# Is a field that never embeds twice in another field necessarily a prime field?

Call a field $k$ unrepeatable$^1$ if for every field $L$ there are either zero or one field homomorphisms $k \to L$. Then the prime fields $\mathbb{Q}$ and $\mathbb{F}_p$ for $p$ prime are clearly unrepeatable and it seems very likely to me that those are the only ones. Is that true?

Notice that an unrepeatable field cannot have a non-identity automorphism, and must have a unique embedding into its algebraic closure (so for example $\mathbb{R}$ is not unrepeatable despite having no nontrivial automorphisms).

$^1$ I made up a term for it because the only name I know for these is the rather unwieldy "subterminal object in the opposite of the category of fields".

• Non-prime finite fields are not unrepeatable, since the Frobenius $x\mapsto x^p$ gives a nontrivial automorphism of every $\Bbb F_{p^k}$ with $k\ge2$. Indeed, I believe this shows that no non-prime field of positive characteristic is unrepeatable. – Greg Martin Feb 28 '18 at 5:03
• I definitely should have mentioned that, @GregMartin. (Minor correction: Frobenius endomorphism: it isn't always surjective. It's existence and non-identity-ness still shows unrepeatable positive characteristic fields are prime as you said.) – Omar Antolín-Camarena Feb 28 '18 at 5:21
• A natural extension of the definition is to fix a field extension $k_0\to k$ and to say that $k$ is "unrepeatable" over $k_0$ if for every extension $k_0\to L$ there is at most one $k_0$-embedding $k\to L$. Then when $k_0$ is perfect, $k$ unrepeatable forces $k=k_0$. More generally this holds iff $k$ is a purely inseparable extension of $k_0$ (in the sense that $k$ belongs to the smallest perfect extension of $k_0$). I think this generality better emphasizes what happens (in the answer to the original question, prime fields being perfect, the role of perfectness doesn't show up explicitly). – YCor Feb 28 '18 at 10:43

Proof: Let $k$ be unrepeatable and $F\subseteq k$ denote the prime field of $k$. Let $T\subseteq k$ be a transcendence base of $k/F$ and let $G=F(T)$. If $T\neq\emptyset$, then $G/F$ has non-trivial automorphisms (say take one element $t\in T$ to $t+1$). Since $k/G$ is algebraic this extends to a different embedding of $k$ into $\overline G$. Therefore, we see that $k/F$ must be algebraic.
It follows that $F\subseteq k\subseteq \overline{F}$ and since $k$ is unrepeatable (in $\overline F$), $k$ has to be fixed by all automorphisms in the absolute Galois group of $F$. But then $F=k$ and we are done.
• How does the step of extending the automorphism of $G$ to one of $k$ work? I don't think it's true in general that you can extend automorphisms to algebraic extensions: the automorphism you mentioned of, say, $F(t)$ given by $t \mapsto t+1$, doesn't extend to $F(\sqrt{t})$, does it? – Omar Antolín-Camarena Feb 28 '18 at 6:01
• (I guess that specific example does extend if $F$ has characteristic 2 because then $(\sqrt{t}+1)^2 = t+1$, but I'm pretty sure $t+1$ is not a square in $F(\sqrt{t})$ for other charactertistics.) – Omar Antolín-Camarena Feb 28 '18 at 6:04
• Oh, I guess that step can be replaced by constructing two different homomorphisms $k \to \bar{G}$ to an algebraic closure of $G$! I think I'm happy with the proof after that change (or after an explanation of why my objection was wrong ;)). – Omar Antolín-Camarena Feb 28 '18 at 6:32
• @OmarAntolín-Camarena: you're right, what I had in mind was to extend that inside $\overline G$, but I did not write it down very well. I'll make the change. Thanks. – Sándor Kovács Feb 28 '18 at 6:35