Exercise 13, § 2
Commutative Algebra, V, 365
Bourbaki marks this one as harder than the rest.
(a) Let be a field of characteristic 0 and the ideal of the polynomial ring generated by . Show that is prime and that the integral domain is not integrally closed: if are the images of in , the element of the field of fractions of is integral over but does not belong to . Let be the subring of , which is integral over . Consider in the prime ideal generated by and and the maximal ideal generated by . Consider on the other hand in the maximal ideal generated by and ; show that lies above but that there exists no prime ideal of lying above . (Consider a homomorphism from to an extension field of and show that the image of under such a homomorphism is necessarily 0.)
(b) Let be the quotient ring of the polynomial ring by the ideal generated by and ; let be the image of in ; is integral over but the number 2 is a divisor of 0 in . Let be the prime ideal of generated by 2 and ; then ; if is the prime ideal (0) in , show that there exists no prime ideal contained in which lies above .
(c) Let be a field, the ideal of the polynomial ring generated by and , the quotient ring , the images of in and the subring of ; is integrally closed in its total ring of fractions and is integral over . Consider in the prime ideal generated by and the maximal ideal generated by and . Consider on the other hand in the maximal ideal generated by ; show that lies above but that there exists no prime ideal lying above (same method as in (a)).
Interpret the examples in (a) and (c) geometrically. \*
Solution
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