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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).

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    $\begingroup$ 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. $\endgroup$ May 6, 2023 at 18:56
  • $\begingroup$ @R. van Dobben de Bruyn Then is there such a map, whether canonical or not? $\endgroup$
    – UVIR
    May 6, 2023 at 18:59
  • $\begingroup$ 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. $\endgroup$ May 6, 2023 at 19:14
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    $\begingroup$ @R. van Dobben de Bruyn Can you clarify a little bit more, maybe write an answer, or give me a reference? Thanks! $\endgroup$
    – UVIR
    May 6, 2023 at 19:24
  • $\begingroup$ @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). $\endgroup$
    – LSpice
    May 6, 2023 at 21:21

2 Answers 2

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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.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – UVIR
    May 7, 2023 at 15:10
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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.

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    $\begingroup$ 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.) $\endgroup$
    – YCor
    May 7, 2023 at 7:01
  • $\begingroup$ Thank you very much! $\endgroup$
    – UVIR
    May 7, 2023 at 15:22

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