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Let $A$ be a graded ring and $M$ a graded $A$-module. By $P(M,t)$ we denote the Poincaré series for $M$.

In Atiyah and Macdonald, theorem 11.1 claims $P(M,t)=\dfrac{f(t)}{\prod _{i=1}^n (1-t^{k_i})}$ where $k_i$ are degrees of homogeneous elements generating $A$ as $A_0$-algebra ($A_0$ is the zero graded part).

The pole at $1$ of $P(M,t)$, they denote by $d(M)$.

Everything works just fine but I don't understand if it is always defined, i.e., can $d(M)$ be negative (meaning that $P(M,t)$ is divisible by $1-t$, when $f(t)$ is divisible by a bigger power of $(1-t)$)?

Also 11.3 says that $d(M/xM)=d(M)-1$. What if $d(M)=0$?

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  1. $d(M)$ can't be negative. Whenever $P(M,t)$ is a polynomial, $d(M)=0$.

  2. Here the authors should have added the assumption $d(M)\ge1$. (Maybe they had in mind the usual case when $A_0$ is a field. In this case $d(M)=\dim M$, and if $d(M)=0$ then $M$ is artinian, so there is no non-zerodivisor on $M$.)

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  • $\begingroup$ thanks, for 2 apparently this has been already discussed, it is listed in mathoverflow.net/questions/42241/errata-for-atiyah-macdonald as one of the errors, one person suggested $P(M/xM,t) \neq 0$ as another needed condition in order that theorem works $\endgroup$ – ralleee Feb 4 '16 at 21:10
  • $\begingroup$ @drekavac I know that tread because I've also contributed. I can't see why we need that condition: it means $M/xM\ne0$. But if $M=xM$ then $M=0$ by graded Nakayama lemma and this is impossible if one assumes $d(M)\ge1$. $\endgroup$ – user26857 Feb 4 '16 at 22:10

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