TITLE: Ring of Invariants of $A_3$
QUESTION [9 upvotes]: It is well know that the ring of invariants of the permutation group of three elemetns $S_3$ is given by the elementary symmetric polynomials, i.e. $$\mathbb C[x,y,z]^{S_3} = \mathbb C[s_1,s_2,s_3]$$ where $$s_1 = x+y+z$$ $$s_2=xy + yz + zx$$ $$s_3= xyz.$$
Which is the ring of invariants of $A_3$, the alternating subroup of $S_3$, made of even permutations? 
Of course it contains $s_1, s_2, s_3$. It also contains $$d=(x-y)(y-z)(z-x).$$ They are not algebrically independent because $d^2 \in \mathbb C[s_1,s_2,s_3]$.
How can we prove that these $4$ polynomials generate the whole ring of invariants? Which is a basis of algebrically independent generators, if any?

REPLY [4 votes]: Let $a_n$ be the dimension of the space of polynomials of degree $n$ invariant by the action of $A_3$.
The formal power series $\sum_{n \ge 0} a_n t^n$ is called the Molien series of $A_3$.
You can compute it with Molien's formula, or by comparing it with the Molien series of $S_3$.
Since the ring of invariants of $S_3$ has an algebraically independant  generating set given by the elementary symmetric polynomials of degrees $1,2,3$, its Molien series is $\frac 1{(1-t)(1-t^2)(1-t^3)}$.
Since the ring of invariants of $A_3$ is obtained by $S_3$'s by adding $d$ of degree $3$ whose square is invariant by $S_3$, we then have that $A_3$'s Molien series is $\frac{1+t^3}{(1-t)(1-t^2)(1-t^3)} = \frac{1-t+t^2}{(1-t)^2(1-t^3)}$.
However, a ring of polynomial with a generating set that is algebraically independent will have a Molien series of the form $\prod (1-t^{d_i})^{-1}$ where the $d_i$ are the degrees of the generators. In particular, those rational fractions never vanish for $t \in \Bbb C$.
The Molien series for $A_3$ has a zero at the roots of $1-t+t^2$, so it doesn't have an algebraically independent generating set.