Consider the affine space $\mathbb{A}^n$ where $\mathbb{A}^n$ is the set of $n$-tuples $(a_1,\dots,a_n)$ with $a_i\in k$, and $k$ is an algebraically closed field.
Next, consider the map $f:\mathbb{A}^1\to \mathbb{A}^3$ given by $$t\mapsto(t,t^2,t^3).$$ I want to show that this map is continuous with respect to Zarisky topology (closed sets are given as a vanishing set of some polynomials).
In other words, I want to show that the preimage of every closed set is closed.
- We can show that the image of $f$ is a closed set i.e. $f(\mathbb{A})=Z(I)$ for some $I\in k[x,y,z]$.
- Next, if $X$ is any closed set of $\mathbb{A}^3$ with $X\cap Z(I)=\emptyset$, then $f^{-1}(X\cap Z(I))=\emptyset$ is closed. So, I am only interested in closed sets that intersect $f(\mathbb{A})$ nontrivially.
- Consider a closed set $X$ of $\mathbb{A}^3$ with $X\cap f(A)\neq\emptyset$. Since $f(\mathbb{A})=Z(I)$, then $X\cap f(\mathbb{A})$ is closed. So, $$f^{-1}(X\cap f(\mathbb{A}))=f^{-1}(X)\cap \mathbb{A}$$ as $f$ is a bijection with it's image $f(\mathbb{A})$. Now, we want to show that $f^{-1}(X)$ is closed.
Can I think about $f^{-1}(X)$ as $Z(f^*(J))$ where $X=Z(J)$ as $X$ is closed subset and $f^*$ is the dual map to $f:\mathbb{A}^1\to \mathbb{A}^3$ on the level on rings i.e. $$f^*:k[x,y,z]\to k[t]$$ where $x=t$, $y=t^2$, and $z=t^3$? In that case, I can see that $f^{-1}(X)=Z(f^*(J))$ is closed.
For the identity $f^{-1}(X) = Z(f^*(J))$, Observe that both $p \in f^{-1}(X)$ and $p \in Z(f^{*}(J))$ are equivalent to $f(p) \supset J$.
More concretely, let $p=(t-a)$ and $J = (g_1(x,y,z),g_2(x,y,z)..,g_n(x,y,z))$.
$p \in f^{-1}(X)$
⇔ $f((t-a)) \in X$
⇔ $(x-a,y-a^2,z-a^3) \supset (g_k(x,y,z))$ for every $k$.
⇔ $g_k(a,a^2,a^3) = 0$ for every $k$.
$p \in Z(f^{*}(J))$
⇔ $(t-a) \supset (f^{*}(g_k(x,y,z)))$ for every $k$.
⇔ $(t-a) \supset (g_k(t,t^2,t^3))$ for every $k$.
⇔ $g_k(a,a^2,a^3) = 0$ for every $k$.