Gordans Finiteness Theorem (Hilberts proof slightly modernized)

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1 Gordans Finiteness Theorem Hilberts proof slightly modernized Klaus Pommerening April 1975 Let k be an infinite entire ring, V = k 2, the free k-module of rank 2, G, the group SLV. Then G acts in a canocical way on S n V, the module of binary forms of degree n, and on its affine coordinate ring P := SS n V = SS n V. The ring I := P G of invariants is the classical ring of invariants of a binary form of degree n. Let k := Z [ 1 n! ] = Z [ 1 p p n prime. Theorem 1 Gordan 1868 Let n! be a unit in k. Then I = Ik is a finitely generated k-algebra, and Ik = Ik k k as a graduated k-algebra. Gordan s original proof [1] works for k a field of characteristic and provides an explicit system of generators. Here we reproduce Hilbert s proof [2] that gives the theorem in the stated generality. The restriction that n! is a unit will be needed only in step 3, and we also have a somewhat weaker non-explicit version without this restriction that however uses Emmy Noether s general result on finite generation of invariants under finite groups and therefore is anhistoric. Corollary 1 If k is an infinite noetherian entire ring not necessarily n! a unit, then I is a finitely generated k-algebra. Historical remark. Except for the main theorem on symmetric polynomials Gordan s theorem is the earliest result on finite generation of invariant rings. The proof starts with two lemmas. Lemma 1 Let n a ij x j = for j = 1,... m with a ij Z for all i, j j=1 be a system of diophantine equations. Then the semigroup of non-negative solutions i. e. solution vectors x = x 1,..., x n N n is finitely generated. 1 ]

2 This is a corollary of Dickson s lemma for which we have a separate note [3]. Lemma 2 Let σ 1,..., σ N k[x 1,..., X N ] = k[x] be the elementary symmetric polynomials. Then for each i the powers X j i, j < N, generate the k[σ]-module k[σ][x i]. More explicitely, for each natural number p N there are polynomials g 1,..., g N k[σ 1,..., σ N ] such that X p i = g 1 X N 1 i + + g N 1 X i + g N for all i = 1,... N. Proof. If p N 1, the assertion is obvious. If p = N, then X 1,..., X N are zeroes of the polynomial T X 1 T X N = T N σ 1 T N 1 + ± σ N k[x][t ], whence Xi N = σ 1 Xi N 1 σ N for i = 1,..., N. If p > N, we use induction and assume X p 1 i i = 1,..., N. Then = g 1 σxn 1 i + + g N σ for X p i = g 1σX i N + + g NσX i [ ] = g 1σ σ 1 Xi N 1 ± σ N + g 2σX i N g NσX i = [g 1σσ 1 + g 2σ }{{} =:g 1 σ as was to be shown. ]Xi N [g Nσ g 1σσ }{{ N 1 ]X } i ± g 1σσ N, }{{} =:g N 1 σ g N σ The proof of the theorem involves several rings. First let S := T n SV = SV SV = k[α 1, β 1,..., α n, ], the polynomial ring in 2n indeterminates. This k-algebra is canonically N n -graded where the homogeneous parts have the form S d = S d 1 V S dn V for d = d 1,..., d n N n. The group G = SLV acts on S in a homogeneous way; let R := S G be the algebra of invariants. It has the induced grading R = d N n R d where R d = S d R. Examples for homogeneous invariants are 1. p ij := α i β j α j β i for 1 i < j n R d where i j d =,..., 1,..., 1,...,. 2

3 2. For an N-valued symmetric matrix M = m ij with zero diagonal, X M := i<j p m ij ij R d where d i = Let us denote this multi- degree by d =: DM. n m ij. For d N let d = d,..., d N n, hence S d = S d V S d V. Then S := d N S d j=1 is an N-graduated subalgebra of S. Also on this algebra G operates homogeneously, and the ring of invariants is R := S G = d N R d. Furthermore we have the operation α i α πi, β i β πi for π S n, of the symmetric group S n on S; it induces a homogeneous operation on S. The operations of S n and of G commute elementwise. Therefore the direct product S n G acts on S homogeneously. Lemma 3 The graduated ring P = SS n V is isomorphic with S Sn. Proof. A short consideration will motivate the proof. Let {x, y} be a basis of the dual space V. Decompose the general binary form of degree n, f = n u j x n j y j as a product of linear factors over a suitable ring extension k k: f = j= n β i x + α i y i=1 1 this corresponds to the decomposition of the polynomial x f = u n j y x j into linear factors y x + β i α i. Then f = β i x + α i y = β 1 x + α i β i y = β 1 x n j σ j α 1 β 1,..., α n y j. Therefore u = β 1, u 1 = β 1 α 1 β α n,..., u n = α 1 α n. The general formula is u j = α M β M M P j {1,...,n} 3

