# Group homomorphism between multiplicative groups of fields

Let $$\mathbb{K}$$ and $$\mathbb{F}$$ be two (algebraically closed) fields.

I don't know what a group homomorphism $$\mathbb{K}^*\to \mathbb{F}^*$$ would look like. It is easy to see that $$x\mapsto x^n$$ with $$n\in \mathbb{Z}$$ is a group homomorphism from $$\mathbb{K}^*$$ to $$\mathbb{F}^*$$. Could you give me some other examples, like group homomorphism $$\mathbb{C}^*\to \mathbb{C}^*$$?

When $$\mathbb{K}=\mathbb{F}$$, and if we restrict to the algebraic group homomorphism, we know that $$x\mapsto x^n$$, $$n\in \mathbb{Z}$$ are all the homomorphisms.

My question is:

What will a general group homomorphism $$\mathbb{K}^*\to \mathbb{F}^*$$ would look like? Could you give me some other examples, like group homomorphism $$\mathbb{C}^*\to \mathbb{C}^*$$?

How to get a group homomorphism $$\mathbb{K}^*\to \mathbb{F}^*$$ with "nice property"? For example, we can restrict to algebraic group homomorphisms when $$\mathbb{K}=\mathbb{F}$$. But how to restrict group homomorphism when $$\mathbb{K}\neq \mathbb{F}$$?

• Complex conjugation defines a homomorphism $\Bbb C^{*}\rightarrow \Bbb C^{*}$. Commented Jun 18 at 18:15

Well, there are loads of homomorphisms between any two such groups. It helps to first figure out their structure: First and foremost, any such group $$G=\mathbb{K}^{\times}$$ must be divisible. Divisible groups are classified - they are all of the form $$G\simeq \mathbb{Q}^{(I)}\oplus \bigoplus_{p \text{ prime}}C_{p^{\infty}}^{(I_p)}$$ for some index sets $$I, I_p$$. Here, $$A^{(J)}$$ denotes the direct sum and $$C_{p^{\infty}}$$ is the ($$p$$-) Prüfer group, i.e. the $$p$$-component of $$\mathbb{Q}/\mathbb{Z}$$.
However, we can say more. Since any finite subgroup of $$G$$ must be cyclic, each $$I_p$$ is at most $$1$$. In fact, apart from perhaps one prime (the characteristic of $$\mathbb{K}$$), each $$I_p$$ must be exactly $$1$$, so that the torsion part must be either $$\bigoplus_{p}C_{p^{\infty}}(=\mathbb{Q}/\mathbb{Z})$$ or $$\bigoplus_{p\neq \chi(\mathbb{K})}C_{p^{\infty}}$$ in positive characteristic.
This alone guarantees the existence of continuum many distinct homomorphisms between any two such groups: Even assuming minimal "compatibility" (which happens for the algebraic closure of two finite fields of different characteristics $$q_1$$ and $$q_2$$), we find a subgroup of the homomorphisms isomorphic to $$\mathrm{Hom}\left(\bigoplus_{p\neq q_1}C_{p^{\infty}}, \bigoplus_{p\neq q_2}C_{p^{\infty}}\right)\simeq \prod_{p\neq q_1,q_2}\mathbb{Z}_p,$$ which is an uncountable group of size continuum.
Moreover, if the fields is not the algebraic closure of a finite field, the first summand $$\mathbb{Q}^{(I)}$$ does not vanish. In fact, its rank must always be infinite (as can be easily checked by considering the possible characteristic of the field). This gives many more homomorphisms. For the specific case of $$\mathbb{C}^{\times}$$, we find a multiplicative group isomorphic to $$\mathbb{Q}^{(2^\mathbb{N})}\oplus \mathbb{Q}/\mathbb{Z}$$, which leads to a total of $$2^{2^\mathbb{N}}$$ many endomorphisms of this group.