Fibres of the map $Spec\mathbb{Z}[x] \rightarrow Spec\mathbb{Z}$ I am trying to understand what is a spectrum of the ring $\mathbb{Z}[x]$. I have read Spectrum of $\mathbb{Z}[x]$ but because of my very restricted knowledge of schemes I do not understand the definition of fibre of a morphism of schemes in full generality.
I understand the idea that we have to consider the map $\pi: \text{Spec }\mathbb{Z}[x] \rightarrow \text{Spec }\mathbb{Z}$ induced by standard inclusion $\mathbb{Z} \rightarrow \mathbb{Z}[x]$. Then we just have to describe what is $\pi^{-1}((p))$ and $\pi^{-1}((0))$. I already know the answers $\pi^{-1}((p))$ turns out to be in bijection with $\text{Spec }\mathbb{F}_p[x]$ and $\pi^{-1}((0))$ is just $\text{Spec }\mathbb{Q}[x]$. But I cant understand why this is the case. Could someone explain me why is that true? In particular, I cant understand how I can describe a preimage if I dont even know anything not only about the map $\pi$ but also about $\text{Spec }\mathbb{Z}[x]$?! Any help is appreciated.

For variety...
The $\operatorname{Spec}$ functor maps colimits in $\mathbf{CRing}$ to limits in $\mathbf{Sch}$.
Fibers are pullbacks, such as:
$$ \require{AMScd}
\begin{CD}
(\operatorname{Spec} \mathbb{Z}[x])_p @>>> \operatorname{Spec} \mathbb{Z}[x]
\\ @VVV @VVV
\\ \operatorname{Spec} \mathbb{F}_p @>>> \operatorname{Spec} \mathbb{Z}
\end{CD}
$$
Since the schemes involved are affine, the fiber is the spectrum of the pushout
$$ \require{AMScd}
\begin{CD}
\mathbb{F}_p[x] \cong \mathbb{F}_p \otimes_{\mathbb{Z}} \mathbb{Z}[x] @<<< \mathbb{Z}[x]
\\ @AAA @AAA
\\ \mathbb{F}_p @<<< \mathbb{Z}
\end{CD}
$$
and pushouts of rings are computed by tensor products.
