TITLE: Chevalley–Shephard–Todd theorem
QUESTION [15 upvotes]: The wikipedia article claims that the theorem "was first proved by G. C. Shephard and J. A. Todd (1954) who gave a case-by-case proof. Claude Chevalley (1955) soon afterwards gave a uniform proof". I read the paper by Chevalley and it seems that he only proves the implication: "If the group is generated by pseudo-reflections, then the ring of invariants is polynomial". I wonder whether there is a uniform proof of the inverse implication? Where is it written?

REPLY [10 votes]: Torsten's argument is of course beautiful, but it might be worth recording that there is also a slick combinatorial argument, in case you need to teach this to students without algebraic geometry. (After rereading, it looks like Josh Swanson is linking to the same argument, so upvote his answer if you like this.) Let $G$ be any finite subgroup of $GL(V)$ for $V$ a vector space of characteristic $0$; let $R = \mathrm{Sym}(V)$ and $S = R^G$. 
For any finite dimensional vector space $U$ with an action of $G$, we have 
$$\dim U^G = \frac{1}{|G|} \sum_{g \in G} \mathrm{Tr}(g: U \to U).$$
So, working degree by degree,
$$\sum_{k=0}^{\infty} (\dim S_k) t^k = \frac{1}{|G|} \sum_{g\in G} \sum_{k=0}^{\infty} t^k Tr(g: Sym^k V \to Sym^k V).$$
If the eigenvalues of $g$ on $U$ are $\lambda_1(g)$, ..., $\lambda_n(g)$, we deduce
$$\sum_{k=0}^{\infty} (\dim S_k) t^k= \frac{1}{|G|} \sum_{g\in G} \prod_{j=1}^n \frac{1}{1-\lambda_j(g) t} \quad (\ast). $$
This is called Molien's formula. 
In particular, if $S$ is a polynomial ring with generators in degrees $e_1$, ..., $e_n$, then
$$\prod_{i=1}^n \frac{1}{1-t^{e_i}} = \frac{1}{|G|} \sum_{g\in G} \prod_{j=1}^n \frac{1}{1-\lambda_j(g) t} \quad (\dagger).$$
Let $T$ be the set of pseudoreflections. Expanding both sides in Laurent series around $t=1$, we have
$$\prod_{i=1}^n \left( \frac{1}{e_i (1-t)} + \frac{e_i-1}{2 e_i} + O(1-t) \right)$$
$$= \frac{1}{|G|} \left( \frac{1}{(1-t)^n} + \sum_{g \in T} \frac{1}{(1-t)^{n-1} (1-\lambda_n(g))} + O((1-t)^{-n+2}) \right)$$
 where $\lambda_n(g)$, for $g \in T$, is taken to be the eigenvalue which is not $1$. Grouping together elements of $T$ which generate the same subgroup and matching coefficients, one concludes:
$$|G| = \prod e_i \ \mbox{and}\ |T| = \sum (e_i-1). \quad (\S)$$
As in Torsten's proof, let $H$ be the subgroup of $G$ generated by $T$. By the reverse direction of CST, $R^H$ is a polynomial ring, say with generators in degrees $d_1$, ... $d_n$. Since the first generator of $R^G$ must involve at least one generator of $R^H$, the first two generators of $R^G$ must involve at least two generators of $R^H$ etc we see that, after reordering $d_1 \leq d_2 \leq  \cdots \leq d_n$ and $e_1 \leq e_2 \leq \cdots \leq e_n$, we have $e_i \geq d_i$.
But $\sum (e_i -1) = \sum (d_i-1) = |T|$. So $e_i = d_i$. Then $|G| = \prod e_i = \prod d_i = |H|$, so $G=H$.