Bourbaki

Exercise 10, § 2

Commutative Algebra, V, 364

(a) Let A′A' be the polynomial ring Z[X,Y]\mathbf{Z}[X, Y] in two indeterminates and let σ\sigma be the automorphism of A′A' leaving YY invariant and such that σ(X)=−X\sigma(X) = -X; σ\sigma and the identity automorphism form a group G\mathscr{G} of automorphisms of A′A' and the ring of invariants of G\mathscr{G} is A=Z[X2,Y]A = \mathbf{Z}[X^2, Y]. Let p′p' be the prime ideal A′(Y−2X)A'(Y - 2X), p=A∩p′p = A \cap p'; the decomposition group GZ(p′)\mathscr{G}^{\mathbf{Z}}(p') is reduced to the identity element. Show that A/pA/p and A′/p′A'/p' are not isomorphic (consider the tensor products of these rings with 2/222/22).

(b) Let KK be a field of characteristic ≠2\neq 2, A′A' the polynomial ring K[X,Y]K[X, Y] and G\mathscr{G} the automorphism group of A′A' consisting of the identity and the K[Y]K[Y]-automorphism σ\sigma of A′A' such that σ(X)=−X\sigma(X) = -X; the ring of invariants AA of G\mathscr{G} is K[X2,Y]K[X^2, Y]. Let p′p' be the prime ideal A′(X3−Y)A'(X^3 - Y) of A′A' and p=A∩p′p = A \cap p'; the decomposition group GZ(p′)\mathscr{G}^{\mathbf{Z}}(p') is reduced to the identity element but A/pA/p and A′/p′A'/p' are not isomorphic.

Solution

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Transcribed from the printed volume. Not translated.