Bourbaki

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 a,a′a, a' be two elements of A and s, s' two elements of S such that sa∈Asa \in A and s′a′∈As'a' \in A. Show that one of the two elements sa′,s′asa', s'a belongs to A. Deduce that the set p1p_1 of elements a∈Aa \in A for which there exists s∈Ss \in S such that sa∈Asa \in A is a prime ideal of A.

(b) Show that the set p0p_0 of a∈Aa \in A such that sa∈Asa \in A for all s∈Ss \in S is a prime ideal of R and A; it is the greatest ideal of R contained in A and p0⊂p1p_0 \subset p_1.

(c) The set n of elements x∈Rx \in R such that there exists s∈Ss \in S for which sx=0sx = 0 is both an ideal of R and an ideal of A and n⊂p0n \subset p_0.

(d) The ring A is integrally closed in R.

(e) If R is a field, A is a valuation ring of R, p0=(0)p_0 = (0) and p1p_1 is the maximal ideal of A.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.