(\Z/p\Z)^{\times}$ acting on $\Z/p\Z$ via the canonical identification %of finite groups $$(\Z/p\Z)^{\times}\simeq \Aut(\Z/p\Z).$$ This group is also isomorphic with the group $$\{x\mapsto ux+v\ : u\in \F_p^{\times},\ v\in \F_p\}$$ of affine transformations of the $\F_p$-line. Classifying the subgroups of this group is a good exercise in group theory, and is left to the reader. %If $H$ is any subgroup of $G$, then $H\cap \Z/p\Z$ is a subgroup of $\Z/p\Z$ hence %is either 0 or $\Z/p\Z$. Likewise, the image of $H$ under the quotient map %$G\twoheadrightarrow (\Z/p\Z)^{\times}$ is a subgroup of the cyclic group $(\Z/p\Z)^{\times}$, %hence is cyclic of order dividing $p-1$. \item Let $F$ be a finite field of size $\#F$, with $K/F$ a finite extension of degree $d$. Prove that $K/F$ is Galois and that $\Gal(K/F)$ is a cyclic group of order $d$ with generator the automorphism of $K$ given by $$\alpha\mapsto \alpha^{\#F}.$$ (This automorphism is called the {\em arithmetic Frobenius} map of $F$). \noindent {\bf Solution:} Set $q:=\#F$ and let $p$ be the characteristic of $F$, so $q=p^r$ for some $r$. We have seen that $K^{\times}$ is cyclic of order $q^d-1$, and hence every $\alpha$ in $K$ is a root of $X^{q^d}-X$, which is a separable polynomial with $F$ (even $\F_p$) coefficients. Thus, $K/F$ is the splitting field of a separable polynomial and hence Galois. Consider the map $\varphi_F:K\rightarrow K$ given by $\varphi_F(\alpha) = \alpha^q$. It is straightforward to check that this is an automorphism of $K$ which fixes $F$ pointwise as every $x\in F$ is a root of $X^q-X$. The iterates $\{\varphi_F^i\}$ for $i=0,1,\ldots, d-1$ are all distinct, since otherwise we would have $\varphi^j = 1$ for some $1\le j \le d-1$, whence every element of $K$ would be a root of $X^{q^j}-X$ which is impossible as this polynomial has at most $q^j$ roots and $K$ has $q^d$ elements. On the other hand, the Galois group of $K/F$ has order $d$ as $[K:F]=d$ and we conclude by size considerations that $\Gal(K/F)$ is cyclic of order $d$, generated by $\varphi_F$. \item This exercise gives Artin's proof of the fundamental theorem of Algebra. Let $F$ be a field not of characteristic 2 and assume that all odd degree polynomials in $F[X]$ have a root in $F$. Let $K$ be a quadratic extension of $F$ with the property that every element of $K$ has a square root in $K$. \begin{enumerate} \item Prove that any finite extension of $K$ has degree a power of $2$. (Hint: Reduce to the Galois case and then consider the fixed field of the 2-Sylow subgroup of the Galois group).\label{2gp} \item Prove that $K$ has no non-trivial finite extensions which are Galois over $F$, and conclude that $K$ is algebraically closed. (Hint: Use the fact that a non-trivial 2-group has an index 2 normal subgroup).\label{algclos} \item Let $F=\R$ and $K=\R[X]/(X^2+1)$. Explain (using the intermediate value theorem) why $F$ satisfies the hypotheses above, and using explicit formulae, show that $K$ also satisfies the hypotheses. Conclude that $\C:=K$ is algebraically closed (this is the Fundamental Theorem of Algebra). \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item Let $L$ be any finite extension of $K$. To show that $L/K$ has degree a power of $2$ it suffices to show that this holds for any finite extension $L'$ of $F$ which contains $L$ and