Bourbaki

Exercise 8, § 1

Commutative Algebra, VI

Bourbaki marks this one as harder than the rest.

Given a ring AA, one says that a polynomial P(X1,…,Xn)P(X_1, \ldots, X_n) with coefficients in AA is dominated if it is of the form Xα+∑βcβXβX^\alpha + \sum_\beta c_\beta X^\beta, with β<α\beta < \alpha in the ordered product set Nn\mathbf{N}^n for all the β\beta such that cβ≠0c_\beta \neq 0. One says that, in a ring RR, a subring AA is paravaluative if, for every dominated polynomial P∈A[X1,…,Xn]P \in A[X_1, \ldots, X_n] (nn arbitrary), one has P(s1,…,sn)≠0P(s_1, \ldots, s_n) \neq 0 whatever the elements si∈R−As_i \in R - A may be (1⩽i⩽n1 \leqslant i \leqslant n).

a) Now prove that if AA is paravaluative in RR, the product of two elements of S=R−AS = R - A belongs to SS.

b) Let AA be a subring of a ring RR such that the product of two elements of S=R−AS = R - A belongs to SS; let p0\mathfrak{p}_0 be the prime ideal of AA and of RR formed by the a∈Aa \in A such that sa∈Asa \in A for every s∈Ss \in S (exerc. 6 b)). In order that AA be paravaluative in RR, it is necessary and sufficient that A/p0A/\mathfrak{p}_0 be paravaluative in R/p0R/\mathfrak{p}_0.

c) Let RR be a ring, h:R→R′h : R \to R' a homomorphism of rings. If A′A' is a paravaluative subring in R′R', A=h−1(A′)A = h^{-1}(A') is paravaluative in RR. Deduce that if (A,p)(A, \mathfrak{p}) is a maximal pair for RR (exerc. 7), AA is paravaluative in RR (use b) and exerc. 7 d) and e)).

d) Let R′R' be a ring, RR a subring of R′R'. If AA is a paravaluative subring of RR, now prove that there exists a paravaluative subring A′A' in R′R' such that A′∩R=AA' \cap R = A. (If S=R−AS = R - A, and S0=S∪{1}S_0 = S \cup \{1\}, consider the ring T′=S0−1R′T' = S_0^{-1}R' and the canonical homomorphism h:R′→T′h : R' \to T'. Let BB be the subring of T′T' generated by h(A)∪(h(S))−1h(A) \cup (h(S))^{-1} and b\mathfrak{b} the ideal of BB generated by (h(S))−1(h(S))^{-1}. Prove that b≠B\mathfrak{b} \neq B; consider a maximal pair (A′′,p′′)(A'', \mathfrak{p}'') for T′T' such that B⊂A′′B \subset A'' and b⊂p′′\mathfrak{b} \subset \mathfrak{p}'', and take A′=h−1(A′′)A' = h^{-1}(A'')).

Conclude from this that if RR is an integral ring, the paravaluative subrings of RR are the intersections of RR and the valuation rings for the field of fractions of RR.

Solution

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Machine translation of the English, by gpt-5-6-mini, gpt-5-6. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.