Let $C$ be a subring of integral domains $A,B$ and let $C',A',B'$ denote their field of fractions respectively.
Can we always embed $A\otimes_CB$ in $A'\otimes_{C'}B'$ by $a\otimes b\mapsto a/1\otimes b/1$ as a ring?
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The answer is negative.
Let $C=K[X^2,X^3]$, and $A=B=K[X]$. Then $A'\otimes_{C'}B'=K(X)\otimes_{K(X)}K(X)=K(X)$ is a field, while $A\otimes_CB$ is not an integral domain. In order to show the last claim notice that $$[X\otimes(X+1)-1\otimes X(X+1)][X\otimes(X+1)+1\otimes X(X+1)]=0$$ and use this answer where I've proved that the factors are both non-zero. (Well, I've proved that the first one is non-zero, but the same argument applies to the second, too.)
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An simpler example of zero-divisors in $A\otimes_C B=k[x,y]/(x^2-y^2,x^3-y^3)$ is $(x-y)(x+y)=x^2-y^2=0$ (here $x=X\otimes 1$ and $y=1\otimes X$). – Eric Wofsey Aug 24 '15 at 21:59
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@EricWofsey I'm not sure I can understand why you think that $A\otimes_C B=k[x,y]/(x^2-y^2,x^3-y^3)$. – user26857 Sep 14 '15 at 16:14
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Well, $A\otimes_C B$ is the pushout of the diagram $k[x]\leftarrow k[x^2,x^3]\to k[x]$. That is, it is generated by $x$ and $y$ with relations that identify the subring $k[x^2,x^3]$ with $k[y^2,y^3]$. – Eric Wofsey Sep 14 '15 at 19:25
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I'm not thinking about generating $k[x]$ as a $k[x^2,x^3]$-algebra; I'm just generating everything as a $k$-algebra. For any diagram of $k$-algebras $A\stackrel{f}\leftarrow C\stackrel{g}\to B$, you can describe $A\otimes_C B$ as the quotient of $A\otimes_k B$ by the relations that say that that $f(c)=g(c)$ for each $c\in C$. In this case, $x^2$ and $x^3$ generate $C$ as a $k$-algebra, so it suffices to mod out $x^2-y^2$ and $x^3-y^3$. – Eric Wofsey Sep 14 '15 at 21:54