is Galois over $F$ (using that $[K:F]$ has degree $2$). Such an extension exists, as we could take the Galois closure of $L$ over $F$. Thus, we may assume that $L$ is Galois over $F$ with group $G$. Let $H$ be the Sylow 2-subgroup of $G$ and $E=L^H$ the corresponding fixed field, so $[E:F]=[G:H]$ has odd degree. If $\alpha\in E$ then $m_{\alpha,F}$ must have odd degree, as this degree divides $[E:F]$ and our hypothesis on $F$ ensures that $m_{\alpha,F}$ has a root in $F$ and hence that $\alpha\in F$. Thus, $E=F$ and $H=G$ by Galois theory (as $F=L^G$). We conclude that $G$ is a 2-group, whence $[L:F]$ is a power of 2. It follows that $[L:K]$ is also a power of 2. \item Suppose $L/K$ is a finite extension which is Galois over $F$. Then $L$ is Galois over $K$, with Galois group $G$, say, which must be a 2-group by part (\ref{2gp}). If $G$ is nontrivial, then there exists a normal index 2 subgroup $H$ of $G$, or what is the same thing, a nontrivial degree 2 (Galois) extension of $K$. Such an extension is obtained by adjoining a square-root of an element of $K$ (as the characteristic of $K$ is not 2). But every element of $K$ has a square root in $K$, so any such extension must be trivial and $G$ is the trivial group. Hence $L=K$. This implies that $K$ is an algebraic closure of $F$, since any polynomial with $F$ coefficients has splitting field over $K$ that is finite Galois over $F$ and hence must be trivial as an extension of $K$; that is, any polynomial over $F$ has all root in $K$. \item Any odd degree polynomial $f$ with real coeffecients has a real root by the intermediate value theorem since $$\lim_{x\rightarrow \pm \infty} f(x) = \pm \infty.$$ On the other hand, let $a+bX\in K:=\R[X]/(X^2+1)$ be arbitrary (so $a,b\in \R$). Let $D:=\sqrt{a^2+b^2}$ be the unique positive (real) square root of the positive real number $a^2+b^2$, and note that $D\pm a > 0$. We set $$s:=\sqrt{\frac{1}{2}\left(D+a\right)}\qquad t:=\sqrt{\frac{1}{2}\left(D-a\right)}.$$ It is straightforward to check that for $u:=\mathrm{sgn}(b)=\pm 1$ $$(s+utX)^2 = (s^2-t^2)+2stuX = a+bX,$$ so every element of $K$ has a square root in $K$. Thus $K$ is algebraically closed by part (\ref{algclos}). \end{enumerate} \item Determine the Galois group of the splitting field (over $\Q$) of $X^4-14X^2+9$, and write down the lattice of subgroups and corresponding subfields. Which subfields are Galois over $\Q$? \noindent {\bf Solution:} Observe first that $f:=X^4-14X^2+9$ is irreducible: the rational root test shows it has no rational roots, and if $f$ were to factor as a product of two quadratics, then since the $X^3$ and $X$ coeffieicnts are zero, it is not hare to see that such a factorization must take the form $$(X^2+aX+b)(X^2-aX+b) = X^4 + (2b-a^2)X^2 + b^2.$$ This would force $b=\pm 3$ whence we must have $\pm 6-a^2 = -14$. But neither $8$ nor $20$ are perfect rational squares, so this last equation has no rational solutuons. Using the quadratic formula, one sees that the roots of $f:=X^4-14X^2+9$ in $\C$ are $$ \alpha:= \sqrt{7 +2\sqrt{10}},\quad \beta:=-\alpha,\quad \gamma:=\sqrt{7 -2\sqrt{10}},\quad \delta:=-\gamma.