Bourbaki

Lemma 1

Tag 0ASU, permanent. Cite it as [Bourbaki, Tag 0ASU].

Lie Groups and Lie Algebras, chapter V, § 5. Invariants in the symmetric algebra

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(z_\lambda)_{\lambda \in \Lambda} consisting of homogeneous elements. In the set of systems of generators of the ideal R+R_+ of R consisting of homogeneous elements, choose a minimal element (α1,…,αs)(\alpha_1, \ldots, \alpha_s). Assume that, for all i, the degree kik_i of αi\alpha_i is coprime to the characteristic exponent p of K. Then the αi\alpha_i 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\alpha_i is equivalent to saying that they are homogeneous and that their images in the K-vector space R+/(R+)2R_+/ (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)(\alpha_1, \ldots, \alpha_s) 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 (\beta_i)_{1 \leq i \leq s}, \quad (y_k)_{1 \leq k \leq r}, \quad (d_{ik})_{1 \leq i \leq s, 1 \leq k \leq r}

of homogeneous elements of S with the following properties:

βi∈R for all i, and the βi are not all zero;deg⁡yk>0 for all k;αi=∑k=1rdikyk for all i;∑i=1sβidik=0 for all k. \begin{align*} &\beta_i \in R \text{ for all } i, \text{ and the } \beta_i \text{ are not all zero;} \\ &\deg y_k > 0 \text{ for all } k; \\ &\alpha_i = \sum_{k=1}^r d_{ik} y_k \text{ for all } i; \\ &\sum_{i=1}^s \beta_i d_{ik} = 0 \text{ for all } k. \end{align*}

Let X1,…,XsX_1, \ldots, X_s be indeterminates and give K[X1,…,Xs]K[X_1, \ldots, X_s] with the graded algebra structure for which XiX_i has degree kik_i. There exist non-zero homogeneous elements H(X1,…,Xs)H(X_1, \ldots, X_s) of K[X1,…,Xs]K[X_1, \ldots, X_s] such that H(α1,…,αs)=0H(\alpha_1, \ldots, \alpha_s) = 0; choose H to be of minimum degree. If ∂H/∂Xi≠0\partial H / \partial X_i \neq 0, the polynomial ∂H∂Xi(α1,…,αs)\frac{\partial H}{\partial X_i}(\alpha_1, \ldots, \alpha_s) is a non-zero homogeneous element of R; if p≠1p \neq 1, H is not of the form H1pH_1^p with H1∈K[X1,…,Xs]H_1 \in K[X_1, \ldots, X_s]. Take

βi=ki∂H∂Xi(α1,…,αs). \beta_i = k_i \frac{\partial H}{\partial X_i}(\alpha_1, \ldots, \alpha_s).

Since K is perfect, the polynomials ∂H/∂Xi∈K[X1,…,Xs]\partial H / \partial X_i \in K[X_1, \ldots, X_s] are not all zero (Algebra, Chap. V, § 1, no. 3, Prop. 4); in view of the assumption made about the kik_i, neither are the βi\beta_i.

On the other hand, S can be identified with a graded algebra of polynomials K[x1,…,xr]K[x_1, \ldots, x_r] for appropriate indeterminates x1,…,xrx_1, \ldots, x_r with suitable degrees mi>0m_i > 0. Let DkD_k be the partial derivative with respect to xkx_k 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))D_k(H(\alpha_1, \ldots, \alpha_s)). On the other hand, if we put y1=m1x1,…,yr=mrxry_1 = m_1 x_1, \ldots, y_r = m_r x_r, the equality (9) follows from the equality (5) of no. 1.

2) Let bb be the ideal of RR generated by the βi\beta_i; there exists a subset JJ of

I={1,2,…,s} I = \{1, 2, \ldots, s\}
such that (βi)i∈J(\beta_i)_{i \in J} is a minimal system of generators of bb. Then J≠∅J \neq \varnothing since b≠0b \neq 0. We shall deduce from (9) and (10) that, if i∈Ji \in J, αi\alpha_i is an RR-linear combination of the αj\alpha_j for j≠ij \neq i, which will contradict the minimality of (α1,…,αs)(\alpha_1, \ldots, \alpha_s) and will complete the proof.

There exist homogeneous elements γji\gamma_{ji} of RR (i∈J,j∈I−Ji \in J, j \in I - J) such that

βj=∑i∈Jγjiβi(j∈I−J). \beta_j = \sum_{i \in J} \gamma_{ji} \beta_i \quad (j \in I - J).
Taking (11) into account, formula (10) gives
∑i∈Jβi(dik+∑j∈I−Jγjidjk)=0. \sum_{i \in J} \beta_i \left( d_{ik} + \sum_{j \in I - J} \gamma_{ji} d_{jk} \right) = 0.
Put
uik=dik+∑j∈I−Jγjidjk. u_{ik} = d_{ik} + \sum_{j \in I - J} \gamma_{ji} d_{jk}.
Thus
∑i∈Jβiuik=0. \sum_{i \in J} \beta_i u_{ik} = 0.
Write uik=∑λ∈Λδikλzλu_{ik} = \sum_{\lambda \in \Lambda} \delta_{ik\lambda} z_\lambda, where the δikλ\delta_{ik\lambda} belong to RR. Relation (14) implies that
∑i∈Jβiδikλ=0 for all k and λ. \sum_{i \in J} \beta_i \delta_{ik\lambda} = 0 \text{ for all } k \text{ and } \lambda.
If one of the δikλ\delta_{ik\lambda} had a non-zero homogeneous component of degree 0, the preceding equality would imply that one of the βi\beta_i (i∈Ji \in J) is an RR-linear combination of the others, contradicting the minimality of (βi)i∈J(\beta_i)_{i \in J}. Thus δikλ∈R+\delta_{ik\lambda} \in R_+ and consequently uik∈R+Su_{ik} \in R_+ S for all ii and kk. Thus, there exist uikh∈Su_{ikh} \in S such that uik=∑h=1suikhαhu_{ik} = \sum_{h=1}^s u_{ikh} \alpha_h, in other words, by (13),
dik+∑j∈I−Jγjidjk=∑h=1suikhαh. d_{ik} + \sum_{j \in I - J} \gamma_{ji} d_{jk} = \sum_{h=1}^s u_{ikh} \alpha_h.
Multiply both sides of (15)ik_{ik} by yky_k and add for ii in JJ fixed and k=1,2,…,rk = 1, 2, \ldots, r; in view of (9), we find that
αi+∑j∈I−Jγjiαj=∑h=1s∑k=1ruikhykαh. \alpha_i + \sum_{j \in I - J} \gamma_{ji} \alpha_j = \sum_{h=1}^s \sum_{k=1}^r u_{ikh} y_k \alpha_h.
Take the homogeneous components of degree kik_i of both sides. Since deg⁡yk>0\deg y_k > 0, αi\alpha_i is an SS-linear combination of the αj\alpha_j for j≠ij \neq i. Since SS is free over RR and α1,…,αs∈R\alpha_1, \ldots, \alpha_s \in R, it follows that αi\alpha_i is an RR-linear combination of the αj\alpha_j with j≠ij \neq 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

Transcribed from the printed volume. Not translated.