Bourbaki

Exercise 8, § 1

Commutative Algebra, VI, 443

Bourbaki marks this one as harder than the rest.

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

(a) Show 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 p0p_0 be the prime ideal of AA and RR consisting of the a∈Aa \in A such that sa∈Asa \in A for all s∈Ss \in S (Exercise 6 (b)). For AA to be paravaluative in RR, it is necessary and sufficient that A/p0A/p_0 be paravaluative in R/p0R/p_0.

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

(d) Let R′R' be a ring and RR a subring of R′R'. If AA is a paravaluative subring of RR, show that there exists a paravaluative subring A′A' of 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 bb the ideal of BB generated by (h(S))−1(h(S))^{-1}. Show that b≠Bb \neq B; consider a maximal ordered pair (A′,p′′)(A', p'') in T′T' such that B⊂A′′B \subset A'' and b⊂p′′b \subset p'' and take A′=h‾(A′′)A' = \overline{h}(A'').)

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

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.