How many different field homomorphisms can we have between any two given fields?

Question

How many field homomorphisms can we have between any two given fields? That is, if $A$ and $B$ are fields, what can we say about the cardinality of $\text{Hom}(A,B)$? I don't know much abstract algebra so I apologize if the answer is obvious. I got curious when thinking about whether the only field homomorphism $\mathbb{R} \rightarrowtail \mathbb{C}$ is the obvious one.

Answer 1

The only reasonable upper bound I know in general is the cardinality of the vector space of linear maps between the fields. (They must have the same characteristic for there to be any maps between them, so must both be vector spaces over $\mathbb{Q}$ or $\mathbb{F}_p$ for some prime $p$.)

For the case of $\mathbb{R}$ and $\mathbb{C}$ there are $2^{2^{\aleph_0}}$ automorphisms of $\mathbb{C}$, and the only automorphisms that fix $\mathbb{R}$ are the identity and complex conjugation, so any pair of automorphisms that aren't equal or aren't related by complex conjugation will send the usual copy of $\mathbb{R}$ to two different copies of $\mathbb{R}$. Hence there are (at least) $2^{2^{\aleph_0}}$ homomorphisms $\mathbb{R} \to \mathbb{C}$, which incidentally is also the cardinality of the vector space of ($\mathbb{Q}$-)linear maps between them.

Answer 2

1. $A$ and $B$ must have the same characteristic.

2. An element $f\in Hom(A,B)$ is injective, thus $B$ is an extension of $A$.

3. Suppose that $B$ is an algebraic extension of $A$, they have the same algebraic closure $C$, each morphism $f:A\rightarrow B$ can be extended to a morphism $f':C\rightarrow C$ https://en.wikipedia.org/wiki/Isomorphism_extension_theorem

So if $k$ is $Z/p$ if the characteristic is $p$ or $Q$ if the characteristic is zero. For an algebraic extension $ B$ of $A$, the elements of $Gal(C:k)$ which fixes $B$ acts transitively on $Hom(A,B)$.