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Consider the semicubical parabola $ C \subset \mathbb{C} $ given by,

$ C:= \{ (x,y) \in \mathbb{A}^2_{\mathbb{C}} \ | \ y^2-x^3=0 \} $. Show that the map

$ \varphi: \mathbb{A}^1_{\mathbb{C}} \to \mathbb{C} \\ t \mapsto(t^2,t^3) $

is a homeomorphism with respect to the Zariski topology.

What I have done so far:

Need to prove that for any closed $ U, \varphi^{-1}(U) = \{ x \in \mathbb{A}^2_{\mathbb{C}} \ | \ (x^2,x^3) \in U \} $ is also closed. Since $ x^2,x^3 \in U, $ there is some $ f_1, f_2 $ such that $ f_1(x^2)=0 $ and $ f_2(x^3)=0 $.

How can I prove that there is an $ f $ such that $ f(x)=0? $

pseudogeek
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1 Answers1

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In order to prove that $\varphi:\mathbb{A}^1_{\mathbb{C}}\rightarrow C$ given by $\varphi(t)=(t^2,t^3)$ is a homeomorphism, we need to prove the following three statements:

  1. $\varphi$ is bijective
  2. $\varphi$ maps closed sets to closed sets
  3. Preimages of closed sets under $\varphi$ are closed.

Proof of statement 1: Given $(x,y)$ such that $x^3=y^2$ there exists an inverse $t$ such that $\varphi(t)=(x,y)$; if $t=\sqrt x$ doesn't work then try $t=-\sqrt x$.

Proof of statement 3: Given a set $S$ of polynomials which determines a closed set in $C$, we can take each polynomial $P(x,y)\in S$ and substitute $x=t^2,y=t^3$ to make a polynomial $f(t)$. The set $S'$ of polynomials $f(t)$ determines a closed set in $\mathbb{C}$.

Proof of statement 2: We need to map a set $S'$ of polynomials $f(t)$ to a set $S$ of polynomials $P'(x,y)$ so that $\varphi$ maps the zeroes of $S'$ to the zeroes of $S$.

Any integer $n>2$ satisfies $n=2a+3b$ for some non-negative integers $a,b$. So any polynomial $f(t)=a_0+a_1t+a_2t^2+...$ can be rewritten in the form $f(t)=a_0+a_1y/x+P(x,y)$ by rewriting $t^n$ as $x^ay^b$. This can be further rewritten as $f(t)=P'(x,y)/x$. $P'(x,y)=0$ whenever $f(t)=0$ so we are done.

Angela Pretorius
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