Homeomorphism that is not an isomorphism of varieties. Hartshorne exercise I 3.2

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.

• $\mathbb{A}^1_k$ is nonsingular but the other variety is singular. Commented Jul 21, 2022 at 18:10
• I do not know the notion of 'sigularity'. It has not been defined yet in the book. From what I just read, it looks like the analytic 'smoothness'. I imagine that there is a property like "morphism preserves the nonsigularity " that kills the question. Since the exercise is given before that, I assume that it can be solved without using theses notions. That's why my question is : Do the polynomial claimed by the solutions I found online exists ? And if so, how to prove it ? Commented Jul 21, 2022 at 19:00

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.
• Okay thank you a lot ! If I understood well, we use the fact that the covariant functor $A$, as any functor, send an isomorphism into an isomorphism. And in this proof we do not need the theorem saying that this functor is fully faithful. In fact, this theorem tells me that the converse is also true : An affine variety is isomorphic to $\mathbb{A^n}$ if and only if its coordinate ring is the polynomial ring in $n$ variables. Commented Jul 21, 2022 at 22:55
• This brings another question to me. In my proof, the fact that $(0,0)$ is "weird" (in some sense I do not know yet) is important. Here, the crucial fact seems to be that $k[t^2,t^3]$ is not isomorphic to a polynomial ring. How theses two facts are related ? Commented Jul 21, 2022 at 22:56
• One way to think about that last comment is that the map $t\mapsto (t^2,t^3)$ is an isomorphism everywhere except $(0,0)$: it is a homeomorphism and induces an isomorphism of local rings everywhere except at $t=0$. This is the "weirdness" you're mentioning. Another exercise in Hartshorne (I.3.3(b)) shows that a morphism is an isomorphism iff it's a homeomorphism and it induces isomorphisms on all the local rings - so that's one concrete reason why the issue at $t=0$ implies that $k[t^2,t^3]$ is not isomorphic to a polynomial ring. (There are of course other answers, but that focuses on $t=0$.) Commented Jul 22, 2022 at 2:02