Hi, NonCommAlg.

Thank you for your answer. I know about the converse and it's easy to prove.

I've been thinking about your hint the whole day today and I'm afraid I'm still very far from seeing it. I noticed instead that free modules are very different from vector spaces. I was shocked to notice that

doesn't have any linearly independent subset when viewed as a submodule of the (free) module

This is awful.

The first thing I would like to ask is if it's obvious for some reason that taking the free module as

has to work given that we know that the theorem is true. In other words, why should I have thought of the free module when I was trying to prove it?

The second thing is, is it enough to prove this for the elements of a basis of

? I don't even see this.

OK, so I know that

What I want to know is that

You told me to assume that

has a basis. To me it just means, "Write

in the basis." OK. Let

be a basis of

Let

where

If I knew that can "push" the summands into

would I know that I can "push" the sum too? If I had

such that

for

would I be able to find

such that

If the ring were an integral domain, I could use the product

but it's not...