1

Show that the map $F: \Bbb A^1 \to \Bbb V(y-x^2) \subset \Bbb A^2, t \longmapsto (t,t^2)$ is an isomorphism from the line to a parabola.

I've learned about morphisms between affine varieties and the pullback and this is an example problem where I think I should use the fact that if $V$ and $W$ are affine varieties in $\Bbb A^n$, then $V \cong W \iff \Bbb C[V] \cong \Bbb C[W]$, here $\Bbb C [V]$ and $\Bbb C[W]$ are the coordinate rings of $V$ and $W$.

So if I define the pullback $F^\#: \Bbb C[x,y]/(y-x^2) \to \Bbb C[t]$ such that $x \longmapsto t, y \longmapsto t^2$ I have an surjective algebra homomorphism with kernel $\ker F^\# =\{p \in \Bbb C[x,y]/(y-x^2) \mid F^\#(p)=p(t,t^2)=0 \}$.

I think I should use the isomorphism theorem for algebras here to get something like $$\Bbb C[x,y]/(y-x^2) \Big / \ker F^\# \cong \Bbb C[t]$$ but I don't know what this kernel is and similarly I don't know how to handle this "double quotient". Any hints what to do here?

Jonas
  • 29
  • 5
  • Can you write down the inverse of $F$? Where should a point $(x,y)$ on the parabola be mapped to? You can certainly show that the induced map on coordinate rings is an isomorphism, but it basically amounts to the same thing. – Viktor Vaughn Apr 12 '22 at 18:24
  • I think I can just use the projection to the first coordinate, but I would like to figure out what this quotient $\Bbb C[x,y]/(y-x^2) \Big / \ker F^# $ turns out to be. – Jonas Apr 12 '22 at 18:26
  • But you know that $F^#$ is supposed to be an isomorphism, so you should show that the kernel is $0$. Once you do that, then your isomorphism just follows from the fact that $\frac{\mathbb{C}[x,y]}{(y-x^2)} \cong \mathbb{C}[t]$. – Viktor Vaughn Apr 12 '22 at 18:29
  • That makes sense. I'm not sure why the kernel would be $0$ though. This would be the case if there doesn't exists any $p$ such that $p(t,t^2)=0$? – Jonas Apr 12 '22 at 18:39

1 Answers1

0

First, here's what I'd call the "right" way to do this problem. Define \begin{align*} G: \mathbb{V}(y-x^2) &\to \mathbb{A}^1\\ (x,y) &\mapsto x \, . \end{align*} $G \newcommand{\Ft}{\tilde{F}}$ is a polynomial map, hence is a morphism, and one can easily check that $F$ and $G$ are mutually inverse, using the fact that a point on the parabola is of the form $(x, x^2)$. Thus $F$ (and $G$) is an isomorphism.

If you'd prefer to work with coordinate algebras, we can similarly define the map \begin{align*} G^\#: \mathbb{C}[t] &\to \mathbb{C}[x,y]/(y-x^2)\\ t &\mapsto x \end{align*} and show that $F^\#$ and $G^\#$ are mutually inverse.

To pursue your desired approach and show that $\ker(F^\#) = 0$: consider the composition \begin{align*} \Ft : \mathbb{C}[x,y] \overset{\pi}{\to} \mathbb{C}[x,y]/(y-x^2) \overset{F^\#}{\to} \mathbb{C}[t] \end{align*} with the quotient map $\pi$. So $\Ft$ is just the evaluation map $x \mapsto t, y \mapsto t^2$. We now apply the following lemma. (See here for a proof.)

Lemma. Let $R$ be a unital commutative ring and $R[x]$ be the one-variable polynomial ring over $R$. For $\alpha \in R$, let \begin{align*} \varphi = \text{eval}_\alpha: R[x] &\to R\\ x &\mapsto \alpha \end{align*} be the evaluation homomorphism. Then $\ker(\varphi) = (x-\alpha)$. Moreover, the induced map $\overline{\varphi}: R[x]/(x-\alpha) \to R$ is an isomorphism.

Considering $\mathbb{C}[x,y]$ as $(\mathbb{C}[x])[y]$, we get that \begin{align*} \frac{\mathbb{C}[x,y]}{(y-x^2)} &\overset{\sim}{\to} \mathbb{C}[x]\\ y &\mapsto x^2 \end{align*} is an isomorphism, which, after changing $x$ to $t$, is exactly the map $F^\#$.

One can also show that the kernel is $(y-x^2)$ more directly, as done in this answer.

Viktor Vaughn
  • 19,278