Let $F$ be a field of characteristic not 2 and let the symmetric group $S_{n}$ act on the polynomial ring $F\left[X_{1}, \ldots, X_{n}\right]$ by permuting the variables, for $n \geq 2$. Let $A=\left(F\left[X_{1}, \ldots, X_{n}\right]\right)^{A_{n}}$ and $B=\left(F\left[X_{1}, \ldots, X_{n}\right]\right)^{S_{n}}$ be the fixed subrings, where $A_{n}<S_{n}$ is the alternating group.
(a) Show that $A$ is an integral extension of $B$.
(b) Show that $A=B[\delta]$ for some $\delta \in A$ such that $\Delta:=\delta^{2}$ belongs to $B$.
For part (a) It suffices to show that any element in $A$ is the root of a monic polynomial in $B$. Suppose $g(X_1,...,X_n) \in A$ so that $g$ remains the same when the input variables are permuted by any even permutation. But I'm having trouble figuring out which polynomial I should use.
If $\sigma$ is any permutation, let us denote by $\sigma\cdot f$ its action on any polynomial $P$.
Pick any transposition $\tau$. Assume that $g$ is fixed by all the even permutations. Then $S=\tau\cdot g+g$ and $P=(\tau\cdot g)g$ are fixed by $\tau$.
I claim that $S$ and $P$ are fixed by any element of $S_n$. Since $A_n$ and $\tau$ generate the group $S_n$, it is enough to prove that they are also fixed by any even permutation.
Let $\sigma\in A_n$, then $\sigma\cdot S=\sigma\tau\cdot g+\sigma\cdot g=\tau(\tau^{-1}\sigma\tau)\cdot g+\sigma\cdot g$. Now, $\tau^{-1}\sigma\tau$ and $\sigma$ are even, so $\sigma\cdot S=\tau\cdot g+g$. We prove in the same way that $P$ is fixed by $\sigma$, and we are done.
Now $0=(g-g)(g-\tau(g))=g^2-Sg+P$, which reaches the desired conclusion.
Note that, as Alex Wertheim mentioned, we have a lemma:
Lemma 1. If a finite group $G$ acts on a ring $A$ by automorphisms, $A$ is integral over $A^G$.
The idea is the same as before. Let $a\in A$, and let $Q(T)=\prod_{g\in G}(T-g\cdot a)$. Now $Q(a)=0$, and clearly $Q$ has coefficients in $A^G$.
So we could apply Lemma $1$ to the ring $A=F[X_1,\ldots,X_n]^{A_n}$ and $G=S_n$, noting that $A^{S_n}=(F[X_1,\ldots,X_n]^{A_n})^{S_n}=F[X_1,\ldots,X_n]^{S_n}$.
However, this would give a polynomial $Q$ of degree $n!$, while the first proof yields a polynomial of degree $2$.
The underlying idea in the first proof to get a polynomial of degree $2$ is a combination of two ideas: Lemma $1$ above, and the following result:
Lemma 2. If $G$ acts by automorphisms on a ring $A$, and $H$ is an normal subgroup acting trivially on $A$, then:
the restriction on the action of $G$ restrict to an action on $A^H$
the action of $G$ on $A^H$ induces an action of $G/H$ on $A^H$
We have $A^G=(A^H)^{G/H}$.
Proof. We first prove that $G$ acts on $A^H$. If $a\in A^H$, and $g\in G$, then for all $h\in H$, we have $h\cdot (g\cdot a)=g\cdot(g^{-1}hg\cdot a)=g\cdot a$, since $g^{-1}hg\in H$ by normality and $a\in A^H$. Hence $g\cdot a\in A^H$
The second point is clear since $H$ acts trivially on $A^H$, and the third one is an easy exercise (it is just about going through the definitions).
In particular, we get $A^{S_n}=(A^{A_n})^{S_n/A_n}$ and the proof of Lemma $1$ yields an integral relation of degree $2$. In fact, this is exactly the first proof I gave.