Bourbaki

Proposition 1

Tag 0D06, permanent. Cite it as [Bourbaki, Tag 0D06].

Commutative Algebra, chapter X, § 1. Profondeur

Let AA be a ring, JJ an ideal of AA and 0→M′→M→M′′→00 \to M' \to M \to M'' \to 0 an exact sequence of AA-modules. Put

p′=prof⁡(J;M′),p=prof⁡(J;M),p′′=prof⁡(J;M′′). p' = \operatorname{prof}(J; M') \quad , \quad p = \operatorname{prof}(J; M) \quad , \quad p'' = \operatorname{prof}(J; M'') .

Then one is in one of the following three cases, which are mutually exclusive:

p′=p⩽p′′,p=p′′<p′,p′′=p′−1<p. p' = p \leqslant p'' \quad , \quad p = p'' < p' \quad , \quad p'' = p' - 1 < p .

Consider the exact sequence of extension modules associated with A/JA/J and with the exact sequence above (A, X, p. 92, th. 2). Exclude the case p=p′=p′′=+∞p = p' = p'' = +\infty; there then exists in this sequence a first nonzero module, and the following module is also nonzero. This gives the following three possibilities:

a) The first nonzero module is Ext⁡Ap′(A/J,M′)\operatorname{Ext}_A^{p'}(A/J, M'). Then p′=p⩽p′′p' = p \leqslant p''.

b) The first nonzero module is Ext⁡Ap(A/J,M)\operatorname{Ext}_A^p(A/J, M). Then p=p′′<p′p = p'' < p'.

c) The first nonzero module is Ext⁡Ap′′(A/J,M′′)\operatorname{Ext}_A^{p''}(A/J, M''). Then p′′+1=p′⩽pp'' + 1 = p' \leqslant p.

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

Cited by

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Cites

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Translated with machine help, not yet reviewed by a person.