0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.