I am looking for a local integral domain $(D, m)$ with $Spec(D)=\{0,m\}\cup\{ P\_i\}\_i$ such that $P\_i's$ are incomparable (that is, $P\_i\not\subseteq P\_j$ and $P\_j\not\subseteq P\_i$ for $i\not= j$) and $\cap\_i P\_i=0$ and $\cup\_i P\_i\not=m$, where $Spec(D)$ is the set of all prime ideals of $D$ ($D$ is not Noetherian in general).



An example of a local integral domain with special spectrum
Let $k$ be a field. Put $R=k[[x,y]]$ and let $D\subset R$ be the subring $k+xR$. It consists of power series without any term $ay^n$ ($a\in k^\times$, $n>0$), or (equivalently) series $f(x,y)$ such that $f(0,y)\in k$. It is easy to see that $D$ is local with maximal ideal $m=\{f\in D\mid f(0,0)=0\}$.


I claim that $D$ is as required.


First, $D[x^{-1}]\to R[x^{-1}]$ is an isomorphism: it is clearly injective by localization, and surjective because for $g\in R$ and $m\in\mathbb{N}$, we have $x^{-m}g=x^{-m-1}(xg)$ and $xg\in D$. Thus, the nonzero primes $P\_i$ of $D$ not containing $x$ correspond bijectively to the primes of height one in $R$ distinct from $xR$, and there are infinitely many of these. On the other hand, we have $m^2\subset xD$ (check!), so the only prime of $D$ containing $x$ is $m$. Since $x$ is not in any of the $P\_i$'s, their union is not $m$.


Here is a geometric explanation: we can view $D$ as the fibre product ring $R\times\_{R/xR}k$. It follows that the natural diagram
$$\begin{array}{rcl}
\mathrm{Spec}(R/xR) & \longrightarrow & \mathrm{Spec}(R) \\
\downarrow\;&&\;\downarrow\\
\mathrm{Spec}(k) & \longrightarrow & \mathrm{Spec}(D) 
\end{array}$$
is a pushout of ringed spaces; this is Theorem 5.1 in [this paper](http://www.numdam.org/item/10.24033/bsmf.2455.pdf) by Ferrand. In other words, $\mathrm{Spec}(D)$ is obtained from $\mathrm{Spec}(R)$ by ``crushing'' the closed subset $\mathrm{Spec}(R/xR)\cong \mathrm{Spec}(k[[y]])$ to the point $\mathrm{Spec}(k)$.


