Show that $\{Z(p):p\in \mathscr P\}$ is a basis for the closed sets of some topology (Called the Zariski topology) on $\mathbb R^n$.

Question

Let $n\in \mathbb N$, and let $\mathscr P$ denote the collection of all polynomials in $n$ variables. For $p\in \mathscr P$, let $Z(p)=\{(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0\}$. Show that $\{Z(p):p\in \mathscr P\}$ is a basis for the closed sets of some topology (Called the Zariski topology) on $\mathbb R^n$.

My attempt:-

I used the result for Basis for the closed sets of topology on X.

(I) We need to prove that $\bigcap_{p\in \mathscr P}Z(p)=\emptyset.$ $Z(p(x_1,x_2,...,x_n)=1)=\{(x_1,x_2,...,x_n)|p(x_1,x_2,...,x_n)=0\}=\emptyset$. Which implies $\bigcap_{p\in \mathscr P}Z(p) \subset Z(p(x_1,x_2,...,x_n)=1)=\emptyset \implies \bigcap_{p\in \mathscr P}Z(p)=\emptyset.$

(II) if $Z(p_0),Z(p_1)\in\mathscr X$ and $(x_1,x_2,...,x_n) \notin Z(p_0)\cup Z(p_1)$, then there is a $Z(p_2)\in\mathscr X$ such that $(x_1,x_2,...,x_n) \notin Z(p_2)\supseteq Z(p_0)\cup Z(p_1)$.

$(x_1,x_2,...,x_n) \notin Z(p_0)\cup Z(p_1)\implies p_0(x_1,x_2,...,x_n)\neq0$ and $p_1(x_1,x_2,...,x_n)\neq 0.$ We can choose $p_2=p_0 p_1$ such that $(x_1,x_2,...,x_n)\notin Z(p_2)$. We know that $Z(p_0)\cup Z(p_1) \subset Z(p_2)$.

Have I done the proof correctly?

Answer 1

$Z(p) \cup Z(q) = Z(pq)$.
Let 1 be the polynomial of $n$ variables with constant value 1.
$Z(1)$ is empty.

Answer 2

The set of $\{Z(I): I \subseteq \mathcal{P}\}$, where $Z(I)=\bigcap_{p \in I} Z(p)$ is in fact the whole topology, not just a closed base for it, as I showed here. The basis fact immediately follows, by the definition of a closed base (you can write all closed sets as intersection of base elements).