Invariant ring of the alternating group I suspect that the invariant ring of $A_n$'s action on $K[x_1, ..., x_n]$ are the "alternating polynomials" - ie the symmetric polynomials adjoined with the Vandermonde polynomial 
$$\prod_{i < j} (x_j - x_i)$$
and I know that the symmetric polynomials as well as the Vandermonde polynomial are fixed by $A_n$. I am having difficulty showing this generates all the invariants, and would like to know some good ways to tackle the problem. Currently, I am trying to say that a polynomial $f \in K[x_1, ..., x_n]^{A_n}$ "either flips signs or doesn't flip signs when its inputs are transposed" though I am having difficulty saying why. 
 A: With a classmate's permission, this is their solution to the problem (G is $A_n$ here):
Let $K$ be a field of characteristic $\neq 2$. Let $G$ be the alternating group $A_n$. Let $K\left[\boldsymbol{x}\right]$ be the polynomial ring $K\left[x_1, x_2, \ldots, x_n\right]$.
We claim that the invariant ring $K[\boldsymbol{x}]^G$ is generated by symmetric polynomials (so we may take the Schur polynomial basis) and the Vandermonde polynomial $P  = \prod_{i < j} (x_j - x_i)$. Note that symmetric polynomials are invariant under $S_n$ (simply by definition) and also the Vandermonde polynomial is invariant under $A_n$ since an even number of transpositions swaps some two variables an even number of times, so the sign of $P$ also changes an even number of times and so it is invariant. We now show that these polynomials indeed generate the whole invariant ring.
It suffices to show that any polynomial $f$ that is invariant under the action of $A_n$ can be expressed as a $K[\boldsymbol{x}]^{S_n}$-linear combination of a symmetric polynomial and $P$. To do this, define $g(x_1, x_2, \dots, x_n) = f(x_2, x_1, x_3, \dots, x_n)$, i.e. $g$ is simply a polynomial $(1\;2)f$, the image of $f$ under the action of the transposition $(1\;2)$. In fact, it is irrelevant which odd permutation we pick in place of $(1\;2)$ while defining $g$, as any transposition can be written as a product of any other transposition and some even permutation (which we know fixes $f$ by assumption). Observe now that for any permutation $\pi\in S_n$ (and $\sigma$ any transposition), we get
$$\pi(f + g) = \pi f + \pi g = \pi  f + \pi (\sigma  f) = \pi f + \sigma' \pi f $$
where $\sigma'$ is a transposition. Note that if $\pi$ is even then $\pi (f + g) = f + g$, and similarly if $\pi$ is odd then $\pi (f + g) = g + f$, thus proving that $f + g$ is symmetric. Moreover, note that if $\pi$ is an odd permutation, we have $$\pi (f - g) = \pi f - \pi \sigma f = g - f$$ so $f-g$ changes sign when acted on by an odd permutation. This means that for any polynomial $f$ invariant under $A_n$, we may write
$$f = \frac{1}{2}(f+g) + \frac{1}{2}(f-g),$$ and so $f$ can be written as a $K[\boldsymbol{x}]^{S_n}$-linear combination of a symmetric polynomial and an alternating polynomial. It now suffices to show that $P$ divides any alternating polynomial to conclude the proof (since the quotient will be a symmetric polynomial).
Note that for  $\sigma = (i\; j)$ and an alternating polynomial $h$, we have $\sigma h = - h$. Also, whenever $x_i = x_j$ we have $\sigma h(\boldsymbol{x}) = h(\boldsymbol{x})$. This means that $h(\boldsymbol{x}) = 0$ when $x_i = x_j$, and so $(x_i - x_j) \vert h$. This holds for any $i \neq j$ and the polynomial ring is a UFD, so we get that $P\vert h$, as desired.
A: I'll assume $K$ does not have characteristic two, so that $-1\ne1$.
The $A_n$ invariant polynomials then split as a direct sum of two
components: the symmetric functions, and the alternating functions
where a permutation $\sigma$ of the variables multiplies a function
by the sign of $\sigma$.
To prove this, consider a function $f$ fixed by $A_n$. Then let
$f^*(x_1,\ldots,x_n)=f(x_2,x_1,\ldots,x_n)$. Then $f+f^*$ is a symmetric
function, while $f-f^*$ is alternating: it is fixed by $A_n$
but any odd permutation of its argument multiplies it by $-1$. So
$f=\frac12(f+f^*)+\frac12(f-f^*)$ is a sum of a symmetric and an alternating
function.
For alternating function $f$,
$$f(x_2,x_1,x_3,\ldots,x_n)=-f(x_1,x_2,x_3,\ldots,x_n)$$
and so
$$f(x_1,x_1,x_3,\ldots,x_n)=0.$$
Therefore $f$ is divisible as a polynomial by $x_1-x_2$. Likewise it
is divisible by all $x_i-x_j$ and so, by unique factorisation, by
$\prod_{i<j}(x_i-x_j)$. The quotient will be symmetric.
