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We define the multiplicity of an $ R $-module $ M $ of dimension $ d > 0 $ to be $$ \text{mult}(M) \stackrel{\text{df}}{=} \text{LC}(P_{M}) \cdot (d - 1)!, $$ where $ P_{M} $ denotes the Hilbert polynomial of $ M $ and $ \text{LC} $ stands for ‘leading coefficient’. Equivalently, $$ \text{mult}(M) \stackrel{\text{df}}{=} {Q_{M}}(1), $$ where $ {HP_{M}}(z) = \dfrac{{Q_{M}}(z)}{(1 - z)^{d}} $ and $ HP_{M} $ denotes the Hilbert-Poincaré series of $ M $.

I can show that if $ I = \langle f_{1},\ldots,f_{r} \rangle $ with $ \deg(f_{i}) = d_{i} $, and $ (f_{1},\ldots,f_{r}) $ is an $ M $-regular sequence, then $$ \text{mult}(M / I M) = (d_{1} \cdots d_{r}) \cdot \text{mult}(M). $$

Consider $ R = \mathbb{k}[x_{1},\ldots,x_{n}] $ and $ I = \langle f_{1},\ldots,f_{r} \rangle $ with $ f_{i} \in R $ homogeneous and $ \deg(f_{i}) = d_{i} \geq 2 $. Does the reverse implication hold, i.e., does $$ \text{mult}(R / I) = d_{1} \cdots d_{r} \quad \Longrightarrow \quad (f_{1},\ldots,f_{r}) ~ \text{is an $ R $-regular sequence}? $$

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Yes, but this is quite nontrivial : see Bourbaki, Commutative Algebra VIII, exercise 4 of §7.

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  • $\begingroup$ Does VIII denote the chapter? Because in my Bourbaki, Commutative Algebra, there are only 7 chapters. $\endgroup$
    – Ella Smith
    Aug 20, 2014 at 12:30
  • $\begingroup$ @Ella: Take the french version (consisting of 10 chapters). $\endgroup$ Aug 20, 2014 at 13:20
  • $\begingroup$ Well, we have to adapt that exercise to the graded case, but I think, like usual, everything can be done smoothly. $\endgroup$
    – user26857
    Aug 20, 2014 at 20:05
  • $\begingroup$ The statements of the exercise are about the associated graded ring, so I think it applies directly to the situation of the OP. $\endgroup$
    – abx
    Aug 21, 2014 at 5:14

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