# Can “Taking algebraic closure” be made into a functor?

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 $$\pm=-$$.

• 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. – Asaf Karagila Jul 23 at 8:44
• 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. – Parthiv Basu Jul 24 at 0:47

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.