Bourbaki

Exercise 5, § 5

Lie Groups and Lie Algebras, V

*Let KK be a field and S=K[X1,…,Xn]S = K[X_1, \ldots, X_n] a graded polynomial algebra over KK, generated by algebraically independent elements XiX_i, homogeneous of degrees >0> 0.

a) Let Y1,…,YnY_1, \ldots, Y_n be homogeneous elements of SS of degrees >0> 0, and let R=K[Y1,…,Yn]R = K[Y_1, \ldots, Y_n] be the subalgebra of SS generated by these elements. Prove the equivalence of the following properties:

(i) (Y1,…,Yn)(Y_1, \ldots, Y_n) is an SS-regular sequence.

(ii) SS is integral over RR.

(iii) The ideal of SS generated by Y1,…,YnY_1, \ldots, Y_n is of finite codimension in SS.

(iv) For any extension K‾\overline{K} of KK, the system of equations Yi(x1,…,xn)=0Y_i(x_1, \ldots, x_n) = 0 (1≤i≤n,xi∈K‾1 \leq i \leq n, x_i \in \overline{K}) has only the trivial solution (0,…,0)(0, \ldots, 0).

((i) ⟺\Longleftrightarrow (iii)) follows from a theorem of Macaulay (Commutative Algebra); (iii) ⟺\Longleftrightarrow (iv) follows from the Zeros Theorem (Commutative Algebra, Chap. V, §3, no. 3, Prop. 2); (ii) ⟺\Longleftrightarrow (iii) is easy.)

If these properties are satisfied, show that the YiY_i are algebraically independent over KK, and that SS is a free RR-module of rank equal to

∏ideg⁡(Yi)∏ideg⁡(Xi). \frac{\prod_i \deg(Y_i)}{\prod_i \deg(X_i)}.

b) Let GG be a finite group of automorphisms of the graded algebra SS, let SGS^G be its subalgebra of invariants, and let Y1,…,YnY_1, \ldots, Y_n be elements of SGS^G satisfying conditions (i),..., (iv) above. Show that SG=K[Y1,…,Yn]S^G = K[Y_1, \ldots, Y_n] if and only if

Card⁡(G)=∏ideg⁡(Yi)∏ideg⁡(Xi).∗ \operatorname{Card}(G) = \frac{\prod_i \deg(Y_i)}{\prod_i \deg(X_i)}.*

Solution

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Transcribed from the printed volume. Not translated.