Question on prime ideals of ${\mathbb Z}[x]$

Question

I was thinking this question myself:

Consider the topological space $(\text{Spec}(\mathbb{Z}[x]),T )$ where open sets $D_I$ in $T$ are given as indexed by ideals $I$ in $\mathbb{Z}[x]: \; D_I =\{p\in \text{Spec}(\mathbb{Z}[x])\}$ such that $I$ doesn't belong to $p$.

Is it true that for $a,b \in \text{Spec}({\mathbb Z}[x])$ there exist $U,V \in T$ such that $a\in U,b\in V$ but $a \notin V, b\notin U$?

Answer 1

No, it is false. Note $X$ the underlying topological space of your affine scheme $\textrm{Spec}\left( \mathbf{Z}[X] \right)$, and let $\eta$ be the point corresponding to the zero ideal. Then as every ideal contains the zero ideal you have $\overline{\{\eta\}} = X$, which means that any open subset of $X$ will have a non-empty intersection with $\{\eta\}$, that is, will contain $\eta$.

Answer 2

This condition is basically asking if $Spec(R)$ is $T_1$. My favorite equivalent definition of this is "all points are closed."

But it is easy to see that if $P, Q$ are distinct primes such that $P\subset Q$, then $Q$ is in the closure of $\{P\}$.

So, necessarily, a ring must have Krull dimension 0 to be $T_1$, but the ring you are talking about is 2 dimensional.