# A ring map from algebraic integers to algebraic closure of $\mathbb F_p$

Let $$p$$ be a prime and $${\mathbb F}_p$$ the finite field with $$p$$ elements. There is a canonical ring map $${\mathbb Z} \to {\mathbb F}_p \cong {\mathbb Z}/ p {\mathbb Z}$$. Denote the image of $$n$$ by $$[n]_p$$.

Now consider the set of algebraic integers $$\overline {\mathbb Z} \subset {\mathbb C}$$, which is the set of roots of monic polynomials with integer coeffiecients. Let $$\overline{\mathbb F}_p$$ be the algebraic closure of $${\mathbb F}_p$$.

Question: Is there a ring map $$\overline{\mathbb Z} \to \overline{\mathbb F}_p$$ which extends the natural map $${\mathbb Z} \to {\mathbb F}_p$$? For example, such a map should send $$a + b \sqrt{2}$$ to $$[a]_p + [b]_p \sqrt{2}$$ (if this ever makes sense).

• Every prime ideal $\mathfrak p \subseteq \overline{\mathbf Z}$ dividing $p\overline{\mathbf Z}$ has $\overline{\mathbf Z}/\mathfrak p \cong \overline{\mathbf F}_p$, but there is no preferred choice (neither of $\mathfrak p$ nor of this isomorphism). In fact, even saying the algebraic closure of $\mathbf F_p$ (or the set of algebraic integers) doesn't make sense; it is only defined up to non-canonical isomorphism. May 6, 2023 at 18:56
• @R. van Dobben de Bruyn Then is there such a map, whether canonical or not?
– UVIR
May 6, 2023 at 18:59
• Yes, $\operatorname{Hom}_{\text{Ring}}(\overline{\mathbf Z},\overline{\mathbf F}_p) \neq \varnothing$. In fact this set has a natural continuous action of $\operatorname{Gal}(\overline{\mathbf Q}/\mathbf Q)$ by precomposition and by $\operatorname{Gal}(\overline{\mathbf F}_p/\mathbf F_p)$ by postcomposition. I think that the $\operatorname{Gal}(\overline{\mathbf Q}/\mathbf Q)$-action is transitive, and the stabiliser of a homomorphism is the inertia group of its kernel. May 6, 2023 at 19:14
• @R. van Dobben de Bruyn Can you clarify a little bit more, maybe write an answer, or give me a reference? Thanks!
– UVIR
May 6, 2023 at 19:24
• @UVIR, re, use Zorn to let $A$ be a maximal integral extension of $\mathbb Z$ such that there is a ring map $A \to \overline{\mathbb F_p}$, then show that, if $A$ is not all of $\overline{\mathbb Z}$, we can extend any given ring map $A \to \overline{\mathbb F_p}$ to a larger ring. \\ TeX note: \mathbb takes a following argument (unlike, e.g., \bf), so, e.g., {\mathbb F} is the same as {\mathbb{F}}, and you might as well drop the outer braces for \mathbb{F} (or \mathbb F). May 6, 2023 at 21:21

This is basic ramification theory that you can find in any textbook on algebraic number theory; for instance [Neukirch]. As in my comments, I will show slightly more than nonemptiness of $$\operatorname{Hom}(\overline{\mathbf Z},\overline{\mathbf F}_p)$$, namely determine this set via Galois theory.

