HW3

Example. Define ${\mathbb{A}}^2\setminus \{(0,0)\}\to {\mathbb{A}}^2,(x,y)\mapsto (x,y)$, then $\Gamma ({\mathbb{A}}^2)\stackrel{\not\simeq }{\to }\Gamma ({\mathbb{A}}^2\setminus \{(0,0)\})$.

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For a field $k$ with $\mathrm{char}k=p\ne 0$, define $F:{\mathbb{A}}_1^1\to {\mathbb{A}}_2^1,t\mapsto t^p$, which is bijective. Show that $F$ is a homeomorphism. See here.

\begin{proof} To show $F$ is a homeomorphism, it suffices to show both $F$ and $F^{-1}$ are continuous. Since $X_f=\{x:f(x)\ne 0\}$ forms a set of topological basis, it is enough to show $F(X_f)$ and $F^{-1}(X_f)$ are open for any $f\in k[x]$.

Note that $F(X_f)=\{x^p:f(x)\ne 0\}=X_g$ where $g(X)=f(X^{1/p})=k[X]$, so $F(X_f)$ is open. Similarly, $F^{-1}(X_f)=\{x^{1/p}:f(x)=0\}=X_h$ where $h(X)=f(X^p)\in k[X]$, so $F^{-1}(X_f)$ is open. Therefore, $F$ is a homeomorphism. \end{proof}

Definition

$X$ is irreducible iff one of the followings hold:

• if $X=Z_1\cup Z_2$ with $Z_i$ closed, then there exists $Z_i=X$;
• for any open sets $U_1,U_2$, there is $U_1\cap U_2\ne \varnothing$;
• any open set is dense.

\begin{proof} i)→ii) Assume that there exists open sets $U_1,U_2$ such that $U_1\cap U_2=\varnothing$, then we have $U_1^c\cup U_2^c=X$ and $U_i^c$ are closed, which is impossible.

ii)→iii) If there exists open $U$ such that $\overline{U}\ne X$, then ${\overline{U}}^c$ is an open set and $U\cap {\overline{U}}^c=\varnothing$, leading to a contradiction.

iii)→i) If there exist closed set $Z_1$ and $Z_2$ such that $X=Z_1\cup Z_2$ and WLOG suppose $Z_1\ne X$, then $Z_1^c\subseteq Z_1\cup Z_2$ and so $Z_1^c\subseteq Z_2$. Since $Z_1^c$ is open, there is $\overline{Z_1^c}=X$. It deduces that $\overline{Z_2}=Z_2=X$. \end{proof}