Bourbaki

Exercise 10, § 2

Tag 002U, permanent. Cite it as [Bourbaki, Tag 002U].

Algebra, VIII, p. 41

Bourbaki marks this one as harder than the rest.

Let pp be a prime number, qq a power of p,Gp, G a finite group of order qq, and A a commutative ring. Let (eg)(e_g) be the canonical basis of the algebra A[G] of the group G over A.

a) The set a\mathfrak{a} of elements ∑ageg\sum a_ge_g of A[G] such that ∑ag=0\sum a_g= 0 is a two-sided ideal of A[G]. If the ring A is of characteristic pp (V, §1, No. 2, p. 2), then we have xq=0x^q= 0 for every x∈ax\in \mathfrak{a} (use induction on qq; if q≠1q\not= 1, choose an element hh of order pp of the center of G (I, §6, No. 5, p. 77, Corollary of Proposition 11) and deduce from the induction hypothesis that there exists a y∈A[G]y\in A[G] such that xq/p=(1−eh)yx^{q/p}= (1-e_h)y).

b) Suppose that A is a local ring, with maximal ideal m\mathfrak{m}, and that the field A/mA/\mathfrak{m} is of characteristic pp. Prove that the ring A[G] is local and that its maximal ideal is the set of elements ∑ageg∈A[G]\sum a_ge_g\in A[G] such that ∑ag∈m\sum a_g\in \mathfrak{m}.

(Let S be a simple A[G]-module; prove that S is annihilated by m\mathfrak{m} and then, by applying Proposition 11 of I, §6, No. 5, p. 76, to the Z\mathbf{Z}[G]-submodule generated by a nonzero element of S, that S contains an element invariant under G.)

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Cites

  • I, §6, No. 5, p. 77, Corollary of Proposition 11 not in this corpus
  • Proposition 11 of I, §6, No. 5, p. 76 not in this corpus
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