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I am now confused with such problem as title goes. To be exact, the problem is

Does there exist a functor from $A:\mathsf{Field}\to \mathsf{Field}$ with a natural transformation from identity functor $\iota: \operatorname{id}\to A$ such that for each $F$, $A(F)$ is the algebraically closure of $F$ through $\iota_F:F\to A(F)$?

It is not easy rather than first glimpse. Let me explain.

Note that, the existence of algebraic closure only ensures that there exist a map from $\operatorname{Obj}(\mathsf{Field})$ to itself. Since the "extension" property is not unique, it is not generally true that we can extend the map to $\operatorname{Mor}(\mathsf{Field})$ for arbitrary choice of algebraic closure.

  1. For example, consider the fields $$\begin{array}{ccc} \mathbb{Q}[\sqrt[3]{2}, \sqrt{2}] &\to & \mathbb{Q}[\omega\sqrt[3]{2}, \sqrt{2}]\\ \uparrow &&\uparrow \\ \mathbb{Q}[\sqrt[3]{2}] & \to & \mathbb{Q}[\omega\sqrt[3]{2}] \end{array}$$ If we choose the algebraically closure of $\left[\begin{matrix}\mathbb{Q}[\sqrt[3]{2}, \sqrt{2}] & \mathbb{Q}[\omega\sqrt[3]{2}, \sqrt{2}]\\ & \mathbb{Q}[\omega\sqrt[3]{2}]\end{matrix}\right]$ by inclusion to $\overline{\mathbb{Q}}$, and the closure of $\mathbb{Q}[\sqrt[3]{2}]\to \overline{\mathbb{Q}}$ by $\sqrt[3]{2}\mapsto \omega\sqrt[3]{2}$. We cannot extend a well-defined functor. Similar problem exists for transcendental extension, for example, square like this $$\begin{array}{ccc} \mathbb{C}[X,Y] &\to & \mathbb{C}[X^2,Y]\\ \uparrow &&\uparrow \\ \mathbb{C}[X] & \to & \mathbb{C}[X^2] \end{array}$$ A reasonable method is to avoid phenomenon above is as follow. Fix an algebraically closed field $F$, and take all of its subfields as "skeleton", then fix an isomorphism to a subfields of $F$ from all fields whose algebraic closure is $F$ up to an isomorphism. The isomorphic class of algebraically closure are completely dependen by its characteristic and the transcendental dimension over prime field $\mathbb{Q}$ or $\mathbb{F}_p$.

  2. Now the problem is how to naturally chose extensions for endmorphisms. But unfortunately, the choice is fragile. For instance, consider the following diagram $$\begin{array}{ccccl} \mathbb{Q}[\sqrt{3}, \sqrt{2}] &\to & \mathbb{Q}[\sqrt{3}, \sqrt{2}] &: &\sqrt{3}\mapsto -\sqrt{3},\sqrt{2}\mapsto \pm \sqrt{2}\\ \uparrow &&\uparrow \\ \mathbb{Q}[\sqrt{3}] & \to & \mathbb{Q}[\sqrt{3}] &:&\sqrt{3}\mapsto -\sqrt{3} \end{array}$$ There is no suitable choice such that $$\begin{array}{ccccl} \overline{\mathbb{Q}} &\to & \overline{\mathbb{Q}} &: &\sqrt{3}\mapsto -\sqrt{3},\sqrt{2}\mapsto \pm \sqrt{2}\\ \parallel &&\parallel \\ \overline{\mathbb{Q}}& \to & \overline{\mathbb{Q}} &:&\sqrt{3}\mapsto -\sqrt{3} \end{array}$$ commutes for both $\pm=+$ and $\pm=-$.

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  • $\begingroup$ I am sure that this came up somewhere recently. Either here or on MathOverflow. Maybe not in this exact formulation, but the essence was the same. $\endgroup$ – Asaf Karagila Jul 23 at 8:44
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    $\begingroup$ A simple example of why this is not possible is also given in (ncatlab.org/nlab/show/algebraically+closed+field). One considers the equalizer diagram $\mathbb{R} \to \mathbb{C} \xrightarrow[\text{conj}]{\text{id}} \mathbb{C}$ and sees that it would not to extend to a commutative diagram under such a functor. $\endgroup$ – Parthiv Basu Jul 24 at 0:47
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No, this is not possible. For instance, let $K$ be any field with a automorphism $f:K\to K$ whose order is finite and greater than $2$. Then $A(f):A(K)\to A(K)$ would be an automorphism of the same order extending $f$. But no such automorphism exists: by the Artin-Schreier theorem, any finite-order automorphism of an algebraically closed field has order at most $2$.

Or without using any big theorems, you can find problems just looking at finite extensions. For instance, if $f$ is the Frobenius automorphism of $\mathbb{F}_{p^2}$ then $F(f)$ is an extension to an algebraic closure which still has order $2$. Since $\mathbb{F}_{p^4}$ is normal over $\mathbb{F}_{p}$, $F(f)$ restricts to an automorphism of $\mathbb{F}_{p^4}$, which must be the Frobenius squared in order to have order $2$. But the Frobenius squared does not restrict to $f$ on $\mathbb{F}_{p^2}$, so this is a contradiction.

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