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?

  • $\begingroup$ 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. $\endgroup$ Mar 11, 2022 at 11:35
  • $\begingroup$ I edited my question before receiving your comment. I worked with $K=\mathbb{Q}$ $\endgroup$ Mar 11, 2022 at 11:37
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    $\begingroup$ 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. $\endgroup$ Mar 11, 2022 at 11:37
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    $\begingroup$ 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. $\endgroup$ Mar 11, 2022 at 11:43
  • $\begingroup$ Measuring this non-uniqueness is one of the main purposes of Galois theory, no? $\endgroup$
    – Arthur
    Mar 11, 2022 at 12:20

1 Answer 1


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.


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