Bourbaki

Exercise 13, § 2

Commutative Algebra, V, 365

Bourbaki marks this one as harder than the rest.

(a) Let KK be a field of characteristic 0 and nn the ideal of the polynomial ring K[X,Y,Z]K[X, Y, Z] generated by Y2−X2−X3Y^2 - X^2 - X^3. Show that nn is prime and that the integral domain A=K[X,Y,Z]/nA = K[X, Y, Z]/n is not integrally closed: if x,y,zx, y, z are the images of X,Y,ZX, Y, Z in AA, the element t=y/xt = y/x of the field of fractions LL of AA is integral over AA but does not belong to AA. Let A′⊇AA' \supseteq A be the subring K[x,t,z]K[x, t, z] of LL, which is integral over AA. Consider in AA the prime ideal pp generated by xz−yxz - y and z2−1−xz^2 - 1 - x and the maximal ideal q⊃pq \supset p generated by x,y,z−1x, y, z - 1. Consider on the other hand in A′A' the maximal ideal q′q' generated by x,t+1x, t + 1 and z−1z - 1; show that q′q' lies above qq but that there exists no prime ideal p′⊂q′p' \subset q' of AA lying above pp. (Consider a homomorphism from A′/pA′A'/pA' to an extension field of KK and show that the image of q′q' under such a homomorphism is necessarily 0.)

(b) Let A′A' be the quotient ring of the polynomial ring Z[X]\mathbf{Z}[X] by the ideal nn generated by 2X2X and X2−XX^2 - X; let xx be the image of XX in A′A'; A′A' is integral over Z\mathbf{Z} but the number 2 is a divisor of 0 in A′A'. Let q′q' be the prime ideal of A′A' generated by 2 and x−1x - 1; then q′∩Z=2Zq' \cap \mathbf{Z} = 2\mathbf{Z}; if pp is the prime ideal (0) in Z\mathbf{Z}, show that there exists no prime ideal p′⊂q′p' \subset q' contained in A′A' which lies above pp.

(c) Let KK be a field, nn the ideal of the polynomial ring K[X,Y,Z]K[X, Y, Z] generated by X2,XY,XZ,YZ−YX^2, XY, XZ, YZ - Y and Z2−ZZ^2 - Z, A′A' the quotient ring K[X,Y,Z]/nK[X, Y, Z]/n, x,y,zx, y, z the images of X,Y,ZX, Y, Z in A′A' and AA the subring K[x,y]K[x, y] of A′A'; AA is integrally closed in its total ring of fractions and A′A' is integral over AA. Consider in AA the prime ideal pp generated by xx and the maximal ideal qq generated by xx and yy. Consider on the other hand in AA the maximal ideal q′q' generated by x,y,zx, y, z; show that q′q' lies above qq but that there exists no prime ideal p′⊂q′p' \subset q' lying above pp (same method as in (a)).

Interpret the examples in (a) and (c) geometrically. \*

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.