**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.
$\square$

## Comments (2)

Comment #7009 by Xiaolong Liu on

Comment #7233 by Johan on

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