4 with suggestive notation. Now let s prove that S Sn = P. Let α i and β i be indeterminates, and u j for j = 1,..., n be given by formula. Then k[u,..., u n ] = SS n V = P is the coordinate algebra of the binary forms of degree n. We have S d = {F k[α, β] F homogeous of degree d in each pair α i, β i }. Therefore u,..., u n S 1, hence P = k[u] is a graduated subring of S. Moreover all Sn u j S because for π S n we have πu j = π α M β M = α πm β πm = u j. M P j M P j Sn Therefore P S even as a graduated subring. For the opposite inclusion let F S d be an S n -invariant, say F = c m α m1 1 βd m 1 1 αn mn βn d mn, m N n where c πm = c m for all π S n. This means that F/β1 d βd n is a symmetric polynomial in α 1 β 1,..., αn of degree d, hence a polynomial in the elementary symmetric polynomials: 1 1 F = β1 d F = Gσ 1 α 1,..., α n,..., σ n α 1,..., α n = G u 1,..., u n βd n β 1 β 1 u u u d with degg d, and therefore F k[u]. The k-algebra whose finite generation the theorem asserts is therefore I = P G = S Sn G = S Sn G = S G Sn = R Sn. The following diagram shows the relations of all these rings. S = d S d S = d S d P = S Sn R = S G = d R d R = S G = d R d I = P G = R Sn Now we prove in three steps that the k-algebras R, R, and I are finitely generated. 4

5 Step 1. Lemma 4 For each d N n we have R d R d = R d k = R d Z Z k. = X M DM = d as a k-module, and In particular as a k-algebra R is generated by the n 2 elements pij, 1 i < j n, and R = Rk = RZ Z k as a graduated k-algebra. We start the proof of Lemma 4 with the special case n = 1: Lemma 5 SV G = k. Proof. Let {α, β} be a basis of V and F = r i= c iα r i β i with r 1. If F is invariant 1 under G, then c 1 r i = 1 r i c i for i = 1,..., r. Now let F be also invariant 1 λ under G for all λ k. Then 1 F = = r c i α + λβ r i β i = i= r i r j r i i= j= r r i r i c i α r i j λ j β j+i j j= i= c j λ i j α r i β i. This gives equations for the coefficients c i, for example r r j c r = c j λ r j = c λ r + + c r j= for all λ k. Because k is an infinite entire ring, we conclude that c =... = c r 1 = and c r = 1 r c =, hence F =. Remark. For Lemma 5 we really need the condition that k is an infinite entire ring. As an illustration we give three counterexamples for weaker conditions. Example 1. Let k be the finite entire ring F 2. Then F = α 2 + αβ + β 2 S 2 V is invariant under and that together generate G. 1 1 Example 2. Or let k be F 3. Then F = α 6 + α 4 β 2 + α 2 β 6 + β 6 S 6 V is invariant 1 ±1 1 1 under,, and, that together generate G Example 3. For an example where k is infinite, but has zero-divisors, take A, an infinite F 2 -module, and k = F 2 A with the multiplication m, an, b = mn, ma + nb. This is the well-known adjunction of 1 to the F 2 -Algebra A with -multiplication. Then F = α 4 + α 2 β 2 + β 4 is invariant. 5

