Vakil Exercise 3.2.Q

Question

I was looking at this exercise from Professor Vakil's book.

3.2.Q. EXERCISE: PICTURING $\mathbb{A}^n_\mathbb{Z}$ . Consider the map of sets $\pi: \mathbb{A}^n_{\mathbb{Z}} → Spec \mathbb{Z}$, given by the ring map $\mathbb{Z} \to \mathbb{Z}[x_1 , . . . , x_n ]$. If $p$ is prime, describe a bijection between the fiber $\pi^{−1}([(p)])$ and $\mathbb{A}^n_{\mathbb{F}_p}$ .(You won’t need to describe either set! Which is good because you can’t.) This exercise may give you a sense of how to picture maps (see Figure 3.7), and in particular why you can think of $\mathbb{A}^n_\mathbb{Z}$ as an “$\mathbb{A}^n$-bundle” over $Spec \mathbb{Z}$.(Can you interpret the fiber over [(0)] as $\mathbb{A}^n_k$ for some field $k$?)

Are we supposed to prove anything more than the isomorphism $\mathbb{Z}[x_1, ..., x_n]/(p)\mathbb{Z}[x_1,\ldots,x_n] = \mathbb{Z}/p[x_1, ..., x_n]$? Is there something deeper going on?

In general, I think the fiber is given by the ideal generated by the image of the given prime ideal. That's why I got the above computation.

But I think I'm missing something, because I don't understand the last part. For the prime ideal $(0)$, I just get $\mathbb{A}^n_\mathbb{Z}$ with the above approach. Or is the answer to the parenthetical "no"?

Answer 1

Primes in the fibre of $p$ are by definition those primes $\mathfrak p$ of $\mathbb Z[x_1, \dotsc, x_n]$ that intersect $\mathbb Z$ at $(p)$, which is equivalent to $(p) \subset \mathfrak p$.

Hence $\pi^{-1}(p) = V((p)) = \operatorname{Spec} \mathbb Z[x_1, \dotsc, x_n]/(p) = \operatorname{Spec} \mathbb F_p[x_1, \dotsc, x_n]$.

So yes, basically this is the isomorphism $\mathbb Z[x_1, \dotsc, x_n]/(p) \cong \mathbb F_p[x_1, \dotsc, x_n]$.

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But you are wrong for the generic fibre:

$\pi^{-1}((0))$ consists of primes $\mathfrak p$ that have trivial intersection with $\mathbb Z$. It turns out that $$\pi^{-1}((0)) = \operatorname{Spec} \mathbb Q[x_1, \dotsc, x_n].$$

For the bijection, you can just take a prime ideal with $\mathfrak p \cap \mathbb Z=0$ and show that $\mathfrak p\mathbb Q[x_1, \dotsc, x_n]$ is a prime ideal. This gives you a map $\pi^{-1}((0)) \to \operatorname{Spec} \mathbb Q[x_1, \dotsc, x_n]$.