Bourbaki

Exercise 1, § 2

Commutative Algebra, IV, 290

(a) Show that in the Noetherian polynomial ring A=Z[X]A = \mathbf{Z}[X] in one indeterminate over Z\mathbf{Z}, the ideal m=2A+AXm = 2A + AX is maximal and that the ideal q=4A+AXq = 4A + AX is m-primary, but is not equal to a power of m.

(b) In the ring B=Z[2X,X2,X3]⊂AB = \mathbf{Z}[2X, X^2, X^3] \subset A which is Noetherian, show that the ideal p=2BX+BX2p = 2BX + BX^2 is prime, but that p2p^2 is not p-primary, although its radical is equal to p.

(c) Let K be a commutative field and A the quotient ring of K[X,Y,Z]K[X, Y, Z] by the ideal generated by Z2−XYZ^2 - XY; let x, y, z be the canonical images of X, Y, Z in A. Show that p=Ax+Azp = Ax + Az is prime, that p2p^2 is not primary and that p2=a∩b2p^2 = a \cap b^2 is a primary decomposition of p2p^2, where a=Axa = Ax and b=Ax+Ay+Azb = Ax + Ay + Az.

Solution

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Transcribed from the printed volume. Not translated.