The Chevalley-Shephard-Todd Theorem

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updates: 10 November 2011

The Chevalley-Shephard-Todd theorem

Let V be a finite dimensional vector space over a field 𝔽. Let W be a finite subgroup of GL(V). If S(V)W is a polynomial algebra then W is generated by reflections.

Proof. Let

I=f S(V)W | f(0)=0
be the ideal in S(V) generated by polynomials without constant term. Let e1,, er be homogeneous generators of I (which exist, by Hilbert).

Step 1. Every f S(V)W is a polynomial in e1,, er.

Proof. The proof is by induction on the degree of f. Assume f is homogeneous and deg(f) >0. Since fI,

f= i=1r piei, with piS(V),
and so f= 1|W| wW wf = i=1r ( 1|W| wW wpi ) ei, and since the internal sum has lower degree it can be written as a polynomial in e1,, er.

Step 2. r=dim(V).

Proof. Let n=dim(V), let x1,, xn be a basis of V and let (x1,, xn) be the field of fractions of S(V)=[ x1,,xn ]. Since xi is a root of mi(t) = wW (t-wxi) S(V)W [t], the variable xi is algebraic over (e1,, er), the field of fractions of S(V)W. Thus 0= trdeg( (x1,, xn) (e1,, er) ) = trdeg( (x1,, xn) ) = trdeg( (e1,, er) ) =n-r.

Step 3. The Jacobian of a map φ: V V x (φ(x), ,φ(x)) is Jφ(x) = det( φi xj ). If φ is linear then there are φij such that φi(x) = j=1n φijxj and Jφ=det(φ ij). The chain rule is the identity Jθφ = Jθ(φx) Jφ(x) . Let θ: V V x (e1(x), ,en(x)) and w: V V x wx for wW. Then θw=θ and so Jθ(x) = Jθw(x) = Jθ(wx) Jw(x) = Jθ(wx) det(w) = det(w) (w-1 Jθ)(x) . Thus Jθ is W-alternating and so Jθ is divisible by Δ= αR+ αsα-1 . Since deg(Jθ) =i=1n (di-1) =Card(R+) , it follows that Jθ= λΔ for some λ .

Step 4. The polynomials e1,, en are algebraically independent if and only if Jθ0.

Proof. : Assume e1,, er are algebraically independent. Since trdeg( (x1,, xn) (e1,, er) ) = trdeg( (x1,, xn) ) = trdeg( (e1,, er) ) n-r, x1,, xn are algebraic over (e1,, er) if and only if 0n-r, that is, if and only if n=r.

Let mi(t) S(V)W [t] be the minimal polynomial of xi over (e1,, er), the field of fractions of S(V)W . Then mi xk = j=1r mi ej ej xk + mi t t xk and 0= mi(xi) xk = = j=1r mi ej (xi) ej xk + mi(xi) δik . Thus det( mi ej (xi) ) Jθ = det(-diag( m1 (x1), mn (xn)) ) = (-1)n i=1r mi (xi) . Since mi(t) is the minimal polynomial of xi, each factor mi (xi)0 and, thus, Jθ0.

: Assume e1,, en are algebraically independent. Let f(y1,, yn) be of minimal degree such that f(e1,, en)=0. Then f yi 0 for some yi, and so gi = f yi ( e1,, en) 0 for somei. But 0= f(e1,, en) xj = i=1n f yi (e1,, en) ei xj , and so i=1n gi ei xj =0. So gi is a solution to the equation (g1,,gn ) (ei /xj )=0 and so Jθ=0.

Notes and References

These notes are a retyping of section 3 of


[St] R. Steinberg, Lectures on Chevalley groups, Notes prepared by John Faulkner and Robert Wilson, Yale University, New Haven, Conn., 1968. iii+277 pp. MR0466335.

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