Let $k$ be a subfield of $\mathbb{C}$, and let $A,B$ be two $k$-algebras equipped with injective $k$-algebra homomorphisms $f : A\rightarrow E$ and $g : B\rightarrow E$, such that $f(A)\cap g(B) = k$.

From this can we deduce that the induced homomorphism $$f\otimes g : A\otimes_k B\rightarrow E.$$

is also injective?

In particular, I'm interested in the case where $A$ is a subring of $k((x))$ containing $k$, and $B = \mathbb{C}$. Surely in this case $f\otimes g$ must be injective, right?

  • $\begingroup$ Note that $\mathrm{id}_E$ is injective, but the canonical map $E\otimes_kE\rightarrow E$ rarely is. $\endgroup$ – Oscar Cunningham Oct 26 '16 at 12:52

A very special case: take $k = \mathbb Q$, and let $A = \mathbb Q(x)$. Then we can embed $A$ into $\mathbb C$, by mapping $x$ to any transcendental complex number.

Now the natural morphism $A \otimes_{\mathbb Q} \mathbb C \to \mathbb C$ is not injective.

So the answer is definitely no. You need to have much finer control of the situation than what you've given in your post.

  • $\begingroup$ Good point. I've added the condition that $f(A)\cap g(B) = k$. Surely this is enough right? $\endgroup$ – oxeimon Oct 26 '16 at 13:19

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