How to show that $\phi: \mathbb{A}^1(\mathbb{C}) \to V(Y^2 - X^3)$ is no isomorphism.

Question

Consider the polynomial function

$$\phi: \mathbb{A}^1(\mathbb{C})\to V(Y^2 - X^3): t \mapsto (t^2, t^3)$$

Show that $\phi$ is a bijective polynomial function, but no isomorphism (= $\phi^{-1}$ is no polynomial function).

I proved that $\phi$ is bijective and $\phi = (X^2, X^ 3)$, thus $\phi$ is a polynomial function. How would I show that $\phi^{-1}$ is no polynomial function?

Answer 1

We use that fact that isomorphic varieties have isomorphic coordinate rings.

Consider the coordinate ring of $\Bbb A^1, \Bbb C[t]$. Note that this is a UFD.

On the other hand, the coordinate ring of $V(Y^2 - X^3)$ is $\Bbb C[x, y]/(Y^2 - X^3)$. It is not a UFD since we have $Y^2 = X^3$.

Therefore $\Bbb A^1 \ncong V(Y^2 - X^3)$.