# Basis for free module of infinite rank

In Corollary 19 of Dummit and Foote in section 10.4 it says:

Let $$R$$ be a commutative ring and let $$M \cong R^s$$ and $$N \cong R^t$$ be free $$R$$-modules with bases $$m_1,m_2,\ldots,m_s$$ and $$n_1,n_2,\ldots n_t$$. Then, $$M \bigotimes_R N$$ is a free $$R$$-module of rank $$st$$ with basis $$m_i \otimes n_j$$, $$1 \leq i \leq s$$, and $$1 \leq j \leq t$$. That is:

$$\begin{equation} R^s \otimes_R R^t \cong R^{st} \end{equation}$$

They give a one line proof based on other corollaries they have proven. I can easily see that the isomorphism is very direct from other material in the section. However, I cannot seem to figure out how it is so clear that $$m_i \otimes n_j$$ is a basis for this tensor product. I would have expected that you needed results like the one I asked here: Spanning lists of the "right length" are a basis for arbitrary $R$-modules. But, they do not prove this anywhere up to this point. How would they expect you to know this if it were not for this result that I mentioned in the post above? Am I missing something extremely obvious?

Moreover, I am wondering whether a similar result holds for modules of infinite rank. Namely, if you have modules $$M$$ and $$N$$ of infinite rank with basis $$\{a_i\}_{i \in I}$$ and $$\{b_j\}_{j \in J}$$,respectively, is a basis the set of simple tensors $$a_i \otimes b_j$$? In an exercise they ask you to prove the the rank of two free modules of arbitrary rank over a commutative ring is free, but the exercise gives no mention of a basis for such a free module.

Any help to either of these questions would be greatly appreciated.

Thank you

• Maybe this helps: math.stackexchange.com/questions/1178004/… – Ruben Jul 3 at 19:30
• @Ruben Okay, thank you for the reference. It is for vector spaces, but it looks like the arguments should work for modules. I just find it weird that they claim that the basis from above is in fact the basis for $M \otimes_R N$. It seems very believable, but I have not found a relatively simple proof of this fact. – Mike Jul 3 at 19:43