Let K be a commutative field, S a graded K-algebra of polynomials, and R a graded subalgebra of S of finite type such that the R-module S admits a basis (zλ)λ∈Λ consisting of homogeneous elements. In the set of systems of generators of the ideal R+ of R consisting of homogeneous elements, choose a minimal element (α1,…,αs). Assume that, for all i, the degree ki of αi is coprime to the characteristic exponent p of K. Then the αi generate the K-algebra R and are algebraically independent over K.
By Algebra, Chap. II, § 11, no. 4, Prop. 7, the assumption made about the αi is equivalent to saying that they are homogeneous and that their images in the K-vector space R+/(R+)2 form a basis of this space. This condition is invariant under extension of the base field; we can thus reduce to the case where the latter is perfect.
The family (α1,…,αs) generates the algebra R by Commutative Algebra, Chap. III, § 1, no. 2, Prop. 1. We argue by contradiction and assume that this family is not algebraically independent over K.
1) We show first of all that there exist families
(βi)1≤i≤s,(yk)1≤k≤r,(dik)1≤i≤s,1≤k≤r
of homogeneous elements of S with the following properties:
βi∈R for all i, and the βi are not all zero;degyk>0 for all k;αi=k=1∑rdikyk for all i;i=1∑sβidik=0 for all k.
Let X1,…,Xs be indeterminates and give K[X1,…,Xs] with the graded algebra structure for which Xi has degree ki. There exist non-zero homogeneous elements H(X1,…,Xs) of K[X1,…,Xs] such that H(α1,…,αs)=0; choose H to be of minimum degree. If ∂H/∂Xi=0, the polynomial ∂Xi∂H(α1,…,αs) is a non-zero homogeneous element of R; if p=1, H is not of the form H1p with H1∈K[X1,…,Xs]. Take
βi=ki∂Xi∂H(α1,…,αs).
Since K is perfect, the polynomials ∂H/∂Xi∈K[X1,…,Xs] are not all zero (Algebra, Chap. V, § 1, no. 3, Prop. 4); in view of the assumption made about the ki, neither are the βi.
On the other hand, S can be identified with a graded algebra of polynomials K[x1,…,xr] for appropriate indeterminates x1,…,xr with suitable degrees mi>0. Let Dk be the partial derivative with respect to xk on S. Take d_{ik} = k_i^{-1} D_k(\alpha_i). Then the equality (10) holds because its left-hand side is Dk(H(α1,…,αs)). On the other hand, if we put y1=m1x1,…,yr=mrxr, the equality (9) follows from the equality (5) of no. 1.
2) Let b be the ideal of R generated by the βi; there exists a subset J of
I={1,2,…,s}
such that (βi)i∈J is a minimal system of generators of b. Then J=∅ since b=0. We shall deduce from (9) and (10) that, if i∈J, αi is an R-linear combination of the αj for j=i, which will contradict the minimality of (α1,…,αs) and will complete the proof.
There exist homogeneous elements γji of R (i∈J,j∈I−J) such that
βj=i∈J∑γjiβi(j∈I−J).
Taking (11) into account, formula (10) gives
i∈J∑βidik+j∈I−J∑γjidjk=0.
Put
uik=dik+j∈I−J∑γjidjk.
Thus
i∈J∑βiuik=0.
Write uik=∑λ∈Λδikλzλ, where the δikλ belong to R. Relation (14) implies that
i∈J∑βiδikλ=0 for all k and λ.
If one of the δikλ had a non-zero homogeneous component of degree 0, the preceding equality would imply that one of the βi (i∈J) is an R-linear combination of the others, contradicting the minimality of (βi)i∈J. Thus δikλ∈R+ and consequently uik∈R+S for all i and k. Thus, there exist uikh∈S such that uik=∑h=1suikhαh, in other words, by (13),
dik+j∈I−J∑γjidjk=h=1∑suikhαh.
Multiply both sides of (15)ik by yk and add for i in J fixed and k=1,2,…,r; in view of (9), we find that
αi+j∈I−J∑γjiαj=h=1∑sk=1∑ruikhykαh.
Take the homogeneous components of degree ki of both sides. Since degyk>0, αi is an S-linear combination of the αj for j=i. Since S is free over R and α1,…,αs∈R, it follows that αi is an R-linear combination of the αj with j=i (Commutative Algebra, Chap. I, § 3, no. 5, Prop. 9 d)).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Cited by
Nothing.
Cites
*Commutative Algebra*, Chap. I, § 3, no. 5, Prop. 9 not yet on this site: AC I
Commutative Algebra, Chap. III, § 1, no. 2, Prop. 1 not yet on this site: AC III