Chevalley–Shephard–Todd theorem 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?  
 A: Old post, but Shephard-Todd's original proof of the "inverse" implication was essentially uniform (and brief) to begin with--see section 8 of their Finite Unitary Reflection Groups. The main tools are Molien's theorem, the "direct" implication, and the Jacobian. (Technically their proof of the direct implication was case-by-case, but plugging in, say, Chevalley's proof makes their proof of the inverse case uniform.)
The same argument with the window dressings changed appears in Stanley's nice old survey Invariants of Finite Groups and their Applications to Combinatorics (1979), section 4. It's also in Humphreys' Reflection Groups and Coxeter Groups, section 3.11.
A: There are indeed many presentations (if I remember correctly Bourbaki has it)
but the proof is very elegant and short so that I find it hard to refrain from giving
it. Let $H$ be the normal subgroup of the finite $G\subset \mathrm{GL}_n$
generated by the pseudo-reflections. By the other direction $X:=\mathbb{A}^n/H$
is again affine space and in particular is smooth. We have an action of $G/H$ on
$X$ and a moment's thought reveals that it acts freely in codimension $1$
 (as a point fixed by a non-identity element would lie below a
reflection hyperplane of $\mathbb{A}^n$ and the fixing element below a
pseudo-reflection). Hence $X \to X/(G/H)=\mathbb{A}^n/G$ is étale in codimension
$1$. If $\mathbb{A}^n/G$ were smooth, purity of the branch locus would imply that
the map were étale. However, that forces $G/H$ to act freely on $X$ but the
image of the origin is fixed by all of $G/H$ and therefore $G=H$.
A: Chevalley was interested in the action of (real) Weyl groups and so a reflection to him had determinant -1 and so was a real reflection, i.e. order 2.  My understanding is that Serre had seen the paper by Shepard and Todd and so he knew that pseudo-reflections were relevant.  He pointed out that Chevalley's proof was valid for pseudo-reflections.
A: 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 $G=H$.
