# Do algebraically closed fields form a reflective subcategory of fields?

Let $${\bf Field}$$ be the category of fields and field homomorphisms and $$\mathfrak{A}$$ the full subcategory of algebraically closed fields. Is it true that $$\mathfrak{A}$$ is a reflective subcategory of $${\bf Field}$$.

The reflector is given by the inclusion maps $$K \hookrightarrow \overline{K}$$. Now, if $$f: K \longrightarrow L$$ is a morphism, with $$L$$ algebraically closed field, does there exist a unique extension $$\tilde{f}: \overline{K} \longrightarrow L$$? My answer is no!

I take the following example: let $$\iota: \mathbb{Q} \hookrightarrow \overline{\mathbb{Q}}$$ and consider the inclusion $$f: \mathbb{Q} \hookrightarrow \mathbb{C}$$. Then both the inclusion $$j: \overline{\mathbb{Q}} \hookrightarrow \mathbb{C}$$ and the conjugation map $$c$$ are such that $$c \circ \iota=f=j \circ \iota$$. So, I have no uniqueness. Is my argument correct?

• Such extensions always exist, but are not unique. Consider $K=\Bbb{R}$, $L=\Bbb{C}$. The usual inclusion $K\hookrightarrow L$ has two extensions $L\to L$. The identity and the complex conjugation. Mar 11, 2022 at 11:35
• I edited my question before receiving your comment. I worked with $K=\mathbb{Q}$ Mar 11, 2022 at 11:37
• An argument for the existence part. Standard. Most likely somebody has done a better job than I did, but that's one I could find quickly. Mar 11, 2022 at 11:37
• Yes, the argument looks correct to me. $\overline{\Bbb{Q}}$ has infinitely many automorphisms, and you can precompose $j$ by any of them. Finding two is, of course, sufficient to prove non-uniqueness. Mar 11, 2022 at 11:43
• Measuring this non-uniqueness is one of the main purposes of Galois theory, no? Mar 11, 2022 at 12:20

You are correct, such an extension is not unique. In fact, we can turn this into a proof by contradiction to show that the full subcategory of algebraically closed fields is not reflective. Indeed, assume $$F\colon\mathbf{Field}\rightarrow\mathfrak{A}$$ is left adjoint to the inclusion functor. Let $$K=\mathbb{Q}$$ and $$L=FK$$. Then, the adjunction yields $$\mathrm{Hom}(K,L)\cong\mathrm{Hom}(L,L)$$, where both $$\mathrm{Hom}$$s are in $$\mathbf{Field}$$. Since $$\mathrm{id}_L\in\mathrm{Hom}(L,L)$$, $$\mathrm{Hom}(K,L)\neq\emptyset$$, which forces $$K$$ and $$L$$ to have the same characteristic, whence $$K$$ is the prime field of $$L$$. Since $$K$$ is prime, it follows that $$\mathrm{Hom}(K,L)$$ is a one-element set, whence so is $$\mathrm{Hom}(L,L)$$. Choose a transcendence basis $$S$$ of $$L/K$$, so that $$K(S)/K$$ is purely transcendental and $$L/K(S)$$ is algebraic. Since $$L/K(S)$$ is algebraic and $$L$$ is algebraically closed, $$L=\overline{K(S)}$$ is an algebraic closure of $$K(S)$$. If $$S\neq\emptyset$$, there is a non-trivial automorphism of $$K(S)$$ sending an element of $$S$$ to its multiplicative inverse, which extends to a non-trivial endomorphism of $$\overline{K(S)}=L$$, contradiction. Thus, $$S=\emptyset$$ and $$L=\overline{K}=\overline{\mathbb{Q}}$$. However, $$\overline{\mathbb{Q}}$$ has non-trivial automorphisms, e.g. complex conjugation, contradiction.