Lemma 10.59.10. Let $R$ be a Noetherian local ring. Let $I \subset R$ be an ideal of definition. Let $0 \to M' \to M \to M'' \to 0$ be a short exact sequence of finite $R$-modules. Then

1. if $M'$ does not have finite length, then $\chi _{I, M} - \chi _{I, M''} - \chi _{I, M'}$ is a numerical polynomial of degree $<$ the degree of $\chi _{I, M'}$,

2. $\max \{ \deg (\chi _{I, M'}), \deg (\chi _{I, M''}) \} = \deg (\chi _{I, M})$, and

3. $\max \{ d(M'), d(M'')\} = d(M)$,

Proof. We first prove (1). Let $N \subset M'$ be as in Lemma 10.59.3. By Lemma 10.59.9 the numerical polynomial $\chi _{I, M'} - \chi _{I, N}$ has degree $<$ the common degree of $\chi _{I, M'}$ and $\chi _{I, N}$. By Lemma 10.59.3 the difference

$\chi _{I, M}(n) - \chi _{I, M''}(n) - \chi _{I, N}(n - c)$

is constant for $n \gg 0$. By elementary calculus the difference $\chi _{I, N}(n) - \chi _{I, N}(n - c)$ has degree $<$ the degree of $\chi _{I, N}$ which is bigger than zero (see above). Putting everything together we obtain (1).

Note that the leading coefficients of $\chi _{I, M'}$ and $\chi _{I, M''}$ are nonnegative. Thus the degree of $\chi _{I, M'} + \chi _{I, M''}$ is equal to the maximum of the degrees. Thus if $M'$ does not have finite length, then (2) follows from (1). If $M'$ does have finite length, then $I^ nM \to I^ nM''$ is an isomorphism for all $n \gg 0$ by Artin-Rees (Lemma 10.51.2). Thus $M/I^ nM \to M''/I^ nM''$ is a surjection with kernel $M'$ for $n \gg 0$ and we see that $\chi _{I, M}(n) - \chi _{I, M''}(n) = \text{length}(M')$ for all $n \gg 0$. Thus (2) holds in this case also.

Proof of (3). This follows from (2) except if one of $M$, $M'$, or $M''$ is zero. We omit the proof in these special cases. $\square$

Comment #7009 by Xiaolong Liu on

Maybe at the proof of (2) when $M'$ have finite length, we should use Artin-Rees lemma at '$M/I^nM\to M''/I^nM''$ with kernel $M'$' instead of at '$I^nM\to I^nM''$' (this is just a NAK).

Comment #7233 by on

@#7009. The map $I^nM \to I^nM''$ is surjective as $M \to M''$ is surjective, but I think to see that it is injective you use that $I^nM \cap M' = 0$ for $n \gg 0$ by Artin-Rees. No?

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