Exercise 8, § 1
Commutative Algebra, VI
Bourbaki marks this one as harder than the rest.
Given a ring , one says that a polynomial with coefficients in is dominated if it is of the form , with in the ordered product set for all the such that . One says that, in a ring , a subring is paravaluative if, for every dominated polynomial ( arbitrary), one has whatever the elements may be ().
a) Now prove that if is paravaluative in , the product of two elements of belongs to .
b) Let be a subring of a ring such that the product of two elements of belongs to ; let be the prime ideal of and of formed by the such that for every (exerc. 6 b)). In order that be paravaluative in , it is necessary and sufficient that be paravaluative in .
c) Let be a ring, a homomorphism of rings. If is a paravaluative subring in , is paravaluative in . Deduce that if is a maximal pair for (exerc. 7), is paravaluative in (use b) and exerc. 7 d) and e)).
d) Let be a ring, a subring of . If is a paravaluative subring of , now prove that there exists a paravaluative subring in such that . (If , and , consider the ring and the canonical homomorphism . Let be the subring of generated by and the ideal of generated by . Prove that ; consider a maximal pair for such that and , and take ).
Conclude from this that if is an integral ring, the paravaluative subrings of are the intersections of and the valuation rings for the field of fractions of .
Solution
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