I started studying algebraic geometry few days ago using Hartshorne's book. Today, I did the exercise 3.2 of chapter I in which there is an example of a homeomorphism that is not an isomorphism of varieties. The morphism is defined by :
$$\begin{array}{ll}
\phi : & \mathbb{A}^1 & \longrightarrow &Z(y^2-x^3) \subset\mathbb{A}^2 \\
& t &\mapsto &(t^2,t^3)
\end{array}$$
I managed to do the exercise but when I looked at solutions online they all did something a bit different than me that I don't really understand. To do the exercise thay all just say that if $\phi^{-1}$ is a morphism then there exists a polynomial $f\in k[X,Y]$ such that $t=f(t^2,t^3)$ for all $t$. I understand the absurity of such a polynomial but I do not understand why it should exists. This conclusion seems stronger that what I managed to show..
This is how I did the exercise :
I had no problem to prove that $\phi$ is a bijective bicontinuous morphism. Now, I need to show that its inverse $\phi^{-1} :(a,b)\in \mathbb{A}^2 \mapsto \frac{b}{a} ~~if~ ~a\neq 0 ~~or~~ 0~~ if ~a=0$ is not a morphism.
I suppose that it is. Applying the definition of morphism of Hartshorne with $f=id:\mathbb{A}^1 \rightarrow k$ tells me that $\phi^{-1}$ regarded as a function to $k=\mathbb{A}^1$ is regular. In particular, there exists an open subset $U$ of $Z(y^2-x^3)$ containing $(0,0)$ and two polynomials $g,h\in k[X,Y]$ with $h(0,0)\neq0$ such that $\phi^{-1}=g/h$ on $U$.
Thus, for every $t$ in the open set $\phi^{-1}(U)$ I have : $t=\phi^{-1}(\phi(t))=\frac{g(t^2,t^3)}{h(t^2,t^3)}$. So $g(0,0)=0$ which is possible only if $g$ has no constant term.
Thanks to the above equality and the fact that open sets are infinite, I know that in $k[T]$ we have $Th(T^2,T^3)=g(T^2,T^3)$. As I said, g has no constant term so $g(T^3,T^2)$ has a valuation at least equals to 2 which implies that $h(0,0)=0$. Absurdity.
This reasoning might be longer and more tidious but I don't see how to prove the existence of the polynomial claimed by online solutions. Moreover, I think that the point $(0,0)$ is crucial in this story since except at this point, $\phi^{-1}$ seems to be regular. I don't see where it is used in other solutions.
Thank you for you help.
Best Answer
If $\phi^{-1}$ were a morphism of varieties, it would induce a map on coordinate algebras which is inverse to the map induced by $\phi$ on coordinate algebras. Namely, the map $a:k[t^2,t^3]\to k[t]$ induced by $\phi$ would have an inverse $b:k[t]\to k[t^2,t^3]$. In particular, $a$ must be surjective, and so we must have some element $e$ of $k[t^2,t^3]$ which maps to $t$. As elements of $k[t^2,t^3]$ are polynomials in $t^2$ and $t^3$, we can write $e=f(t^2,t^3)$ and that's what these other solutions are doing.
They are absolutely showing something stronger than what you're showing: these other solutions demonstrate that there can be no isomorphism between the two varieties, while you only demonstrate that this particular map has no inverse.