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\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}	
	