# 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?

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An additional question: Is it correct that Chevalley only gave a proof for reflection groups (reflection = pseudo-reflection of order 2) in this paper and that Serre later realized that the proof also works for pseudo-reflection groups? – user717 Jul 19 '10 at 7:52
"Invariant theory of finite groups" by Mara Neusel and Larry Smith contains the proof. Note, however, that while the "inverse" implication holds in general by homological algebra, the "direct" implication fails in positive characteristic. – Victor Protsak Jul 19 '10 at 8:19
Chevalley only gives the proof for reflection groups but he only uses that a reflection fixes a hyperplane. – Roman Fedorov Jul 19 '10 at 9:57
Probably the most thorough coverage, with full references (Serre's improvement of Chevalley's statement, Springer's lecture notes, Borel's historical essays, etc.) is given by G.I. Lehrer and D.E. Taylor in Unitary Reflection Groups (Cambridge, 2009). Like most people they work over $\mathbb{C}$, leaving aside the more delicate question of what remains true over more general fields mentioned here by Victor Protsak. This book treats Shephard-Todd theory and later developments in good detail. – Jim Humphreys Jul 19 '10 at 10:41

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 $G=H$.