6 We prove Lemma 4 by induction on n. For n = 1 Lemma 5 gives R = k, and the assertion is trivial. Therefore let n > 1. We make a further induction on w := d d n. If w =, we have d = and R = k, and we are ready. Now let w >, that is, d. We may assume d n otherwise change enumeration. If all other d i =, we would have S d k[α n, ], and the case n = 1 would apply. Therefore we may assume without loss of generality that d n 1. The substitution homomorphism ϕ: S S = k[α 1, β 1,..., α n 1, 1 ], α n α n 1, 1, is G-equivariant, hence maps R onto R := S G and R d onto R d where d = d 1,..., d n 1 + d n N n 1. By induction R d = R Z k is spanned by the X d M with DM = d. Each such X M has d n 1 + d n factors of type p i,n 1 ; if we replace any d n of these by p in, then we get an inverse image X M of X M under ϕ. Therefore ϕ: R d R is surjective. d 1 Next let us determine the kernel of ϕ. Let F R d with ϕf =. Then F β d 1 1 βdn n [ is a polynomial over k α1 β 1,..., α n 1 Therefore αn α n β d 1 1 βdn n 1 ] in the indeterminate αn, and has α n 1 1 as a zero. F, and p n 1,n F. Thus there is an F R d with d = d 1,..., d n 1 1, d n 1, such that F = p n 1,n F. Therefore ker ϕ = p n 1,n R d as a k-module because R has no zero divisors. This gives the exact sequence R d R d R d. By induction on w the module R d = R dz k is spanned by the X M with DM = d. Because the sequence R d X M DM = d R d is also exact, general nonsense the five lemma gives R d = X M DM = d. The exact sequence R dz R d Z R d Z consists of free Z-modules. The following diagram has exact rows and canonical vertical arrows. The two isomorphisms follow by induction. By the five lemma also the middle arrow is an isomorphism. R dz k R d Z k R d Z k = = R d R d R d 6

7 Step 2. Lemma 6 R is a finitely generated k-algebra, and R = Rk = RZ Z k. Proof. The module R d is spanned by the XM with n j=1 m ij = d for all i = 1,..., n. Therefore R is spanned by the X M with n j=1 m 1j =... = n j=1 m nj. This is a homogeneous system of diophantine equations for the m ij. By Lemma 1 its solutions form a finitely generated subsemigroup of N q where q = nn 1/2. Let {M 1,..., M r } be a system of generators. If M = r l=1 a lm l, then X M = i<j p m ij ij = i<j p m ij 1 ij a 1 = X a 1 M 1 X ar M r. Therefore the X 1 := X M1,..., X r := X Mr generate the k-algebra R. The isomorphism in the lemma follows because by Lemma 4 it holds on all homogeneous components. Step 3. We show that I is finitely generated, if n! is a unit in k, and Ik = Ik k k as a graduated k-algebra. Proof. Consider the linear map µ d : R d R d, µ d := π S n π. Then I d := I R d = µ d R d, because n! is a unit in k. In I there are for example the following elements: a µx a 1 1 Xar r = π S n πx 1 a1 πx r ar where a 1,..., a r N, b the elementary symmetric functions in the πx i for each i = 1,..., r, σ j ρx i ρ Sn = ρx i ρ A A P j S n for j = 1,..., n! because each π S n only permutes the summands. Claim : The finite set Z = {a all a i < n!} {b} generates the k-algebra I. Each element of I is a linear combination of elements of the form F = n! 1 πx 1 a1 πx r ar = g 1 n! 1 j πx 1 j g r j πx r j π S n π S n j=1 j=1 n! 1 = g 1 j 1 g r j r πx 1 j1 πx r jr j 1,...,j r= π S n 7

8 with terms in k[z]. Therefore F k[z] and I = k[z]. For the second statement, the isomorphism, we note that because n! is a unit in k the k[s n ]-module R d is semisimple. Therfore the submodule I d has a direct complement H d, and we have two exact direct sequences H d R d I d = H d k k R d k k I d k k The diagram is commutative. We already know by Lemma 4 that the middle vertical arrow is an isomorphism. Because the rows are exact direct, also the two other vertical arrows are isomorphisms. Therefore I = Ik k as a graduated ring. Step 3a. We finally show that I is finitely generated, if k is noetherian, but n! not necessarily a unit. This simply follows because I is the ring of invariants of the finitely generated k-algebra R under the finite group S n. References [1] P. Gordan: Beweis, dass jede Covariante und Invariante einer binären Form eine ganze Function mit numerischen Coeffizienten einer endlichen Anzahl solcher Formen ist. J. reine angew. Math , [2] D. Hilbert: Über die Endlichkeit des Invariantensystems für binäre Grundformen. Math. Ann , [3] K. Pommerening: A remark on subsemigroups Dickson s lemma. Online: 8