Is $\mathbb{C}[x,y]$ isomorphic to $\mathbb{C}[x]\otimes\_{\mathbb{R}}\mathbb{C}[y]$ as rings?


Generally, in a category $\mathcal{C}$ with fibered product, morphisms $f: X\rightarrow Z$, $g:Y\rightarrow Z$, and $h: Z\rightarrow W$. when can say $X\times\_Z Y\simeq X\times\_W Y$?



Is $\mathbb{C}[x,y]$ isomorphic to $\mathbb{C}[x]\otimes_{\mathbb{R}}\mathbb{C}[y]$ as rings?
Well ${\bf C}[x,y]$ is isomorphic with $({\bf C}[x]) \otimes\_{\bf C} ({\bf C}[y])$. But the tensor products **over ${\bf R}$** cannot coincide, because already ${\bf C} \otimes\_{\bf R} {\bf C}$ has zero divisors. For example, $1 \otimes 1 \neq \pm \phantom. i \otimes i$ but $(1 \otimes 1)^2 = (i \otimes i)^2$, so $1 \otimes 1 \pm \phantom. i \otimes i$ are complementary zero divisors, which can then be found in $({\bf C}[x]) \otimes\_{\bf R} ({\bf C}[y])$ too.


For the general question, I don't know, but you probably want to ask not just about an isomorphism but about one *induced by the morphisms $f,g,h$* — else there can be "coincidental" isomorphisms. For example, in general a complex vector space $V$ need not be isomorphic with $V \otimes\_{\bf R} {\bf C}$, and the obvious map is not an isomorphism unless $V = \lbrace0\rbrace$, but if $V$ is infinite-dimensional there are unnatural isomorphisms. 


