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I'm not seeing a line in the proof of Proposition 11.4 in Atiyah-MacDonald. Here is a link to some notes I found online which contain the proof:

http://folk.uio.no/fredrme/Kommalg.pdf

It is also 11.4 in these notes. Specifically, I don't understand how we conclude that $l(M/M_n)$ is a polynomial. Based on the text it's completely obvious, I guess I'm just not seeing it. I have not obtained any permission to use these notes, I simply found them online. If this is against site policy, please let me know and I will remove them.

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Do you understand the proof of Theorem 11.1 in Atiyah-MacDonald, p.117? –  Makoto Kato Aug 1 '12 at 2:00
    
I'd thought so but perhaps not? I see how the corollary lets us say that $l(M_n/M_{n+1})$ is a polynomial. –  Mike B Aug 1 '12 at 2:52
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2 Answers

up vote 1 down vote accepted

All rings will be commutative.

Let $k \geq 0$ be an integer. We define $B_k(X) \in \mathbb{Q}[X]$ as follows. $B_0(X) = 1$, $B_1(X) = X$. If $k > 1$, $B_k(X) = X(X - 1)\cdots (X - k + 1)/k!$. $B_k(n)$ is an integer for every integer $n$.

Let $f \in \mathbb{Q}[X]$. We define $\Delta f \in \mathbb{Q}[X]$ by $\Delta f(X) = f(X + 1) - f(X)$.

Let $f:\mathbb{Z} \rightarrow \mathbb{Z}$ be a map. We define a map $\Delta f:\mathbb{Z} \rightarrow \mathbb{Z}$ by $\Delta f(n) = f(n+1) - f(n)$.

Lemma 1 Let $f \in \mathbb{Q}[X]$. The following conditions are equivalent.

(1) $f = \Sigma_k a_k B_k(X)$ for some $a_k \in \mathbb{Z}$.

(2) $f(n) \in \mathbb{Z}$ for all $n \in \mathbb{Z}$.

(3) There exists $n_0 \in \mathbb{Z}$ such that $f(n) \in \mathbb{Z}$ for all $n \geq n_0$, $n \in \mathbb{Z}$.

(4) $\Delta f$ satisfies (1) and there exists $n_0 \in \mathbb{Z}$ such that $f(n_0) \in \mathbb{Z}$.

Proof:Left to the readers.

Let $n_0 \in \mathbb{Z}$. Let $\mathbb{Z}_{n\geq n_0}$ be the set {$n \in \mathbb{Z}; n \geq n_0$}. Let $f:\mathbb{Z}_{n\geq n_0} \rightarrow \mathbb{Z}$ be a map. If there exists $g(X) \in \mathbb{Q}[X]$ and a integer $m_0$ such that $f(n) = g(n)$ for all $n \geq m_0$, we say $f$ is polynomial-like. $g(X) \in \mathbb{Q}[X]$ is uniquely determined by $f$. The degree of $g(X)$ is called the degree of $f$. If $f$ is a map $f:\mathbb{Z} \rightarrow \mathbb{Z}$, polynomial-like $f$ is defined similarly.

Lemma 2 The following conditions are equivalent.

(1) $f$ is polynomial-like.

(2) $\Delta f$ is polynomial-like.

(3) There exists $r \geq 0$ and an integer $n_0$ such that $\Delta^r f(n) = 0$ for all $n \geq n_0$.

Proof:Left to the readers.

Let $S$ be a graded ring ($S_d=0$ for $d<0$). We assume that $S_0$ is an Artinian ring and $S$ is generated over $S_0$ by finite elements $x_1,\dots, x_r$ of $S_1$. $S$ is isomorphic to $S_0[X_1,\dots, X_r]/I$, where $S_0[X_1,\dots, X_r]$ is a polynomial ring and $I$ is a homogeneous ideal. Hence $S$ is Noetherian. Let $M$ be a finitely generated graded $S$-module. Each $M_n$ is a finitely generated $S_0$-module. Hence $M_n$ is Artinian $S_0$-module. We denote the length of $M_n$ over $S_0$ by $\chi(M, n)$.

Theorem $\chi(M, n)$ is polynomial-like and its degree $\leq r - 1$.

Proof: We can assume that $S = S_0[X_1,\dots, X_r]$. We use induction on $r$. If $r = 0$, $M$ is finitely generated over $S_0$. Since $M$ is of finite length, $\chi(M, n) = 0$ for large $n$. Suppose $r > 0$.

Let $\lambda:M \rightarrow M$ be the endomorphism induced by the multiplication by $X_r$. Let $K$ = Ker($\lambda$), $L$ = Coker($\lambda$). $K$ and $L$ are graded modules over $S$. We get the following exact sequence.

$0 \rightarrow K_n \rightarrow M_n \rightarrow M_{n+1} \rightarrow L_{n+1} \rightarrow 0$.

Hence $\Delta \chi(M, n) = \chi(M, n+1) - \chi(M, n) = \chi(L, n+1) - \chi(K, n)$.

Since $K_n$ is annihilated by $X_n$, $K$ can be regarded as a graded module over $S_0[X_1,\dots,x_{r-1}]$. By the induction assumption, $\chi(K, n)$ is polynomial-like of degree $\leq r-2$. Similarly $\chi(L, n+1)$ is polynomial-like of degree $\leq r-2$. Hence, by Lemma 2, $\chi(M, n)$ is polynomial-like of degree $\leq r-1$. QED

Let $A$ be a ring. Let $I$ be an ideal of $A$. Let gr($A$) = $\bigoplus_{n \geq 0} I^n/I^{n+1}$. gr($A$) is a graded ring.

