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This question comes from reading Washington's proof of Iwasawa's theorem, and wanting to learn the commutative algebra version of the classification of finitely-generated $\Lambda$-modules. I went to the reference in Serre, and am now hung up on a point in his classification.


Let $A$ be a regular local ring of dimension $2$, and let $M$ be a finitely-generated, torsion-free $A$-module. $M$ is reflexive if $M = M^{**}$, or equivalently, if $M = \cap_P M_P$, where the intersection ranges over all prime ideals of height $1$. The detail that I am hung up on is the following:

$M$ reflexive implies that it is free.


Serre is kind enough to give a proof of this, however, he uses confusing notation which is contained in the lecture notes from a 1955 lecture in Tokyo. For completeness, his proof is as follows (I will try to stay close to the original French, odd word choices are due to the French, and might be insightful): One can plunge $M$ into a free module $N$ of the same rank. $M$ being reflexive signifies that that, in the primary decomposition of $M$ in $N$, the prime ideals that intervene are of height $1$. One then has that $\text{codh}(N/M) \geq 1$, then $\text{dh}(N/M) = \text{dim}(A) - \text{codh}(N/M) \leq 1$; since $N$ is free, one has $\text{dh}(M) = \text{dh}(N/M) - 1 \leq 0$, which signifies that $M$ is free.

My guess on the dh is that it denotes projective dimension (I am sure that it stands for some form of homological dimension, based on the name of the paper). That would enable concluding that $M$ is free if the projective dimension is $0$. The prime decomposition fact is proven in Bourbaki (Commutative Algebra, Chapter 7, Section 4, Subsection 2, Proposition 7 i)). The confusing thing about that interpretation is that it seems to imply that the co-projective dimension is easier to get than the projective dimension. Also, where does the equality between the dimension of $M$ and $N/M$ come from (idea: the projective dimension of $N/M$ is at most one, and this is a start of a projective resolution of $N/M$, so the completion indicates that $M$ has dimension one fewer than $N/M$. However, why does it follow that every projective resolution has the same length, or if that isn't true, why is this a start of a minimal projective resolution?).

Also, I will give a proof that a friend gave me. Use the embedding of $M$ into $N$, and look at a regular sequence for $N$. This gives rise to a regular sequence of $M$. But then, since $N$ is free, the regular sequence for $M$ is of length at least $2$, so the projective dimension of $M$ is at most $0$, which is again what we wanted.

However, that only uses $M$ is torsion-free, and not reflexive. If there is no error in the above proof, then one can easily show that the maximal ideal is free for a sample counter-example to that statement.


Finally, some remarks are in order for how I am using this. I am using this in the case that $A = \mathbb{Z}_p[[T]]$, so a proof specific to that ring wouldn't be too troubling (while it is true in general, I would only be generally disturbed without a complete proof). Also, I only really need that if $M$ is torsion-free, and $M/f$ is finite, then $M$ is trivial (applied to the case $f(T) = ((1+T)^{p^n} - 1)/T$). Finally, I am trying to get around having to understand a long matrix bash (on the order of 5 pages) in order to understand this theorem.

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Erick, did you check out this question: mathoverflow.net/questions/7490/… –  Hailong Dao Apr 29 '10 at 5:49
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"Odd word choices are due to the French"---I would dispute that. I think that odd word choices are due to the fact that it has been badly translated ;-) Just because "plonge" looks a bit like "plunge" it doesn't mean it should be translated as "plunge" in this context! Similarly for "intervient" and "signifie". –  Kevin Buzzard Apr 29 '10 at 11:22
    
I saw that. I am currently trying to understand the proofs that I have seen (in particular, why doesn't the second one work?). –  Erick Knight Apr 29 '10 at 14:09
    
The second one does not work because a reg sequence on N would not become a reg sequence on M (a regular element would, but not sequence in general). –  Hailong Dao Apr 29 '10 at 14:15
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Also, by the way, codh is depth and dh is proj dim in current terminology. –  Hailong Dao Apr 29 '10 at 17:08

4 Answers 4

up vote 1 down vote accepted

Hello,

I guess that $\mathrm{codh}$ actually means $\mathrm{depth}$, that is the length of a maximal regular sequence on a module. Then $\mathrm{codh}(N/M)\geq 1$ is a consequence of the fact that the maximal ideal of $A$ does not annihilate the module due to reflexivity. The next inequality is Auslander-Buchsbaum. Finally $\mathrm{dh}(M)=\mathrm{dh}(N/M)-1$ is a standard fact from homological algebra: if for some module $Q$ $\mathrm{dh}(Q)\leq d$ and $0\rightarrow S\rightarrow P_{d-1}\rightarrow\ldots\rightarrow P_0\rightarrow Q\rightarrow 0$ is exact with projective $P_k$, then $S$ is projective too. Apply this to $0\rightarrow M\rightarrow N\rightarrow N/M\rightarrow 0$.

H

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Is that fact supposed to be obvious/immediate? I just see a lot of potential pathologies going on with projective resolutions. If not, then is there a nice reference that you have in mind? –  Erick Knight Apr 29 '10 at 13:54
    
You mean the last fact about the projectivity of S? See for example Lemma 4.1.6 in Weibel's Introduction to homological algebra. –  Hagen Apr 29 '10 at 14:25
    
That seems to work. Another way of seeing that dh(M) <= dh(N/M)-1 is to show that Ext^1(M,-) = 0, which follows from Mayer-Vietoris applied to 0 -> M -> N -> M/N -> 0. That generalizes for higher dimensions too. Thanks for parsing this proof. –  Erick Knight Apr 30 '10 at 1:02

If you accept the fact that a $2$-dimensional (local) ring has global dimension $2$, the following is a (somewhat) alternative proof. Choose a free f.g. presentation $F_1 \to F_0 \to M^\ast \to 0$ and then dualise giving an exact sequence $0\to M^{\ast\ast}\to F_0^\ast\to F_1^\ast\to K \to 0$, where $K$ is defined as the cokernel. As the $F^\ast_i$ are free and $\mathrm{pd} K\le 2$ we get that $M^{\ast\ast}$ is projective (and then free if the ring is local).

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The fact about the projectivity of $S$ is also known as the "long Schanuel's Lemma".

In the case you are interested in, the "short Schanuel's Lemma", as explained here

http://en.wikipedia.org/wiki/Schanuel's_lemma

suffices. You have two short exact sequences $0 \to M \to N \to N/M \to 0$ and $0 \to P_1 \to P_0 \to N/M \to 0$, where $P_0$ and $P_1$ are projective modules because the projective dimension of $N/M$ is $\leq 1$. Therefore $N \oplus P_1$ is isomorphic to $M \oplus P_0$ by the Lemma. So $M$ is projective, being a direct sum of the projective module $N \oplus P_1$.

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hello. since M is reflexive an exrsize of herzog and bruns says that depth(M) is larger than 2 and therefore equals 2. therefore M is cm and hence M is free because over regular local rings every cm module is free

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