Bourbaki

Lemma 2

Tag 0D0X, permanent. Cite it as [Bourbaki, Tag 0D0X].

Commutative Algebra, chapter X, § 1. Profondeur

Let ρ:A→B\rho : A \to B be a local homomorphism of noetherian local rings, NN a finitely generated BB-module and yy an element of mBm_B. The following two conditions are equivalent:

(i) the AA-module N/yNN/yN is flat and the homothety yNy_N is injective; (ii) the AA-module NN is flat and the homothety yκA⊗Ny_{\kappa_A \otimes N} is injective.

When they are satisfied, the homothety yM⊗ANy_{M \otimes_A N} is injective for every AA-module MM.

Assume that the hypotheses of (i) are satisfied, and let us prove (ii) as well as the last assertion. Let MM be an AA-module. Since the AA-module N/yNN/yN is flat, we deduce from the exact sequence 0→N→yNN→N/yN→00 \to N \xrightarrow{y_N} N \to N/yN \to 0 the exact sequences

0→M⊗AN→uM⊗AN→M⊗A(N/yN)→0 0 \to M \otimes_A N \xrightarrow{u} M \otimes_A N \to M \otimes_A (N/yN) \to 0
0→Tor1A(M,N)→vTor1A(M,N)→0 0 \to \mathrm{Tor}_1^A(M, N) \xrightarrow{v} \mathrm{Tor}_1^A(M, N) \to 0

where u=1M⊗yNu = 1_M \otimes y_N and v=Tor1A(1M,yN)v = \mathrm{Tor}_1^A(1_M, y_N); it follows that the homothety of ratio yy is injective in M⊗ANM \otimes_A N, and bijective in Tor1A(M,N)\mathrm{Tor}_1^A(M, N). Assume moreover that the A-module MM is finitely generated; then the B-module Tor1A(M,N)\mathrm{Tor}_1^A(M, N) is finitely generated (A, X, p. 107, Prop. 6), hence is zero since yy belongs to mBm_B (Nakayama's lemma), which implies that the A-module NN is flat (A, X, p. 74, Theorem 2).

(ii) ⇒\Rightarrow (i): assume that the hypotheses of (ii) are satisfied. Consider the two exact sequences of B-modules

(1)

0→Ker(yN)⟶N→pIm(yN)→0 0 \to \mathrm{Ker}(y_N) \longrightarrow N \xrightarrow{p} \mathrm{Im}(y_N) \to 0

(2)

0→Im(yN)→iN⟶N/yN→0, 0 \to \mathrm{Im}(y_N) \xrightarrow{i} N \longrightarrow N/yN \to 0,

where pp and ii are the canonical homomorphisms. It follows that the homomorphism 1⊗p:κA⊗AN⟶κA⊗AIm(yN)1 \otimes p : \kappa_A \otimes_A N \longrightarrow \kappa_A \otimes_A \mathrm{Im}(y_N) is surjective, and (since NN is flat) that the kernel of the homomorphism 1⊗i:κA⊗AIm(yN)⟶κA⊗AN1 \otimes i : \kappa_A \otimes_A \mathrm{Im}(y_N) \longrightarrow \kappa_A \otimes_A N is isomorphic to Tor1A(κA,N/yN)\mathrm{Tor}_1^A(\kappa_A, N/yN). But the mapping (1⊗i)∘(1⊗p)(1 \otimes i) \circ (1 \otimes p), equal to yκA⊗ANy_{\kappa_A \otimes_A N}, is injective by hypothesis; it follows that 1⊗p1 \otimes p is bijective and 1⊗i1 \otimes i injective, and consequently that one has Tor1A(κA,N/yN)=0\mathrm{Tor}_1^A(\kappa_A, N/yN) = 0. It follows that the A-module N/yNN/yN is flat (III, § 5, No. 2, Theorem 1 and No. 4, Proposition 2).

Since NN and N/yNN/yN are flat over AA, the same is true of Im(yN)\mathrm{Im}(y_N) (exact sequence (2)). One then deduces from exact sequence (1) that κA⊗AKer(yN)\kappa_A \otimes_A \mathrm{Ker}(y_N) is isomorphic to the kernel of 1⊗p1 \otimes p, hence is zero; thus the homothety yNy_N is injective by Nakayama's lemma.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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