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We define multiplicity of a module $M$ of dimension $d>0$ as $$e(M) := \operatorname{lc} (P_M) (d-1)!,$$ where $P_M$ denotes the Hilbert polynomial of $M$ and $\operatorname{lc}(P_M)$ its leading coefficient. Equivalently, we have $e(M) = Q_M(1)$, where $HP_M (z) = \frac{Q_M(z)}{(1-z)^d}$ and $HP_M$ is the Hilbert-Poincaré series of $M$.

I can prove that if $I = (f_1, \dots, f_r)$, with $\deg(f_i)=d_i$ and $f_1, \dots, f_r$ is an $M$-regular sequence, then $$e(M/IM) = d_1 \cdots d_r \cdot e(M)$$

Consider now $R= \mathbb{k}[x_1, \dots, x_n]$, $I = (f_1, \dots, f_r)$, with $\deg(f_i)=d_i \geq 2$ and $f_i$ homogeneous. Does it holds the reverse implication, i.e. $$ e(R/I) = d_1 \cdots d_r \quad \Rightarrow \quad f_1,\dots,f_r \quad \text{is a $R$-regular sequence?}$$

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  • $\begingroup$ How do you prove for $r=2$? Yes, I asked also on MO but no answers. $\endgroup$ – Ella Smith Aug 17 '14 at 17:07
  • $\begingroup$ What you need to observe is how multiplicity behaves under a short exact sequence. When $r = 2$, you could first mod out by $f_1$ and assume $R' = R/ (f)$ is a complete intersection of multiplicity $d_1$. Now use the following short exact sequence $0 \to R'/ (0:f_2 R') \to R' \to R'/ f_2R'$ to argue, under the hypothesis, the image of $f_2$ in $R'$ is regular. Having this and the additional assumption mentioned in @user26857's first comment, you can finish the proof. $\endgroup$ – Youngsu Aug 18 '14 at 1:24
  • $\begingroup$ @user26857 I'd go with the additivity of multiplicity. Assume to the contrary that $f_2$ is not regular on $R'$. From the short exact sequence, one has $e(R'/f_2R') \le e(R')$. What remains to show is $\dim R' /f_2 R' = \dim R'$. Since $f_2$ is not regular , it is in an associated prime of $R'$. Since $R'$ is a complete intersection ring, Min $R'$ = Ass $R'$. I did not point out $r=2$ is easy enough, but I did want to point out your comment was insightful. I'm curious how your proof goes. $\endgroup$ – Youngsu Aug 30 '14 at 18:18
  • $\begingroup$ @user26857 Well, what you said is correct, but that is not what I used. In your example $\dim R/fR \neq \dim R$, but in the proof $\dim R'/ f_2 R' = \dim R'$. In other words, if $e(R/fR) > e(R)$, then $f$ is not in a minimal prime. See Theorem 14.6 of Matsumura's commutative ring theory. $\endgroup$ – Youngsu Aug 31 '14 at 15:15
  • $\begingroup$ @user26857 Then $e(R'/f_2R') > e(R')$ implies that $\dim (R'/f_2 R') < \dim R'$. $\endgroup$ – Youngsu Aug 31 '14 at 15:17

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