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Suppose one has a local ring $(A,\mathfrak{m})$ and a finite length $A$-module $M$ with $\operatorname{supp}(M) = \{\mathfrak{m}\}$. Does $M$ have a composition series consisting only of $A/\mathfrak{m}$'s?

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Does it follow from Eisenbud, Commutative Algebra, Theorem 2.13? –  user5262 Nov 27 '11 at 17:18

1 Answer 1

up vote 4 down vote accepted

Yes, this is true:

a) Since $M$ has finite length, it has by definition a composition series.
This means that there exists a non-refinable filtration $M_0=M \supsetneq M_1 \supsetneq...\supsetneq M_n=0$ , in other words a filtration where each $M_i/M_{i+1}$ is a simple $A$-module.

b) Now, a simple $A$-module $S$ is isomorphic to $A/{\mathfrak m}$ with ${\mathfrak m}\subset A$ maximal.
Indeed choose a non-zero element $s\in S$ (the zero module is not regarded as simple!) and consider the morphism $f:A\to S: a\mapsto as$. Its image is non-zero [because $f(1)=s\neq0$] hence it is $S$ since, by simplicity, $S$ is the only non-zero submodule of $S$ . So $f$ is surjective and $S\simeq A/Ker(f)$. But $A/Ker(f)$ can only be simple if $Ker(f)$ is a maximal ideal, so that $Ker(f)$ is a maximal ideal $\mathfrak m$ and, as announced, $S\simeq A/\mathfrak m$

c) Since your ring is local, $\mathfrak m$ is its only maximal ideal, and your question has an affirmative answer: $M$ has a composition series with quotients isomorphic to $A/\mathfrak m$. As Pierre-Yves very judiciously remarks, you don't need the hypothesis that the support of $M$ is $\lbrace\mathfrak m \rbrace$.

Acknowledgment
My former proof was much more complicated and needed $A$ noetherian. Any merit that the above proof might have should go to QiL and Pierre-Yves Gaillard (see their comments below).

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Thank you very much! –  user5262 Nov 27 '11 at 18:13
    
Dear Georges: Which definition of "finite length module" are you using? –  Pierre-Yves Gaillard Nov 27 '11 at 18:54
    
Dear @Pierre-Yves, in order not to clutter the comments, I'll add a section to my answer giving two equivalent definitions of "finite length". –  Georges Elencwajg Nov 27 '11 at 19:59
    
Dear Georges, isn't it simply the fact that the only simple module $M$ over a local ring $A$ is its residue field ? Such module is automatically generated by a single element because of the simplicity. By Nakayam (the noetherian hypothesis is not needed because $M$ is finitely generated), $\mathfrak m M \neq M$, hence $\mathfrak m M=0$ and $M$ is a vector space over $k$, generated by one element, so $M\simeq k$. –  user18119 Nov 27 '11 at 20:16
    
Dear Georges: Thanks for your edit, but it seems to me, from what you wrote, that the Jordan-Hölder factors of any finite length module, over any (commutative) ring, consists of modules of the form $A/\mathfrak m$, with $\mathfrak m$ maximal. So I don't understand why user5262 makes this support assumption, and why you make this noetherianity assumption... Sorry to bother you, but I'm really confused... [Dear @QiL: I hadn't read your comment while writing mine. It seems to me there are some similarities between our reactions.] –  Pierre-Yves Gaillard Nov 27 '11 at 20:26

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