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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.

Sha Vuklia
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Samarkand
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2 Answers2

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Okay, let's take a prime ideal $I$ of $\mathbb{Z}[x]$ and consider its pre-image under $\iota: \mathbb{Z} \rightarrow \mathbb{Z}[x]$. The result is a prime ideal of $\mathbb{Z}$, so either $(0)$ or $(p)$.

Let's consider the case $\iota^{-1}(I)=p\mathbb{Z}$, i.e. $I\cap \mathbb{Z} = (p)$. This happens if and only if $I$ contains $p\mathbb{Z}[x]$. Now the ideals of $\mathbb{Z}[x]$ containing the latter are in one-to-one correspondence with the ideals of $\mathbb{Z}[x]/p\mathbb{Z}[x] \cong \mathbb{F_p}[x]$ through the quotient map $\mathbb{Z}[x] \rightarrow \mathbb{F}_p[x]$. This shows that the fibre of $\pi: \text{Spec} \mathbb{Z}[x] \rightarrow \text{Spec}{\mathbb{Z}}$ over $(p)$ is in bijection with $\text{Spec}\mathbb{F}_p[x]$.

Now consider the case where $\iota^{-1}(I) = (0)$. In other words $I \cap \mathbb{Z} = (0)$. Let $S$ be the multiplicative set of non-zero integers in $\mathbb{Z} \subset \mathbb{Z}[x]$. Then $I \cap S = \emptyset$, and such ideals $I$ of $\mathbb{Z}[x]$ are in one-to-one correspondence with the ideals of the ring $\mathbb{Z}[x]S^{-1} $ through the localization map $\mathbb{Z}[x] \rightarrow \mathbb{Z}[x]S^{-1}\cong \mathbb{Q}[x]$. This shows the fibre over $(0)$ is in bijection with $\text{Spec}\mathbb{Q}[x]$.

Zavosh
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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.