Fix an algebraic closure $$\overline{\mathbf F}_p$$ of $$\mathbf F_p$$. For a finite Galois extension $$\mathbf Q \to K$$ with ring of integers $$\mathcal O_K$$, consider the left action by precomposition $$\operatorname{Gal}(K/\mathbf Q) \times \operatorname{Hom}_{\text{Ring}}(\mathcal O_K,\overline{\mathbf F}_p) \to \operatorname{Hom}_{\text{Ring}}(\mathcal O_K,\overline{\mathbf F}_p).$$ Now [Neukirch, Prop. I.9.1] says that the action on the set of primes $$\mathfrak p \subseteq \mathcal O_K$$ above $$p$$ is transitive (and this set is nonempty). Given such a prime $$\mathfrak p$$, [Neukirch, Prop. I.9.4] says that $$\operatorname{Stab}_{\operatorname{Gal}(K/\mathbf Q)}(\mathfrak p) \to \operatorname{Gal}((\mathcal O_K/\mathfrak p)/\mathbf F_p)$$ is surjective. But $$\operatorname{Hom}(\mathcal O_K,\overline{\mathbf F}_p)$$ is a torsor under $$\operatorname{Gal}((\mathcal O_k/\mathfrak p)/\mathbf F_p)$$ by Galois theory, so the action above is transitive. The stabiliser of $$\phi \colon \mathcal O_K \to \overline{\mathbf F}_p$$ is the inertia subgroup of $$\ker \phi$$, more or less by definition [Neukirch, Prop. I.9.6].

Now we just need a straightforward limit argument. Recall that $$\operatorname{Gal}(\overline{\mathbf Q}/\mathbf Q)$$ is the limit of $$\operatorname{Gal}(K/\mathbf Q)$$ for all finite Galois extensions $$\mathbf Q \to K$$ (with its natural profinite topology), and likewise $$\operatorname{Hom}(\overline{\mathbf Z},\overline{\mathbf F}_p)$$ is the limit of $$\operatorname{Gal}(\mathcal O_K,\overline{\mathbf F}_p)$$ over all finite Galois extensions $$\mathbf Q \to K$$. Since a cofiltered limit of finite nonempty sets is nonempty [Tag 0A2R] (or a countable cofiltered limit of surjective maps is nonempty), we see $$\operatorname{Hom}(\overline{\mathbf Z},\overline{\mathbf F}_p) \neq \varnothing$$, and a limit of torsors under finite groups is a torsor under the limit group.

References.

[Neukirch] J. Neukirch, Algebraic number theory. Grundlehren der Mathematischen Wissenschaften 322. Springer, 1999. ZBL0956.11021.

• Thank you very much!
– UVIR
May 7, 2023 at 15:10

The other answer is excellent and provides a lot more context, but if you are just after the existence statement, then there is a much more straightforward argument.

Since $$p$$ is not invertible in $$\overline{\mathbb Z}$$, you can find some maximal ideal $$\mathfrak m$$ which contains $$p$$. Then $$\overline{\mathbb Z}/\mathfrak m$$ is a field extension of $$\mathbb F_p$$, and it is enough to argue that it is an algebraic closure.

Indeed, any element of $$\overline{\mathbb Z}$$ satisfies a monic equation over $$\mathbb Z$$ (by definition), and reducing this equation we see any element of $$\overline{\mathbb Z}/\mathfrak m$$ is algebraic over $$\mathbb F_p$$.

Conversely, take a nonconstant polynomial $$f\in\mathbb F_p[x]$$. We may assume it is monic, so that it has a monic lift $$g\in\mathbb Z[x]$$, which has a root in $$\overline{\mathbb Z}$$. Reduction gives a root of $$f$$ in $$\overline{\mathbb Z}/\mathfrak m$$.

This implies $$\overline{\mathbb Z}/\mathfrak m$$ is isomorphic to $$\overline{\mathbb F_p}$$. Composing any such isomorphism with the reduction map $$\overline{\mathbb Z}\to\overline{\mathbb Z}/\mathfrak m$$ gives a desired homomorphism.

• Nice. This answer not only proves the existence: it proves that every quotient field of characteristic $p>0$ of $\overline{\mathbf{Z}}$ is isomorphic to $\overline{\mathbf{F}_p}$. (And there's none of char 0: first $\overline{\mathbf{Z}}$ is not a field, and if one kills a nonzero element of $\overline{\mathbf{Z}}$, one kills the constant term in its minimal monic polynomial, which is a nonzero integer.)
– YCor
May 7, 2023 at 7:01
• Thank you very much!
– UVIR
May 7, 2023 at 15:22