Bourbaki

Exercise 1, § 3

Commutative Algebra, V, 370

Bourbaki marks this one as harder than the rest.

Let AA be a ring, p\mathfrak{p} a prime ideal of AA, BB a finitely generated AA-algebra and let r1⊂…⊂rmr_1 \subset \ldots \subset r_m be a non-empty increasing sequence of ideals of BB lying above p\mathfrak{p}. Show that there exist an element a∈A−pa \in A - \mathfrak{p} and a finite family (yk)1≤k≤q(y_k)_{1 \leq k \leq q} of elements of BB with the following properties:

(1) If we write B′=A[y1,…,yq]B' = A[y_1, \ldots, y_q] and if SS is the set of powers of aa, the ring S−1BS^{-1}B is integral over S−1B′S^{-1}B'.

(2) For all ii such that 1≤i≤m1 \leq i \leq m there exists an integer h(i)≥0h(i) \geq 0 such that rir_i contains y1,…,yh(i)y_1, \ldots, y_{h(i)} and for every polynomial P∈A[Xh(i)+1,…,Xq]P \in A[X_{h(i)+1}, \ldots, X_q] the relation P(yh(i)+1,…,yr)∈riP(y_{h(i)+1}, \ldots, y_r) \in r_i implies that all the coefficients of PP belong to p\mathfrak{p}.

The ideal ri∩B′r_i \cap B' is then generated by p\mathfrak{p} and the yky_k such that k≤h(i)k \leq h(i). (Argue by induction on mm, thus reducing it to the case m=1m = 1. Then let c=r1ac = r_1a; argue by induction on the number nn of generators of the AA-algebra BB. Let (xi)1≤i≤n(x_i)_{1 \leq i \leq n} be such a system of generators and let ss be the ideal of A[X1,…,Xn]A[X_1, \ldots, X_n] consisting of those PP such that P(x1,…,xi)∈cP(x_1, \ldots, x_i) \in c; then s⊇pA[X1,…,Xn]s \supseteq \mathfrak{p}A[X_1, \ldots, X_n]; distinguish two cases according to whether ss is equal or not to pA[X1,…,Xn]\mathfrak{p}A[X_1, \ldots, X_n]. In the second case, there is a polynomial Q∈A[X1,…,Xn]Q \in A[X_1, \ldots, X_n] none of whose coefficients is in p\mathfrak{p} and which belongs to BB. Show that it is possible to determine integers mim_i (2≤i≤n2 \leq i \leq n) such that, if we write xi′=Q(x1,…,xn),xi′=xi−x1mix'_i = Q(x_1, \ldots, x_n), x'_i = x_i - x_1^{m_i} (2≤i≤n2 \leq i \leq n), there exists an element b∈A−pb \in A - \mathfrak{p} for which bx1bx_1 is integral over C=A[x2′,…,xn′]C = A[x'_2, \ldots, x'_n]; apply the induction hypothesis to the AA-algebra

B1=A[x2′,…,xn′] B_1 = A[x'_2, \ldots, x'_n]
and the ideal c∩B1c \cap B_1.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.