Isomorphism of cuspidal cubic and some geometric meaning Consider the integral domain $R:=\mathbb{C}[x,y]/(y^2-x^3)$. Over its field of fraction let $t:=y/x\in R_{frac}$. I want to show $\mathbb{C}[t] \cong R[t]$. How can I show this? and what is geometric meaning over this? I think they are isomorphic because $\mathbb{C}[t]$ is a polynomial over the complex line which maps to a function defined on $V(y^2-x^3)\subset \mathbb{C}^2$. Is this right interpretation? Thank you.
 A: We can start building the isomorphism $R[t] \to \mathbb{C}[t]$ as follows:
$$\mathbb{C}[x, y] \to \mathbb{C}[t],\quad x\mapsto t^2, y \mapsto t^3 $$
This sends $y^2-x^3$ to $t^6-t^6=0$, and so we have a map
$R \to \mathbb{C}[t]$ and this map is injective becuase the kernel is exactly $(y^2-x^3)$. (here's why: writing elements of $R$ as having a representative $p(x) + q(x)y$ (since we can replace $y^2$ by $x^3$) we see that $p(x) + q(x)y$ maps to $p(t^2) + q(t^2)t^3$ and if this is $0$ then by looking at the parity (even/oddness) of powers of $t$ we see that $p(x)=0$ and $q(x)=0$.
Then adjoining $y/x$ we get $R(y/x) \to \mathbb{C}[t]$ makes the map surjective, but we have to check we didn't lose injectivity. Writing any element of $R(y/x)$ as represented by $p(x) + q(x)x (y/x)$ the same argument as before works.
The geometric meaning of this is blowing up the cuspidal cubic at the origin (with ideal $(x,y)$) gives you something which on  one chart is isomorphic to $\mathbb{A}^1$ (and on the other chart where you adjoin $x/y$ is also isom to $\mathbb{A}^1$, together they glue to a $\mathbb{P}^1$.
