Exercise 6, § 1
Commutative Algebra, VI, 442
Bourbaki marks this one as harder than the rest.
Let R be a ring and A a subring of R such that S = R − A is non-empty and the product of two elements of S belongs to S.
(a) Let be two elements of A and s, s' two elements of S such that and . Show that one of the two elements belongs to A. Deduce that the set of elements for which there exists such that is a prime ideal of A.
(b) Show that the set of such that for all is a prime ideal of R and A; it is the greatest ideal of R contained in A and .
(c) The set n of elements such that there exists for which is both an ideal of R and an ideal of A and .
(d) The ring A is integrally closed in R.
(e) If R is a field, A is a valuation ring of R, and is the maximal ideal of A.
Solution
No solution has been written yet.