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?

  • 1
    $\begingroup$ Maybe try to show that the field of fractions of the ring you just produced is a degree two extension of the field of fractions of $\mathbb C[x,y,z]^{S_3}$? Then by Galois theory, you are done. $\endgroup$
    – Arkady
    Apr 15, 2018 at 16:01
  • $\begingroup$ @Ravi what does this theory tell us? If $[H:G]=k$ then the field of fractions of the ring of invariants under $H$ is an extension of that under $G$ of degree $k$? $\endgroup$
    – Maffred
    Apr 15, 2018 at 16:06
  • 1
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – Arkady
    Apr 15, 2018 at 16:16
  • $\begingroup$ What do you mean by "of order 2"? $\endgroup$
    – verret
    Apr 16, 2018 at 1:04
  • 1
    $\begingroup$ Right, the usual terminology is "even permutation". Because the elements of order $2$ in $S_3$ are actually odd permutations. $\endgroup$
    – verret
    Apr 16, 2018 at 4:21

2 Answers 2


Since $A_n$ is normal in $S_n$, the group $S_n$ setwise stabilizes the ring $R=\mathbf{C}[x_1,\dots,x_n]^{A_n}$ of $A_n$-invariants. Writing $S=\mathbf{C}[x_1,\dots,x_n]^{S_n}$ for the ring of symmetric polynomials, we claim that $R=S+S\delta$, where $\delta=\prod_{1 \leq i < j \leq n} (x_i-x_j)$ is the Vandermonde determinant.

Since $S_n$ acts via its quotient $S_n / A_n$ on $R$, there are two isotypes of $S_n$-representations that occur in $R$: the trivial representation and the sign representation. It follows that for $f \in R$ we have $$f=\mathrm{sym}(f)+\mathrm{alt}(f)$$ where we write "sym" and "alt" for the symmetrization and anti-symmetrization of $f$. Since every alternating polynomial is divisible by $\delta$, with quotient a symmetric polynomial, this proves our claim.

Now the ring $S$ of symmetric polynomials is generated by the elementary symmetric polynomials $e_1,\dots,e_n$ (see e.g. Chapter 1 of Macdonald's book Symmetric functions and Hall polynomials), so it follows that $R$ is generated by the Vandermonde $\delta$ and the elementary symmetric polynomials.

Finally, let $\mathfrak{m}$ be the ideal of $R$ generated by $e_1,\dots,e_n$ and $\delta$. Then $\mathfrak{m}$ is a maximal ideal (since the quotient by it is $\mathbf{C}$), but the elements $e_1,\dots,e_n$ and $\delta$ are linearly independent modulo $\mathfrak{m}^2$. It follows that $R$ is singular at $\mathfrak{m}$, and hence not a polynomial ring. Thus is cannot be generated by algebraically independent polynomials.

  • $\begingroup$ I missed the question about algebraically independent generators at first. I edited just now to address it. $\endgroup$
    – Stephen
    Apr 17, 2018 at 17:50

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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .