Question about tensor product of algebras Take two ring homomorphism, $a:R\to A$ and $ b:R\to B$. In general, there is not a field $F$ such that $A,B\hookrightarrow F$, and such that $R\to A \hookrightarrow F$ and $R\to B \hookrightarrow F$ are the same (as ring homomorphisms). However I don't understand why such a field doesn't always exist: if I consider the tensor product of algebras $A\otimes_RB$, and take its quotient by a maximal ideal, isn't the result a field containing a copy of $A$ and $B$, with the inclusions equalizing $a$ and $b$? Thanks for any clarify
 A: The problem is that the inclusions $A \to A \otimes_R B$ and $B \to A \otimes_R B$ need not be injective. Hence $A$ and $B$ are not included in $A \otimes_R B$, let alone a quotient of the former by a maximal ideal.
Under the identification $A \simeq A \otimes_R R$, the inclusion $A \to A \otimes_R B$ can be viewed as $1_A \otimes i$ where $i \colon R \to B$ is the $R$-lineal map $1 \mapsto 1$. In general $A \otimes_R -$ does not preserve monomorphisms. A sufficient condition would be for $A$ to be $R$-flat (in which case $A \otimes_R -$ preserves all monos. It is also worth recalling that projectives and in particular free modules are flat). Even then, we may lose injectivity when dividing by the aforementioned maximal ideal.
Your question can be rephrased as "when are there inclusions $A, B \to D$ to a domain $D$". This is because a field is a domain, and having the result for domains implies the result for field by composing with the injection $D \to \mathsf{Frac}(D)$. Note that as a necessary condition both $A$ and $B$ should be domains.
In the question you cite in the comments, the tensor product that is claimed to be non-zero is taken over a field $L$. Hence the discussion in the comments of this answer applies: if $L$ is a field and $A$ and $B$ are non-trivial $L$-vector spaces so is $A \otimes_L B$.
