"Assume that $X = V((z_1 - 1)z_2 - 1) \hookrightarrow \mathbb{C^2}$ and $f(z_1, z_2) = z_1^2(f : X \rightarrow \mathbb{C})$.
Show that f is closed map(in the Zariski topology)."

The book that I read says "It is sufficient to prove $f(X) = \mathbb{C}$".
But I don't know why!
Can you teach me please.

Thank you in advance!


The map $X \to \mathbb{A}_{\mathbb{C}}^{1} - \{0\}$ defined as $(z_{1}, z_{2}) \to z_{1} - 1$ is an isomorphism, and $\mathbb{A}^{1}_{\mathbb{C}}- \{0\} \subset \mathbb{A}_{\mathbb{C}}^{1}$ is an open subset. Since the only closed subets of $\mathbb{A}^{1}_{\mathbb{C}}$ are finite set of points or $\mathbb{A}_{\mathbb{C}}^{1}$ itself, so is $\mathbb{A}_{\mathbb{C}}^{1}- \{0\}$ and $X$, i.e. closed subsets of $X$ are also finite subsets or itself. Since an image of a point, which is also a point, is clearly a closed subet, we only need to show that $f(X)$ is closed, which is true if $f(X) = \mathbb{A}^{1}_{\mathbb{C}}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.