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A small sanity check related to Question 2 from here: proof of the Krull-Akizuki theorem (Matsumura)

Let $C$ be an $A$-module, with $A$ commutative ring and suppose that there exists a chain of submodules $C=C_0 \supset C_1 \supset \cdots \supset C_m=0$ such that $C_i/C_{i+1} \cong A/m_i$ where $m_i$ is a maximal ideal of $A$. It seems to me that this implies that this chain is actually a compoosition series and so $C$ has finite length, since every quotient is a field and hence a simple $A$-module. Is this assertion correct?

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Manos
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  • You are correct. The $A$-module $A/m$ is simple when $m$ is maximal by the correspondence theorem for rings. The nontrivial submodules of $A/m$ would be the nontrivial ideals between $m$ and $A$, of which there are none. – Jared May 21 '13 at 20:15
  • That they are fields is not directly the reason they are simple as $A$-modules (the multiplication has no direct connection to the structure as an $A$-module). The fact that any submodule would correspond to a submodule of $A$ between $A$ and $m_i$ (which is maximal) is what makes them simple. – Tobias Kildetoft May 21 '13 at 20:15
  • It's important to remember that these composition series are written entirely in terms of a modules, so talking about the quotient of modules being a field is essentially out of scope (and not true in a natural sense, anyway). – rschwieb May 21 '13 at 20:30

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The only thing that's wrong is the "$C_i/C_{i+1}$ is a field" part. That quotient is a simple $A$ module, yes, but it is not necessarily a field. It is module isomorphic to $A/M$, but not ring isomorphic. (Added: of course, the quotient can be given a field structure by idenfitication with $A/M$, but it does not match the multiplication of the rng quotient $M/N$. Since the identification does not match, I discard it as a natural thing.)

For example, you can take a commutative Artinian ring whose ideals are linearly ordered to see why the quotients aren't fields. If $M/N$ was a field for proper ideals $M$ and $N$, that would mean that the identity of $M/N$ lifts to be an identity of $M$, but unfortunately $M$ does not contain any idempotent elements besides 0.

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  • But these particular quotients are in fact fields. – Tobias Kildetoft May 21 '13 at 20:21
  • @TobiasKildetoft What particular quotients do you mean? It's true that the quotients can be identified with a ring structure via the isomorphism $R/M$, but the rng quotient of ideals need not be a field. – rschwieb May 21 '13 at 20:23
  • True, the fact that they have a structure as a field which is also a $A$-algebra is rather unrelated to this (since that structure does not come from the modules being filtered). – Tobias Kildetoft May 21 '13 at 20:25
  • @TobiasKildetoft I'm not sure I follow, but I think the fact that the rng quotient of ideals does not always have identity is convincing enough. – rschwieb May 21 '13 at 20:32