Zariski topology - normal for finite $\Bbb{F}$, not hausdorff for $\Bbb{F}=\Bbb{R}$ I need to prove that for $n\in\Bbb{N}$, and zariski topology on $\Bbb{F}^n$:

*

*Normal space if $\Bbb{F}$ is finite

*Not Hausdorff space for $\Bbb{F}=\Bbb{R}$
What I've tried:

*

*prove that any singleton is close, so because $\Bbb{F}$ is finite I can say that its the discrete topology.

*prove that any open non-empty group is dense in $\Bbb{F}^n$.  And then I can't find 2 disjoint open groups.

Do you think it's correct?
Will appreciate any help
 A: If $\Bbb F$ is finite, then so is $\Bbb F^n$. All singletons are closed in the Zariski topology, as $\{(a_1,\ldots, a_n)\} = V(f)$ where $f(x)=\prod_{i=1}^n(x_i - a_i) \in \Bbb F[x_1,\ldots,x_n]$. As all singleton sets are closed, all finite sets are closed too, and hence all subsets of $\Bbb F^n$ are closed (and so open as well, as the complements are closed too). And so $\Bbb F^n$ is (finite and) discrete and so normal.
Now, let $a=(a_1, \ldots, a_n)$ and $b=(b_1, \ldots, b_n)$ be any distinct points of $\Bbb F^n$, where $\Bbb F$ is an infinite field. Suppose we had two disjoint open subsets $U$ and $V$ in the Zariski topology such that $a \in U$ and $b \in V$. Then $U^\complement$ is Zariski closed, so an intersection of basic $V(f_i)$ closed sets with $f_i \in \Bbb F[x_1, \ldots, x_n]$, $i \in I$. and contains $b$, and $V^\complement$ has a similar form and contains $a$, and also $U^\complement \cup V^\complement = (U \cap V)^\complement = \Bbb F^n$.
It follows we have some $f \in  \Bbb F[x_1, \ldots, x_n]$ such that $f(b)\neq 0$ and some $g \in \Bbb F[x_1, \ldots, x_n]$ such that $g(a) \neq 0$ (so in particular $f,g$ are non-zero polynomials!) and such that $V(f) \cup V(g)= \Bbb F^n$.
But then $V(f\cdot g)=\Bbb F^n$ for a non-zero polynomial $f \cdot g$ too, and this is a contradiction for an infinite field.
