Proposition 11.4. Let $A$ be a Noetherian local ring, $\mathfrak{m}$ its maximal ideal, $\mathfrak{q}$ an $\mathfrak{m}$-primary ideal, $M$ a finitely generated $A$-module, $(M_n)$ a stable $\mathfrak{q}$-filtration of $M$.
From (i) and (ii), we have $l(M/M_n)$ is finite and there exists $g(x)\in \mathbb{Q}[x]$ such that $l(M/M_n)=g(n)$ for all sufficiently large $n$. In the proof of (iii), let $(M'_n)$ be another stable $\mathfrak{q}$-filtration of $M$ and let $g'(n)=l(M/M'_n)$. We see that there exits $n_0\in\mathbb{N}$ s.t. $g(n+n_0)\geq g'(n)$ and $g'(n+n_0)\geq g(n)$ for all sufficiently large $n$. But I don't understand why does the sequence $g(n)/g'(n)=1$ as $n\rightarrow\infty$. Moreover, why do we conclude that $g$ and $g'$ have the same degree and leading coefficient? We have $g(n)/g'(n)$ converges to $1$ as a sequence but $g(x)/g'(x)$ may not converges to $1$ as a function.
Best Answer
$g$ and $g'$ take positive values, from the inequality $g'(n+n_0)/g'(n) \geq g(n)/g'(n)$ you get $$\limsup \dfrac{g(n)}{g'(n)} \leq 1$$ and from the inequality $g(n+n_0)/g'(n+n_0) \geq g'(n)/g'(n+n_0)$ you get $$\liminf \dfrac{g(n+n_0)}{g'(n+n_0)} = \liminf \dfrac{g(n)}{g'(n)} \geq 1$$ so that $$1 \leq \liminf \dfrac{g(n)}{g'(n)} \leq \limsup \dfrac{g(n)}{g'(n)} \leq 1$$ and finally the sequence $\dfrac{g(n)}{g'(n)}$ converges to $1$. For you last question, if $\deg(g) > \deg(g')$, the last sequence should diverge to $+\infty$ ; if $\deg(g') > \deg(g)$, then this time the sequence should converge to $0$, do you see why ?
First edit (this is false because I reversed the order of the filtration but it contains some arguments that I will use in the second edit) : we can assume that $g(n)$ and $g'(n)$ take strictly positive values for $n \gg 0$. In fact, if $l(M/M_p) = 0$ for a certain natural number $p$, then as you said $M = M_p$ and $M = M_{k}$ for $k \geq p$ as $(M_n)$ is a filtration. But now recall that this is a stable $\mathfrak q$-filtration, so for $k \gg 0$ (in particular $k \geq p$), we have $M = M_{k+1} = \mathfrak{q} M_k = \mathfrak{q} M$. Now use the fact that $A$ is a local ring with maximal ideal $\mathfrak m$, that $\mathfrak q$ is $\mathfrak m$-primary and that $M$ is finitely generated. More precisely : $\mathfrak q \subset \mathfrak m = r(\mathfrak q)$, so $M = \mathfrak m M$ and now apply Nakayama's lemma to get $M = 0$. The claim of the Proposition 11.4 is now trivial.
Second edit : as Blade noticed, I made a mistake in my first edit so here is a proof of the claim (that I hope is correct). Let $n_0$ be a natural number such that $\mathfrak q M_n = M_{n+1}$ and $g(n) = l(M/M_n)$ for every $n \geq n_0$. If for every $n \geq n_0$, you can find $k \geq n$ so that $g(k) = 0$, then the filtration $(M_n)$ is equal to the constant filtration $(M)$ (do you see why ?). Now we can apply the above arguments because $(M)$ is a stable $\mathfrak q$-filtration and we obtain $M = 0$ by Nakayama's lemma. This way, we can assume that $g$ and $g'$ take strictly positive values for $n \gg 0$.