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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

Mike
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    Maybe this helps: https://math.stackexchange.com/questions/1178004/basis-for-tensor-product-of-infinite-dimensional-vector-spaces – Ruben Jul 03 '19 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 03 '19 at 19:43

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