3
$\begingroup$

Let $A$ be a commutative ring with identity and $M$ a free $A$-module.

Now suppose that $T$ generates $M$. When does $T$ contain a basis of $M$?

It partly seems like a "dual" to the following result:

Let $A$ be a commutative ring with identity and $M$ a free $A$-module. If $N$ is a free submodule of $M$, then the basis of $N$ can be extended to a basis of $M$ if and only if $M/N$, the quotient module, is free.

So I'm guessing that we can obtain a similar result, that is, a necessary and sufficient condition under which the question is true.

$\endgroup$
  • $\begingroup$ Can you be a bit more precise about what type of conditions are you looking for? Since $2,3$ is a generating set of $\mathbb{Z}$, but neither are, I am not sure what you are looking for and without some idea about that, it is difficult to answer this in any non-trivial way. $\endgroup$ – Mohan Oct 1 '17 at 17:49
  • $\begingroup$ Yes. The only scenario I know for this to be true is when we work in a field. I suppose that the condition can be weaker. The proposition I quoted might provide some hint: If $T$ contains a basis $S$ of $M$, then do we have $M/(T-S)$(or some other quotient) free? Can we add some extra conditions so that the inverse is also true? $\endgroup$ – William Sun Oct 1 '17 at 19:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.