Hy guys. I'm doing some independent analysis which makes use of the tensor product of modules (over commutative rings with unit 1, and ring homomorphisms map $1 \mapsto 1$). Let $\pi: A' \to A$ be a surjective homomorphism of rings which identifies $A$ as an $A'$-algebra (and hence an $A'$-module). Let $M$ be any $A'$-module. The following analysis suggests that the tensor product $M \otimes\_{A'} A$ consists entirely of simple tensors, but I am not sure that I am entirely convinced, and any feedback would be appreciated.


Let $m \otimes a$ be an arbitrary element of $M \otimes\_{A'} A$. By the surjectivity of $\pi$, we can find an element $a' \in A'$ such that $\pi(a') = a$, so


$m \otimes a = m \otimes \pi(a') = a'(m \otimes 1) = a'm \otimes 1$.


Thus, if we have an arbitrary element $\sum\_{i = 1}^n m\_i \otimes a\_i \in M \otimes\_{A'} A$, we can choose pre-images $a\_1',...,a\_n'$ of the $a\_i$ such that $ \sum\_{i = 1}^n m\_i \otimes a\_i = (\sum\_{i=1}^n a\_i'm\_i) \otimes 1 $, so every tensor is simple.



An example of a tensor product consisting of only simple tensors?
Your argument is right. In your situation, $A \cong A^{\prime} / I$ for some ideal $I$ of $A^{\prime}$, and the tensor product $M\otimes\_{A^{\prime}} A$ is isomorphic to $M / \left(IM\right)$, where $IM$ is the submodule of $M$ spanned by all products of the form $im$ with $i\in I$ and $m\in M$. This explains why it consists of only simple tensors.


The isomorphism $M\otimes\_{A^{\prime}} A \cong M / \left(IM\right)$ is actually pretty useful. Three years ago I asked on this site why people [state the Chinese Remainder Theorem only for rings, and not for modules](https://mathoverflow.net/questions/18959/chinese-remainder-theorem-for-rings-why-not-for-modules). Two people answered on the same day that the version for modules follows immediately from that for rings by the above-mentioned isomorphism.


This is indeed better suited to math.stackexchange.com, as was my question. :)


