# Poincaré series of an $A$-module

I'm studying Atiyah and Macdonald's "Introduction to Commutative Algebra" and I'm having some problems computing the Poincaré series of an $A$-module.

Even the Example after $11.3$, which is dealt very quickly on the text, does not appear clear to me:

Why can we conclude that $P(A,t)=(1-t)^{-s}$ (where $A=A_0[x_1,\ldots,x_s]$ with $A_0$ Artin ring and $x_1,\ldots,x_s$ independent indeterminates) ?

Do you know any source of guided excercises and\or examples which can help me to get more confident with these objects?

• Please ask a specific question. Commented Nov 25, 2013 at 10:57
• I've edited my question making it more specific, I hope. Plus, what I'd need is some additional reference, possibly with examples and excercises.
– uldi
Commented Nov 25, 2013 at 11:06

Atiyah-Macdonald explain that $l(A_n)=\binom{s+n-1}{s-1}$ (by writing down an explicit basis consisting of monomials). It follows $$P(A,t) = \sum_{n=0}^{\infty} \binom{s+n-1}{s-1} t^n.$$ We claim that this equals $(1-t)^{-s}$ (as formal power series), and prove this by induction on $s$. The case $s=1$ is the usual geometric series. For the induction step, we start with $$(1-t)^{-s} = \sum_{n=0}^{\infty} \binom{s+n-1}{s-1} t^n$$ and apply $\frac{d}{dt}$ on both sides. It follows $$s (1-t)^{-(s+1)} = \sum_{n=0}^{\infty} \binom{s+n}{s-1} (n+1) t^n$$ The claim for $s+1$ follows from $$\binom{s+n}{s} = \frac{n+1}{s} \cdot \binom{s+n}{s-1}.$$ We can also give a more direct proof as follows: Taking the $s$th power in $(1-t)^{-1} = \sum_{n=0}^{\infty} t^n$, we get $$(1-t)^{-s} = \sum_{n=0}^{\infty} \left(\sum_{k_1+\dotsc+k_s = n} 1 \cdot \dotsc \cdot 1 \right) t^n = \sum_{n=0}^{\infty} \binom{s+n-1}{s-1} t^n.$$

• Can you please point me out where is shown that $l(A_n)=\binom{s+n-1}{s-1}$? Does this work with an arbitrary additive function or does it have to be the length? Atiyah-Macdonald say that $A_n$ is generated by the monomials in $x_1,\ldots,x_s$ of degree $n$, and that they are $\binom{s+n-1}{s-1}$. Does that follow from this assumption? Why? The second part works great, thank you.
– uldi
Commented Nov 25, 2013 at 12:09
• They restrict to the length right before the example. And they prove $l(A_n) = \binom{s+n-1}{s-1}$ in the example. Remember that that the length of a free module is just the rank multiplied with the length of the base ring, which is $0$ because $A_0$ is assumed to be Artinian. Commented Nov 25, 2013 at 14:26
• Could you please provide me some reference about these facts (I mean the length of a free module being the rank multiplied with the length of the base ring)? I can't get the grasp of it and I really feel like I'm missing something important.
– uldi
Commented Nov 25, 2013 at 14:55
• The length is additive. You can find all that in Atiyah-Macdonald. Commented Nov 25, 2013 at 16:23

Since this question concerns a typo in Atiyah-MacDonald which has not been pointed out in the answers nor the comments, I add a new answer for future reference. The Poincaré series in the example should be

$$P(A, t) = \text{len}(A_0) (1-t)^{-s}.$$

This follows from the fact, explained in the example, that $$A_n$$ is free over $$A_0$$ of rank $$\binom{n+s-1}{s-1}$$ and length is additive (in particular $$\text{len}(A_0^k) = k \text{len} (A_0)$$). Hence, $$\text{len}(A_n) = \text{len}(A_0) \binom{n+s-1}{s-1}$$ and $$P(A,t) =\text{len}(A_0) \sum_{n \geq 0} \binom{n+s-1}{s-1}t^n =\text{len}(A_0) (1-t)^{-s}.$$