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.


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 $\text{Spec}\mathbb{Q}[x]$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.