In a broader exercise about the fibered product of schemes, I'm given a ring (commutative and unitary) $R$ and the following ring homomorphism:
$\begin{align*} R[x]&\to R[y]\\ x&\mapsto y^2\\ \end{align*}$
which gives $R[y]$ an $R[x]$-algebra structure. My hope is to prove that, as an $R[x]$-algebra, $R[y]$ is isomorphic to $\frac{R[x][y]}{(x-y^2)}$. Is this true? How can I prove this? It appears obvious to me, but maybe I'm missing something.
Edit: as suggested in the comments, the explicit isomorphism by the two should be as follows:
$ \begin{align*} \frac{R[x,u]}{(x-u^2)} &\rightarrow R[y]\\ x&\mapsto y^2 \\ u&\mapsto y \end{align*} $
this is an $R[x]$-algebra homomorphism (with the obvious $R[x]$-algebra structure on the first ring) and has the following inverse:
$ \begin{align*} R[y] &\rightarrow \frac{R[x,u]}{(x-u^2)}\\ y&\mapsto u \end{align*} $
(I checked that both are $R[x]$-algebra homomorphisms and are in fact one the inverse of the other)
Correct me if I'm wrong or if there was a non-explicit way!
You can show that the kernel of the surjective homomorphism $R[x,u]\to R[y], x\mapsto y^2, u\mapsto y$ is $(x-u^2)$ so that it induces the desired isomorphism.
By thinking of $R[x,u]$ as $R[x][u]$, $u^2-x$ is a monic polynomial and any element of it has a representation $(u^2-x)P(u)+au+b, a,b\in R[x]$. If its image in $R[y]$ is zero, we must have $a=b=0$.