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For an additive function $\lambda$ and an exact sequence of modules

$0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0$,

we have $\lambda(M_2) = \lambda(M_1) + \lambda(M_3)$ by definition. If the modules are graded and every morphism preserves degree, then the Poincaré series are also additive, simply by using the additivity of $\lambda$ on every summand of the power series. What if the first morphism merely carries homogeneous elements of degree $d$ to, say, elements of degree $d+k$ for instance? (and similarly for the second morphism) What would the formula for the Poincaré series be in this case?

I figured that I'd simply have to use additivity on a single sequence

$0 \rightarrow M_{1_n} \rightarrow M_{2_{n+deg(f)}} \rightarrow M_{3_{n+deg(f)+deg(g)}} \rightarrow 0$,

but I'm clueless on how to relate the Poincaré series' from here. (I deleted my old unanswered question since it was bloated and hid the main issue I was trying to point out)

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1 Answer 1

up vote 3 down vote accepted

If the first morphism has degree $k$ and the second has degree $n$, then you can instead work with the genuine exact sequence of graded modules $$0 \to \Sigma^{k+n} M_1 \to \Sigma^n M_2 \to M_3 \to 0$$

where $\Sigma^k M$ denotes $M$, but where elements of degree $d$ in $M$ are interpreted as elements of degree $d+k$. (By "genuine" I mean the morphisms preserve degree.) This has the effect of multiplying the Poincaré series by the appropriate powers of $t$ and the rest is straightforward. More precisely, letting $\chi$ denote the Poincaré series, we get $$t^n \chi(M_2) = t^{k+n} \chi(M_1) + \chi(M_3).$$

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Neat, thanks a bunch! (upvote incoming when I have enough rep. [which I have as I've just seen.]) –  jonny Mar 21 '12 at 19:41
    
No problem. For the sake of completeness, the more general fact is that you can define tensor products of graded modules (say where each piece is finite-dimensional over the base field $F$) and Poincaré series are multiplicative with respect to tensor products. To shift a graded module by $k$ just take the tensor product with the module which is $0$ in all degrees except $k$, where it is equal to $F$. –  Qiaochu Yuan Mar 21 '12 at 19:45

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