This is the theorem the title refers to. In his Basic Algebra I, Jacobson proves it by means of a Lemma:

Lemma Let $D$ be a PID and $K$ be a submodule of $D^{(n)}$ (the free module of rank $n$). Then

  1. $K$ is finitely generated;

  2. $K$ is free of rank $\le n$.

Question What is the relevance of conclusion 2. to the subsequent proof? I guess it is not needed.

Indeed, Jacobson's proof goes as follows. Take a finitely generated module $M$ over the PID $D$ and a generating homomorphism $ \eta\colon D^{(n)} \to M$. Then the kernel $K$ of $\eta$ is finitely generated and we have a relations matrix $A$, whose rows are a set of generators for $K$. We then apply the machinery of Smith normal form to $A$.

It seems to me that we made no use of 2. This point guarantees that we can take $A$ of full rank, but that is something we don't need. Am I wrong?

Thank you.

  • 1
    $\begingroup$ The title doesn't fit to the question. $\endgroup$ – Martin Brandenburg May 9 '13 at 17:57

Before the proof of the lemma, Jacobson says that

.. A first result we shall need is that $K$ is finitely generated. This will follow from the following stronger result.

and later

.. The method we are going to apply will work just as well if we have a finite set of generators, and as a practical matter it is sometimes useful not to have to resort to a base.

And indeed, in Jacobson's presentation 2.) is not needed to prove the structure theorem, we don't have to bother with finding a base for $K$. Of course, it is still a nice to know and proving it does not require that much work.


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