This question was asked in my assignment of Algebraic Geometry and I am struck on this.
Question: Let $Y$ be an affine variety and $f\in A(Y)$. Let $U(f) = Y/Z(f)$.
(a) Show that $U(f)$ is an affine variety.
$U(f)$ is an affine variety means that $U(f)$ is an irreducible and closed subset of $\mathbb{A}^n$. By irreducibility it means that $U(f)$ can't be expressed as the union of $Y_1 $ and $Y_2$ both closed in $Y$ where $Y_1$ and $Y_2$ are both proper subsets. I am not able to prove that $U(f)$ is closed. Conversly, I think that it is open because it is complement of Z(f) which is an algebraic set and in zarsiki topology: complement of algebraic sets are open.
Can you please help me with proving how $U(f)$ is closed and irreducible?
(b) Show that when $f $ varies, the $U(f) \subset Y$ form a basis for Zariski Topology.
I think as $U(f)$ is open , so as f varies union of $Y/Z(f) $ will form the basis if $\bigcup_{i \in I} (Y/Z(f_i)) =Y$.
Y is an affine variety means that Y is an irreducible closed subset of $\mathbb{A}^n$. But I am not able to move forward from this?