Question on maximal spectrum of the polynomial ring

Question

Let $k$ be a field and $A:=k[x_1,x_2,\cdots,x_n]$ be the polynomial ring with $n$ variables. Now let's consider the maximal spectrum of ring $A$, i.e. the subset of $Spec(A)$ consisting of all maximal ideals, we write it as $Spm(A)$. The topology on $Spm(A)$ is the induced topology given by $Spec(A)$. For the simplest case $n=1$, if we consider the polynomial ring with one variable $k[x]$, we note that $Spm(\mathbb{C}[x])$ is not a Hausdorff space. But I'm wonder whether $Spm(\mathbb{F}_p[x])$ is a Hausdorff space or not? (I think $Spm(\mathbb{F}_p[x])$ is not a Hausdorff space.) Here $\mathbb{F}_p$ is a finite field with $p$ elements. And whether $Spm(\mathbb{F}_p[x_1,x_2,\cdots,x_n])$ is a Hausdorff space or not? More generally, I want to find out when $Spm(A)$ can be a Hausdorff space?

Answer 1

Let $A$ be any integral domain in which the intersection of all the maximal ideals is zero, for example a polynomial ring over a field. Let $X = \textrm{Spm}(A)$. The sets $D(a) = \{ \mathfrak m \in X : a \not\in \mathfrak m \}$ for $a \in A$ form a basis of open sets for the topology on $X$.

By hypothesis, $D(a)$ is empty if and only if $a = 0$, and $D(ab) = D(a) \cap D(b)$. You can conclude that the intersection of two nonempty open sets in $X$ is nonempty, which means that $X$ is never Hausdorff, unless $A$ is a local ring.