The prime ideals of $\mathbb{Z} [x] \otimes_{\mathbb{Z}} \mathbb{F}_p$

Question

I'm trying to do exercise 9 of section 15.5 in the book Abstract Algebra written by Dummit and Foote, whose aim is to establish a homeomorphic map between the elements in the fiber over $(p)$ of the Zariski continuous map from $\mathrm{Spec}(\mathbb{Z} [x])$ to $\mathrm{Spec}(\mathbb{Z})$ and the elements in $\mathrm{Spec}(\mathbb{Z} [x] \otimes_{\mathbb{Z}} \mathbb{F}_p)$. But I don't know what does the prime ideals of $\mathbb{Z} [x] \otimes_{\mathbb{Z}} \mathbb{F}_p$ looks like since it's the first time for me to study the primes in a tensor product.

Or generally speaking, given two rings $A$ and $B$, since a ring can be considered as a $\mathbb{Z}$-algebra, so the tensor product $A \otimes_{\mathbb{Z}} B$ of $\mathbb{Z}$-algebras $A$ and $B$ is also a ring. And what is the elements of $\mathrm{Spec}(A \otimes_{\mathbb{Z}} B)$ look like?

Answer 1

For variety, here's a geometric argument.

The contravariant $\operatorname{Spec}$ functor sends colimits of rings to limits of schemes. In the category of rings, $\otimes_R$ is the pushout over $R$; in particular, there is a pushout square

$$ \begin{matrix} \mathbb{Z} &\to& \mathbb{Z}[x] \\ \downarrow & & \downarrow \\ \mathbb{F}_p &\to& \mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{F}_p \end{matrix} $$

Since $\mathbb{Z}$ is the initial ring, this is equivalent to saying that $\otimes_{\mathbb{Z}}$ is the coproduct, so that $\mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{F}_p$ is the coproduct of the rings $\mathbb{F}_p$ and $\mathbb{Z}[x]$.

So, in terms of schemes, this means

$$\begin{align} \operatorname{Spec}(\mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{F}_p) &\cong \operatorname{Spec}(\mathbb{Z}[x]) \times_{\operatorname{Spec}(\mathbb{Z})} \operatorname{Spec}(\mathbb{F}_p) \\&\cong \operatorname{Spec}(\mathbb{Z}[x]) \times\operatorname{Spec}(\mathbb{F}_p) \end{align}$$

Since $\operatorname{Spec}(\mathbb{F}_p)$ is a subscheme of $\operatorname{Spec}(\mathbb{Z})$, this means the fiber product is just the subscheme of $\operatorname{Spec}(\mathbb{Z}[x])$ that lies over $\operatorname{Spec}(\mathbb{F}_p)$.

It may not be clear that this description actually translates to the expected description on points. But by the definition of pullback, it follows that the following are equivalent:

• The $k$-points of $\operatorname{Spec}(\mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{F}_p)$
• A pair of a $k$-point of $\operatorname{Spec}(\mathbb{Z}[x])$ and a $k$-point of $\operatorname{Spec}(\mathbb{F}_p)$ that have the same image in $\operatorname{Spec}(\mathbb{Z})$

Thus, the points of $\operatorname{Spec}(\mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{F}_p)$ really are the points of $\operatorname{Spec}(\mathbb{Z}[x])$ whose image in $\operatorname{Spec}(\mathbb{Z})$ is the image of $\operatorname{Spec}(\mathbb{F}_p)$.

Answer 2

The ring in question can be simplified using the standard isomorphism rules for tensor products: $\mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{F}_5 \cong \mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{Z} / 5 \mathbb{Z} \cong \mathbb{Z} [x] / 5 \mathbb{Z} [x] \cong \mathbb{F}_5 [x]$.

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The prime ideals in a PID are the maximal ideals and $0$, and in this case that means the nonzero prime ideals of $\mathbb{Z}[x] \otimes_{\mathbb{Z}} \mathbb{F}_5$ are in correspondence with irreducible polynomials $p(x) \in \mathbb{F}_5 [x]$.