How to compute tensor product quickly? Suppose $A,B,C$ are rings and there two ring homomorphisms $\varphi: A\to B$ and $\phi: A \to C$, then we have $B\otimes_A C$. My question is: if $A,B,C$ are some special rings and have some special relationships, then how can we compute the tensor product?
For example: $R$ is a ring, consider the ring homorphisms $R[x]\to R: x\to 0$ and $R[x]\to R[y]:x\to y^2$, then what is $R\otimes_{R[x]}R[y]$?
Also could you show me some other examples?
 A: When calculating tensor products quickly, there are usually some properties we can exploit. For instance, we have $A\otimes_A B\cong B$. And for a more complicated example, if $A\subseteq B$, then $B\otimes_A C$ is in some sense the most natural extension (formally, "universal" might be a better word) of the $A$-algebra $C$ into a $B$-algebra.
For instance, $\Bbb C\otimes_{\Bbb R}\Bbb R[x]\cong \Bbb C[x]$, as the most natural extension of the $\Bbb R$-algebra $\Bbb R[x]$ to a $\Bbb C$-algebra (as well as the most natural extension of the $\Bbb R$-algebra $\Bbb C$ into an $\Bbb R[x]$-algebra) is $\Bbb C[x]$.
In your case, we follow this last property, where we may consider $R[x]\subseteq R[y]$ by way of $x = y^2$. If we take the $R[y^2]$-algebra $R$ defined by $y^2\mapsto 0$, and we want to make that $R$ into an $R[y]$-algebra, then we still need to keep $y^2\mapsto 0$. But apart from that, there are no restrictions on $y$. So $y$ becomes its own element in the extended algebra, with as few relations as possible, and we get the result $R[y]/(y^2)$. In short,
$$
R\otimes_{R[x]}R[y]\cong R[y]/(y^2)
$$
Alternatively, we can do actual calculations using $R\cong R[x]/(x)$, which reveals
$$
R\otimes_{R[x]}R[y]\cong R[x]/(x)\otimes_{R[x]}R[y]\\
\cong R[y^2]/(y^2)\otimes_{R[y^2]}R[y]\\
\cong R[y^2]\otimes_{R[y^2]}R[y]/(y^2)
$$
where we can now use the first property mentioned in the first paragraph to simplify. The last isomorphism comes from the fact that $y^2$ exists in all three of $R[y^2]$, $_{R[y^2]}$, and $R[y]$, so by the bilinearity of the tensor product it doesn't matter in which of the two "tensor factor" rings we consider $y^2$ to be $0$.
The gist of it is, there is no single "trick" to calculating the tensor product. Like so many other mathematical constructions, there are a few ways to do it quickly if you're lucky with your exact case, and learning these tricks takes time and practice.
A: If $\psi$ and $\phi$ are surjective then 
$B\otimes_A C\cong A/\ker(\psi)\otimes_AA/\ker(\phi)\cong $
$\cong A\otimes_A A/(\ker(\psi)+\ker(\phi))\cong A/(\ker(\psi)+\ker(\phi))$
If you want I can be more precise.
In your example the map $\phi$ is not surjective but you can say that 
$R\otimes_{R[x]} R[y]\cong R[x]/\ker(\psi)\otimes _{R[x]} R[y]\cong$
$\cong R[x]\otimes _{R[x]}R[y]/ \phi(\ker(\psi))\cong R[y]/(y^2) $