Let $M$ be an $A$-module. Let $(M_n)$, $n \geq 0$ be a sequence of submodules of $M$ such that $M = M_0$ and $M_n \supset M_{n+1}$ for all $n \geq 0$. We say $(M_n)$ is an $I$-filtration if $IM_n \subset M_{n+1}$. We say $(M_n)$ is a stable $I$-filtration if $IM_n = M_{n+1}$ for all large $n$.

If $(M_n)$ is an $I$-filtration, let gr($M$) = $\bigoplus_{n \geq 0} M_n/M_{n+1}$. gr($M$) is a graded gr($A$)-module.

Lemma 3 Let $A$ be a Noetherian ring. Let $I$ be an ideal of $A$. Let $M$ be a finitely generated A-module. Let $I$ be an ideal of $A$. Let $(M_n)$ be a stable $I$-filtration. Then gr($M$) is a finitely generated gr($A$)-module.

Proof: There exists $n_0$ such that $M_{n+n_0} = I^nM_{n_0}$ for all $n \geq 0$. Hence gr($M$) is generated by $\bigoplus_{n \leq n_0} M_n/M_{n+1}$. Hence gr($M$) is a finitely generated gr($A$)-module. QED

Proposition Let $A$ be a Noetherian local ring with the maximal ideal $\mathfrak{m}$. Let $\mathfrak{q}$ be a $\mathfrak{m}$-primary ideal of $A$. Let $x_1,\dots,x_r$ be generators of $\mathfrak{q}$. Let $M$ be a finitely generated $A$-module. Let $(M_n)$ be a stable $\mathfrak{q}$-filtration. Then the following assertions hold.

(1) $M/M_n$ is of finite length for all $n \geq 0$.

(2) leng($M/M_n$) is polynomial-like of degree $\leq r$.

(3) The degree and leading coefficient of leng($M/M_n$) are the same as leng($M/\mathfrak{q}^nM$).

Proof: (1) and (2): Let gr($A$) = $\bigoplus \mathfrak{q}^n/\mathfrak{q}^{n+1}$. Let gr($M$) = $\bigoplus M_n/M_{n+1}$. gr($M$) is a graded gr($A$)-module. Let $\tilde x_i$ be the image of $x_i$ in $\mathfrak{q}/\mathfrak{q}^2$. Then gr($A$) = $(A/\mathfrak{q})[\tilde x_1,\dots,\tilde x_r]$. By Lemma 3, gr($M$) is a finitely generated gr($A$)-module. Hence by Theorem, $\chi(M, n)$ is polynomial-like and its degree $\leq r - 1$. Let $l_n $= leng($M/M_n$) = leng($M_0/M_1$) $+\cdots +$ leng($M_{n-1}/M_n$). Since $l_{n+1} - l_n = \chi(M, n)$, $l_n$ is polynomial-like of degree $\leq r$.

(3): There exists $n_0$ such that $\mathfrak{q}M_n = M_{n+1}$ for $n \geq n_0$.

$\mathfrak{q}^{n+n_0}M \subset M_{n+n_0} = \mathfrak{q}^nM_{n_0} \subset \mathfrak{q}^nM \subset M_n$.

Hence leng($M/\mathfrak{q}^{n+n_0}M$) $\geq$ leng($M/M_{n+n_0}$) $\geq$ leng($M/\mathfrak{q}^nM$) $\geq$ leng($M/M_n$).

Hence the assertion follows. QED

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Thank you for this nice exposition. Once I've had a chance to digest it a little more I plan to accept this answer. –  Mike B Aug 9 '12 at 0:16
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It is given in the proof of Prop 11.4 in A.M. that $l_{n+1} - l_n = f(n)$ where $f$ is a polynomial, for sufficiently large $n$.

An example might help:

Suppose $l_{n+1} - l_n = an^2 + bn + c$ for all $n \geq m$ (I chose a quadratic polynomial for convenience of computation).

Let $l'_n = an^2 + bn + c$, for all $n \geq m$. Then $l'_{n+1} - l'_n = a(2n+1) + b = 2an + (a+b)$.

Let $l''_n = 2an + (a+b)$, for all $n \geq m$. Then $l''_{n+1} - l''_n = 2a$. Thus for all $k \in \mathbb{N}$, $l''_{m+k} = l''_m + 2ak$.

It follows that for all $k \in \mathbb{N}$,

\begin{equation} l'_{m+k} - l'_m = \sum_{0 \leq i \leq k-1} l''_{m + i} = \sum_{0 \leq i \leq k-1} l''_m + 2ai = kl''_m + 2a\frac{k(k-1)}{2}. \end{equation}

Thus, $l'_{m+k} = l'_m + kl''_m + ak(k-1)$, which is clearly a polynomial in $k$.

Then,

\begin{equation} l_{m+k} - l_m = \sum_{0 \leq i \leq k-1} l'_{m + i} = \sum_{0 \leq i \leq k-1} l'_m + il''_m + ai(i-1) = l'_mk + l''_m\frac{k(k-1)}{2} + a (\sum_{0 \leq i \leq k-1} i^2 - i) = l'_mk + l''_m\frac{k(k-1)}{2} + a\frac{(k-1)k(2k - 1)}{6} - a\frac{k(k-1)}{2}. \end{equation}

It follows that for all $k \in \mathbb{N}$,

\begin{equation} l_{m + k} = l_m + l'_mk + l''_m\frac{k(k-1)}{2} + a\frac{(k-1)k(2k - 1)}{6} - a\frac{k(k-1)}{2} \end{equation}

which is again a polynomial in $k$.

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Let me know if I have made an error in any of the computations of the sums. –  Rankeya Aug 1 '12 at 3:02
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