Bourbaki

Exercise 15, § 2

Commutative Algebra, V, 366

Let AA be a ring and pp a prime ideal of AA. Show that, in the polynomial ring B=A[X]B = A[X], the ideal pB=qpB = q is prime; the maximal ideal of the local ring BqB_q is equal to pBqpB_q; if kk is the field of fractions of A/pA/p, Bq/pBqlB_q/pB_q^l is isomorphic to the field of rational fractions k(X)k(X). If AA is integrally closed, so is BB.

Solution

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Transcribed from the printed volume. Not translated.