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.

  • $\begingroup$ Take $A'$ a field, $M$ a vector space of dimension m, $A$ a vector space of dimension n given some ring structure (e.g. product ring). Then we can easily find m,n (I think if both are >1?) where there are non-simple tensors. $\endgroup$ – David Corwin May 14 '13 at 3:21
  • $\begingroup$ @David: there won't be any interesting surjective maps from a field to another ring. I think your argument is fine Chris, even if this question would be better suited to math.stackexchange.com. $\endgroup$ – Jack Huizenga May 14 '13 at 3:27

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. 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. :)

  • $\begingroup$ Thank you. I will keep in mind the distinction between math.stackexchange.com and mathoverflow.net for future questions. $\endgroup$ – Chris May 14 '13 at 4:01

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