Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
The title doesn't fit to the question. –  Martin Brandenburg May 9 '13 at 17:57
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.