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Let $A = \bigoplus_{n\geq 0} A_n$ be a graded ring that is generated as an $A_0$-algebra by a finite collection of elements of $A_1$, where $A_0$ is artinian. I wish to show that if $$ 0 \to M(1) \to M(2) \to \dots \to M(d) \to 0 $$ is an exact sequence of finitely generated graded $A$-modules $M(i)$ with degree $0$ maps, then $$ \sum_{i=1}^d (-1)^i \chi(M(i)) = 0 \text, $$ where each $\chi(M(i))$ is the function from $\mathbb Z_{\geq 0}$ to itself given by $(\chi(M(i))(n) = \ell_{A_0}(M(i)_n)$, the length of the finitely generated $A_0$-module $M(i)_n$.

(Is it correct that $\chi(M(i))$ is called the Hilbert function of $M(i)$?)

I think I have a basic start on this. The result is equivalent to the assertion that for each nonnegative integer $n$, $$ \sum_{i=1}^d (-1)^i \ell_{A_0}(M(i)_n) = 0 \text. $$ I will try to prove this by induction on $d$. For the base case $d = 1$, if I have an exact sequence $$ 0 \to M(1) \to 0 \text, $$ then $M(1)$ is the zero module and thus $M(1)_n$ is also the zero module for each $n$. Since the zero module has length $0$, it follows that for each $n$, $$ \sum_{i=1}^1 (-1)^i \ell_{A_0}(M(i)_n) = -\ell_{A_0}(M(1)_n) = 0\text. $$ For the induction step, I let $d > 1$ and assume that the result holds for $d-1$. Then given an exact sequence $$ 0 \to M(1) \to \dots \to M(d-2) \overset f\to M(d-1) \overset g\to M(d) \to 0 \text, $$ I know that $$ 0 \to M(1) \to \dots \to M(d-2) \overset f\to \operatorname{im} f \to 0 $$ is an exact sequence with one fewer term. It then follows by the inductive assumption that for each nonnegative integer $n$, $$ \sum_{i=1}^{d-2} (-1)^i \ell_{A_0}(M(i)_n) + (-1)^{d-1} \ell_{A_0}((\operatorname{im} f)_n) = 0 \text. $$ But how should I proceed from here to show that $$ \sum_{i=1}^{d} (-1)^i \ell_{A_0}(M(i)_n) = 0 \text? $$

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The given exact sequence is exact on each degree, that is, $$ 0 \to M(1)_n \to M(2)_n \to \dots \to M(d)_n \to 0 $$ is exact for all $n$. Now just use the additivity of length on exact sequences.

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  • $\begingroup$ I suggest you to take a look at this thread. $\endgroup$ – user26857 May 13 '16 at 20:54
  • $\begingroup$ Okay, interesting. I'll have to think about which short exact sequence to apply the additivity result to... The first one that comes to mind is $0\to (\operatorname{im} f)_n \to M(d-1)_n \to M(d)_n \to 0$. $\endgroup$ – justin May 13 '16 at 21:05

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