$$ Observe that $\alpha\gamma = \sqrt{49-40} = 3$, whence $\gamma\in \Q(\alpha)$ so $K:=\Q(\alpha)$ contains all 4 roots of $f$ and is hence a splitting field. By the irreducibility of $f$, we have $[K:\Q]=4$ so $G:=\Gal(K/\Q)$ has order 4. Since $\alpha(\alpha^3-14\alpha) = -9$, we deduce that $$\gamma = \frac{3}{\alpha} = \alpha\frac{14 -\alpha^2}{3}$$ Since $K=\Q(\alpha)$, and $\sigma\in G$ is completely determined by its action on $\alpha$, and since $f$ is irreducible, for any root $\alpha'$ of $f$ there exists $\sigma\in G$ taking $\alpha$ to $\alpha'$. Thus, the elements of $G$ are: $$ 1,\quad \sigma_1:\alpha\mapsto \gamma,\quad \sigma_2: \alpha\mapsto -\alpha, \quad \sigma_3:\alpha\mapsto -\gamma. $$ We compute $$ \sigma_1^2(\alpha) = \sigma_1(\gamma) = \sigma_1\left(\frac{3}{\alpha}\right) = \frac{3}{\gamma} = \alpha $$ so $\sigma_1$ has order $2$. Since $\sigma_1$ obviously has order 2, we deduce that $G$ contains at least 3 elements of order dividing $2$ and hence $G\simeq \Z/2\Z \times \Z/2\Z$ is the Klein 4-group. Thus, $G$ has 3 nontrivial subgroups given by $H_i:=\langle 1,\sigma_i\rangle$ for $i=1,2,3$. Note that $\eta_i:=\alpha\sigma_i(\alpha)$ and $\tau_i=\alpha+\sigma_i(\alpha)$ are fixed by $H_i$ for $i=1,2,3$. We easily compute that $$\eta_1=3,\quad \tau_1 = 2\sqrt{5},\qquad \eta_2 = -7-2\sqrt{10},\quad \tau_2 = 0,\qquad \eta_3 = -3,\quad \tau_3 = 2\sqrt{2}.$$ From this, we conclude that $$K^{H_1} = \Q(\sqrt{5}),\qquad K^{H_2} = \Q(\sqrt{10}),\qquad K^{H_3} = \Q(\sqrt{2})$$ since in each case $K^{H_i}$ is degree 2 over $\Q$ and contains the given quadratic field. Each of these is Galois over $\Q$ since $G$ is abelian so every subgroup is normal. \item Fix a positive integer $n$ and let $K:=\Q(\zeta_n)$ for a primitive $n$ th root of unity $\zeta_n\in \C$. Prove that complex conjugation $\tau\in \Aut(\C)$ restricts to an automorphism of $K$ fixing $\Q$, and show that the corresponding element of $\Gal(K/\Q)$ corresponds to $-1$ under the isomorphism $\Gal(K/\Q)\simeq (\Z/n\Z)^{\times}$. Prove that the fixed field $K^+$ of the subgroup generated by complex conjugation is equal to the intersection $K\cap \R$ taken inside $\C$. We call $K^+$ the {\em maximal real subfield} of $K$. \noindent{\bf Solution:} We suppose $n \ge 3$ as the cases $n=1,2$ are trivial. The primitive $n$ th roots of unity in $\C$ are given by $$e^{2\pi i k/n}=\cos(2\pi k/n)+ i\sin(2\pi k/n)$$ for $k\in (\Z/n\Z)^{\times}$. Thus, if $\tau\in \Aut(\C)$ denotes complex conjugation, we have $$\tau(e^{2\pi i k/n}) = \cos(2\pi k/n) - i \sin(2\pi k/n) = e^{2\pi i k/n}$$ so $\tau(\zeta_n)=\zeta_n^{-1}$. In particular, $\tau$ preserves $K$ and so restricts to an automorphism of $K$ which obviously fixes $\Q$ (any field automorphism automatically fixes the prime subfield). We also see from this that under the isomorphism $\Gal(K/Q)\simeq (\Z/n\Z)^{\times}$ which associates $\sigma\in \Gal(K/\Q)$ to the unique $a_{\sigma}\in (\Z/n\Z)^{\times}$ satisfying $\sigma\zeta_n = \zeta_n^{a_{\sigma}}$ (independent of $\zeta_n$), complex conjugation corresponds to $-1\in (\Z/n\Z)^{\times}$. Now it is clear that $K^+$ contains $K\cap \R$ since $\R$ is fixed by complex conjugation. On the other hand, $b:=\zeta_n+\tau\zeta_n$ and $c:=\zeta_n\tau\zeta_n$ lie in $K\cap \R$ so $\zeta_n$ and $\tau\zeta_n$ are the roots of the degree 2 polynomial $$X^2-bX+c\in K\cap \R[X].