Bourbaki

Example 3

Tag 0D2F, permanent. Cite it as [Bourbaki, Tag 0D2F].

Commutative Algebra, chapter X, § 2. Modules et anneaux macaulayens

Let MM be a finitely generated Macaulay AA-module and (x1,…,xn)(x_1, \ldots, x_n) an MM-regular sequence of elements of AA. Then the AA-module M‾=M/(x1M+⋯+xnM)\overline{M} = M/(x_1M + \cdots + x_nM) is Macaulay. Let indeed mm be a maximal ideal of AA belonging to the support of M‾\overline{M}; one has xi∈mx_i \in m for every ii since xix_i annihilates M‾\overline{M}, and the canonical images of the xix_i in AmA_m form an MmM_m-regular sequence of elements of mAmmA_m. One consequently has (§1§ 1, No. 4, Proposition 7 and VIII, §3§ 3, No. 2, Cor. of Proposition 3) the equalities

profAm(M‾m)=profAm(Mm)−n, \mathrm{prof}_{A_m}(\overline{M}_m) = \mathrm{prof}_{A_m}(M_m) - n,
dim⁡Am(M‾m)=dim⁡Am(Mm)−n, \dim_{A_m}(\overline{M}_m) = \dim_{A_m}(M_m) - n,
whence our assertion.

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

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