A polynomial algebra that is free as an $A$-module

Question

I'm working through some problems when I stumbled across a question asking about conditions for when the polynomial algebra $k[x_1,\ldots,x_n]$ is also a free $A$-module, where $A$ is some $k$-subalgebra of $k[x_1,\ldots,x_n]$. However, I'm blanking on nontrivial examples.

What are some examples of $A$ for which this occurs?
Why are these kinds of algebras important, and where can I look to learn more about this concept?

If A is the subalgebragenerated by any subset pf the variables, that works. — Mariano Suárez-Álvarez, Oct 02 '13 at 04:28
@Mariano Thanks. In this case the rank of $k[x_1,\ldots,x_n]$ as an $A$-module is infinite. Is this always the case? Are there other more unexpected examples? — JHF, Oct 02 '13 at 04:54
of course not: the rank of $k[x_1,\dots,x_n]$ as a module over itself is one! — Mariano Suárez-Álvarez, Oct 02 '13 at 07:38

score 1 · Answer 1 · edited Apr 13 '17 at 12:19

Let $V$ be a finite-dimensional vector space equipped with a faithful linear representation of a finite group $G$, and consider the induced action on the polynomial algebra $k[V]$. The Chevalley-Shepard-Todd theorem asserts that the following conditions are equivalent:

$k[V]$ is a free (finite) $k[V]^G$-module
$k[V]^G$ is a polynomial algebra
$G$ is a "pseudoreflection group."

For example, if $k = \mathbb{C}$ then the condition is that $G$ is a complex reflection group. These include, but are more general than, the finite Coxeter groups, and in particular the finite Weyl groups. (The phrasing of the condition here is slightly misleading. Being a pseudoreflection group is a condition not on a group but on a pair of a group $G$ and a faithful linear representation of it.)

The motivating example here is when $V \cong k^n$ and $G \cong S_n$ acting via permutations. Then $k[V]^G$ is the ring of symmetric polynomials in $n$ variables. It is itself a polynomial ring on the elementary symmetric polynomials. The question of what a basis of $k[V]$ looks like as a $k[V]^G$-module is addressed here.

In general, I think the condition that a map $f : A \to B$ of commutative rings makes $B$ a finite free $A$-module should be thought of as roughly meaning that the corresponding map $\text{Spec } B \to \text{Spec } A$ is a "finite (branched) cover." You can write down some nice examples coming from algebraic curves or, if you have a more number-theoretic bent, from rings of integers in algebraic number fields. The inclusions $k[V]^G \to k[V]$ above look geometrically like quotient maps $V \to V/G$.

A polynomial algebra that is free as an $A$-module

1 Answers1