$$ In particular, $[K:K\cap \R]\le 2$. But $[K:K^+]=2$ since complex conjugation generates an order 2 subgroup of $\Gal(K/\Q)$ (here I'm using that $n\ge 3$), whence $$2=[K:K^+] \le [K:K^+]\cdot [K^+:K\cap \R] = [K:K\cap \R] \le 2$$ so we must have equality throughout, forcing $[K^+:K\cap \R]=1$ so $K^+=K\cap \R$. \item This problem works out a formula for $\cos(2\pi/17)$ in terms of square-root extractions. Let $\zeta:=e^{2\pi i/17}$; it is a primitive $17$ th root of unity. Let $\alpha:=\zeta+\zeta^{-1}=2\cos(2\pi/17)$. Let $\sigma\in \Gal(\Q(\zeta)/\Q)$ be the element determined by $$\sigma \zeta = \zeta^3.$$ \begin{enumerate} \item Show that $\sigma$ generates $\Gal(\Q(\zeta)/\Q)$ \item Define the {\em periods} of $\alpha$ to be \begin{align*} & \eta_1:= \alpha + \sigma^2\alpha + \sigma^4\alpha + \sigma^8\alpha && & \eta_1':=\sigma\eta_1\\ & \eta_2 := \alpha + \sigma^4\alpha && & \eta_2':=\sigma^2 \eta_2 \\ & \eta_3 := \sigma \eta_2 && & \eta_3':=\sigma\eta_2' \end{align*} Prove that $\eta_1,\eta_1'$ are the roots of $X^2+X-4$, that $\eta_2,\eta_2'$ are the roots of $X^2-\eta_1X-1$, that $\eta_3,\eta_3'$ are the roots of $X^2-\eta_1'-1$ and that $\alpha$ and $\sigma^4\alpha$ are the roots of $X^2-\eta_2X+\eta_3$. \item Conclude that $\cos(2\pi/17)$ is equal to \begin{equation*} \frac{1}{16}\left(-1 +\sqrt{17}+\sqrt{2(17-\sqrt{17})} +2\sqrt{17+3\sqrt{17}-\sqrt{2(17-\sqrt{17})}-2\sqrt{2(17+\sqrt{17})}}\right) \end{equation*} \end{enumerate} \noindent{\bf Solution:} This is all straightforward and we refer the reader to Dummit and Foote, pg. \item Let $F$ be a field and $f\in F[X]$ a monic separable polynomial of degree $n$. Fix a splitting field $K$ of $f$ and write $G:=\Gal(K/F)$. \begin{enumerate} \item Prove that $G$ is a subgroup of $S_n$, the symmetric group on $n$ letters. If $f$ is irreducible, prove that $G$ is a {\em transitive} subgroup of $S_n$ with $\#G$ divisible by $n$. (Viewing $S_n$ as the permutations of an $n$-element set $T$, a transitive subgroup $G$ is one which acts transitively on these $n$ elements, i.e. for any $x,y\in T$ there exists $g\in G$ such that $gx=y$). \item Prove that if $n$ is prime and $f$ is irreducible, then $G$ contains an $n$-cycle. (Hint: use Sylow's theorem.) \label{ncycle} \item Suppose that $f$ is irreducible of degree $5$ and has exactly 3 real roots. Prove that $G$ is isomorphic to $S_5$. (Hint: View $K$ as a subfield of $\C$ and consider complex conjugation acting on $K$. Now use (\ref{ncycle}) and some group theory.) \end{enumerate} \noindent{\bf Solution:} \begin{enumerate} \item We have seen that for any $g\in G$ and any root $\alpha$ of $f$ in $K$, we must have $g\alpha$ a root of $f$; in particular, $G$ acts on the set of $n$ distinct roots of $f$ (using separability here) by permutations, and in this way we get a map $G\rightarrow S_n$ which is obviously a homomorphism since the group law in each case is composition. This map is injective since any $g\in G$ is determined by its action on the roots of $f$, as $K$ is generated over $F$ by these roots. Thus, we may view $G$ as a subgroup of $S_n$. If $f$ is irreducible, then we use the following theorem from class: \begin{theorem} Let $F$ be any field and $K,K'$ extensions of $F$. Suppose that $f\in F[X]$ is irreducible over $F$ and has roots $\alpha\in K$ and $\beta\in K'$. Then there exists an isomorphism of fields $F(\alpha)\simeq F(\beta)$ restricting to the identity on $F$ and mapping $\alpha$ to $\beta$. \end{theorem} \begin{proof} By