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.