8

I have some trouble filling in the details of this proof from Atiyah-Macdonald. In this result, the authors assume what follows:

1) $A = \oplus_{n=0}^\infty A_n$ is a Noetherian graded ring, and therefore $A_0$ is Noetherian and $A$ is a finitely generated $A_0$-algebra, generated by $x_1, \ldots, x_s$ of degrees $k_1, \ldots, k_s$ respectively.

2) $M = \oplus_{n=0}^\infty M_n$ is a finitely generated $A$-module, so in particular $M$ is Noetherian and $M_n$ are finitely generated $A_0$-modules.

3) $\lambda$ is an additive function (so $\lambda(M) = \lambda(N) + \lambda(M/N)$ for $N \subset M$) from finitely generated $A_0$-modules to the positive integers.

4) Lastly $P(M,t) = \sum_{n=0}^\infty \lambda(M_n)t^n$.

enter image description here

If $K_n, L_n$ are as in the proof, then I understand that $K = \oplus_n K_n$ and $L = \oplus_n L_n$ is the kernel and cokernel, respectively, of the homomorphism induced by multiplication by $x_s$. If $x_s$ has degree $k_s$, then that would mean $L = \oplus_n \dfrac{M_n}{x_sM_{n-k_s}}$ where $M_{n-k_s} = \{0\}$ for $n < k_s$ and so the first $k_s$ factors in the sum are simply $M_i$. In particular then $$P(L,t) = \sum_{n=0}^{k_s-1} \lambda(M_n)t^n + \sum_{n= k_s}^\infty \lambda(L_n)t^n.$$

After the authors set up the equations $$ \lambda(L_{n+k_s}) - \lambda(K_n) = \lambda(M_{n+k_s}) - \lambda(M_n) = 0$$ they multiply each such equation by $t^{n+k_s}$ and then sum over all $n$, which gives $$\sum_{n=0}^\infty \lambda(L_{n+k_s})t^{n+k_s} - t^{k_s}P(K,t) = \sum_{n=0}^\infty \lambda(M_{n+k_s})t^{n+k_s} - t^{k_s}P(M,t).$$ But now adding to both sides $\sum_{n=0}^{k_s-1} \lambda(M_n)t^n$ would make me obtain $$P(L,t) - t^{k_s}P(K,t) = (1-t^{k_s})P(M,t)$$ which is not correct since the authors obtain a difference of a polynomial $g(t)$. I can't seem to find my mistake. What exactly is $g(t)$ in that equation?

Also, I have another short question, which is related, about proposition 11.3 in the same book. $d(M)$ denotes here the pole at $t=1$ of $P(M,t)$.

enter image description here

The equation (2) becomes $(1-t^{k_s})P(M,t) = P(L,t) + g(t)$. Then I see that the pole of $P(L,t) + g(t)$ is of degree one less than the degree of the pole of $P(M,t)$, but how does one conclude $d(L) = d(M) - 1$ in this case?

Bartosz Malman
  • 1,371
  • There must be something simple that I am missing here. Given sequences $0 \rightarrow K_n \rightarrow M_n \xrightarrow{x_s} M_{n+k_s} \rightarrow L_{n+k_s} \rightarrow 0$ I would have that $L_{n+k_s} = M_{n+k_s}/x_sM_n$ by exactness. But then what do I take for $L_n$ where $n = 0, 1, \ldots, k_s-1$? I understood that $L$ is the cokernel of the map on $M$ given by multiplication by $x_s$. – Bartosz Malman Jul 20 '13 at 19:29

2 Answers2

5

After a careful checking of the proof I've to admit that you are right: $g(t)=0$. If you want more support, please check here or here.

Now the answer to your second question becomes obvious.

  • Thank you very much for these references. I was very confused, as for example Lang seems to give the same proof as AM and there $g(t)$ also appears. I'll go through the proofs in the links you posted in the morning and then accept your answer. – Bartosz Malman Jul 20 '13 at 22:43
  • @Malman Lang has copied the proof of AM word by word. –  Jul 21 '13 at 06:51
  • It depends on the choice of the $L_i$ for $i < k_s$: for $L_i = 0$ the result is that of AM and if $L_i = M_i$ for $i < k_s$ it is $g(t) = 0$. – psl2Z Jul 19 '24 at 09:47
0

If it is $L_n = 0 $ for $n = 0,1, ..., k_s -1$, then $P(L,t) = \sum_{n = k_s}^{\infty} \lambda(L_n)t^{n}$ and NOT $P(L,t) = \sum_{n = 0}^{k_s-1} \lambda(M_n)t^{n} + \sum_{n = k_s}^{\infty} \lambda(L_n)t^{n}$. However with $g(t) = \sum_{n = 0}^{k_s-1} \lambda(M_n)t^{n}$ it is $P(L,t) + g(t) = \sum_{n = 0}^{k_s-1} \lambda(M_n)t^{n} + \sum_{n = k_s}^{\infty} \lambda(L_n)t^{n}$ and thus $$g(t) + P(L,t) - t^{k_s}P(K,t) = (1-t^{k_s})P(M,t)$$ which is what AM said. However, if it is $L_n = M_n$ for $n = 0,1,...,k_s-1$ your version is correct.

psl2Z
  • 5,772