the universal mapping property of polynomial rings, we have a surjective map $F[X]\twoheadrightarrow F(\alpha)$ mapping $X$ to $\alpha$. This map kills $(f)$ so yields a surjective mapping $F[X]/(f)\twoheadrightarrow F(\alpha)$ which must be injective since the source is a field (since $f$ is irreducible, $(f)$ is prime and hence maximal as $F[X]$ is a PID so the quotient ring is a field). Thus, $F(\alpha)\simeq F[X]/(f)$. By the same token, $F[X]/(f)\simeq F(\beta)$. Composing these isomorphisms yields the desired isomorphism. \end{proof} Applying this theorem, we see that in our situation (with $f$ irreducible) for {\em any} roots $\alpha,\beta$ of $f$ in $K$, there exists an automorphism of $K$ over $F$ mapping $\alpha$ to $\beta$; in other words, $G$ acts transitively on the roots of $f$ so via our embedding of $G$ into $S_n$ we see that $G$ is a transitive subgroup of $S_n$. Moreover, since $f$ is irreducible, we have $[F(\alpha):F]=n$ for any root $\alpha$ of $f$ in $K$. Since $\#G=[K:F] = [K:F(\alpha)][F(\alpha):F]$, we see that $n|\#G$. \item If $n=p$ is prime and $f$ is irreducible, then by the previous part we know that $\#G$ is divisible by $p$. By Sylow's theorem, $G$ then contains an element of order $p$, which under the embedding of $G$ into $S_p$ must be a $p$-cycle. \item If $f$ is an irreducible quintic with exactly 3 real roots, then $f$ has exactly 2 complex non-real roots. The action of complex conjugation on the roots of $f$ therefore swaps these two roots, so under the embedding of $G$ in $S_5$ corresponds to a 2-cycle. By the previous part, the image of $G$ in $S_5$ also contains a $5$-cycle. But $S_5$ is generated (as a group) by any 5-cycle and any 2-cycle, so the image of $G$ is all of $S_5$ whence $G\simeq S_5$. \end{enumerate} \item Keep the notation of the previous problem. \begin{enumerate} \item Let $r_1,\ldots,r_n$ be the $n$ distinct roots of $f$ in $K$, and define the {\em discriminant of $f$} to be $$\Delta(f):= \prod_{i\neq j} (r_i-r_j),$$ where the product runs over all pairs $(i,j)\in \Z^2$ with $1\le i,j\le n$. Prove that $\Delta(f)\in F$. \item Prove that $G$ is a subgroup of $A_n$ (the alternating group) if and only if $\Delta(f)$ is a square in $F$. Hint: use the formula for $\Delta(f)$ above and the definition of $A_n$ as the group of even permutations. \end{enumerate} \noindent{\bf Soultion:} \begin{enumerate} \item For an arbitrary element $\sigma\in S_n$, we have $$\prod_{i,j} (r_{\sigma i}-r_{\sigma j}) = \prod_{i,j} (r_{i}-r_{j})$$ which shows that {\em any} permutation of the $r_i$'s leaves $\Delta(f)$ unchanged. In particular, $G$ acts trivially on $\Delta(f)$ so $\Delta(f)\in F$ by Galois theory. \item Observe that $\Delta(f)$ is the square of $$P(r_1,\ldots,r_n):=\prod_{i < j} (r_i-r_j)\in K$$ so that $\Delta(f)$ is a square if and only if $P(r_1,\ldots,r_n)\in F$. If $\sigma\in S_n$, then by definition we have $$\mathrm{sgn}(\sigma) = \frac{P(r_{\sigma 1},r_{\sigma 2},\ldots, r_{\sigma_n})}{P(r_1,r_2,\ldots,r_n)},$$ which shows that $\sigma\in S_n$ fixes $P(r_1,\ldots,r_n)$ if and only if $\sigma$ is an even permutation. Putting these together, we see that $G$ is a subgroup of $A_n$ if and only if $G$ fixes $P(r_1,\ldots,r_n)$, i.e. if and only if $\Delta(f)$ is a square in $F$. \end{enumerate} \end{enumerate} \end{document}