# Given a field $\mathbb{F}$, what is $\text{Aut}(\mathbb{F}^{\ast})$?

There are many well-known problems and results concerning automorphism groups of fields. For example, it is well-known that the automorphism group of each field in $\{ \mathbb{Q}, \mathbb{F}_{p}, \mathbb{R} \}$ is trivial, and it is well-known that the (known) constructions of the 'wild' automorphisms of the field $\mathbb{C}$ require the axiom of choice (or Zorn's lemma).

It thus seems natural to consider the 'easier' problem of evaluating the group $\text{Aut}(\mathbb{F}^{\ast})$ of group automorphisms of the underlying multiplicative group $\mathbb{F}^{\ast}$ of $\mathbb{F}$.

It is clear that the problem of evaluating $\text{Aut}(\mathbb{F}_{p}^{\ast})$ reduces to the closed problem of evaluating the automorphism group of a finite abelian group (see Automorphisms of Finite Abelian Groups), and it is clear that evaluating $\text{Aut}(\mathbb{Q}^{\ast})$ reduces to the problem of evaluating the group of group automorphisms of $C_{2} \oplus \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$ (see this related discussion). It thus seems natural to ask:

(1) What is $\text{Aut}\left(\mathbb{R}^{\ast}\right)$?

(2) What is $\text{Aut}\left(\mathbb{C}^{\ast}\right)$?

(3) What is $\text{Aut}\left(\mathbb{C}(t)^{\ast}\right)$?

(4) What is $\text{Aut}\left(\mathbb{Q}_{p}^{\ast}\right)$, where $\mathbb{Q}_{p}$ denotes the field of $p$-adic numbers?

(5) More generally, given a field $\mathbb{F}$, is there a known way of evaluating $\mathbb{F}^{\ast}$?

• For $(1)$, the group $\mathbb{R}^*$ is isomorphic (through exponential) to the additive group $\mathbb{R}$, so the automorphisms of $\mathbb{R}^*$ are the $\mathbb{Q}$-linear automorphisms of $\mathbb{R}$ (so the group is determined by the $\mathbb{Q}$-dmiension of $\mathbb{R}$, which I guess may depend on the generalized continuum hypothesis or something like that). – Captain Lama Mar 16 '16 at 17:04
• You're working too hard for the case of finite fields: $\mathbb{F}_p^{\times}$ is cyclic of order $p - 1$, so it's much easier to compute its automorphism group than it would be for a general finite abelian group. Its automorphism group is the group of units of $\mathbb{Z}/(p-1)\mathbb{Z}$. – Qiaochu Yuan Mar 21 '16 at 17:13

Before you ask what the automorphism groups of these groups are it's probably a good idea to figure out what these groups themselves are. This is not so bad:

1. $\mathbb{R}^{\times} \cong C_2 \oplus \mathbb{R}$. This comes from looking at the exponential map from $\mathbb{R}$: the $C_2$ factor is $\pm 1$.

2. $\mathbb{C}^{\times} \cong S^1 \oplus \mathbb{R}$. This comes from looking at the exponential map from $\mathbb{C}$.

3. $\mathbb{C}(t)^{\times} \cong S^1 \oplus \mathbb{R} \oplus \bigoplus_{\mathbb{C}} \mathbb{Z}$. This comes from looking at constant terms + unique factorization.

4. $\mathbb{Q}_p^{\times} \cong C_{p-1} \oplus \mathbb{Z}_p \oplus \mathbb{Z}$. The last term is powers of $p$, the first term is looking at constant terms, and the second term comes from the "exponential" map $n \mapsto (1 + p)^n$ from $\mathbb{Z}_p$.

All of these groups have the pleasant property that they split up as a direct sum of torsion and torsion-free parts (although that isn't necessarily the decomposition I wrote above). Whenever an abelian group $A \cong T \oplus F$ has this property, we can write its automorphism group in matrix form as

$$\text{Aut}(A) \cong \text{Aut}(T \oplus F) \cong \left[ \begin{array}{cc} \text{Aut}(T) & \text{Hom}(F, T) \\ 0 & \text{Aut}(F) \end{array} \right].$$

The point is that there can't be any nontrivial homomorphisms $T \to F$, so every endomorphism has this upper triangular form, and then it's an automorphism iff its diagonal components are. This works more generally whenever we can write an abelian group as a direct sum of two subgroups one of which admits no nontrivial homomorphisms to the other. Now let's compute automorphism groups.

Below it will be very convenient to write $I$ to denote the index set of a basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space (we need AC for this). Occasionally I will pretend this is also a basis of something like $\mathbb{R} \oplus \mathbb{R}/\mathbb{Q}$ since (again assuming AC) these have the same dimension.

1. $\text{Aut}(C_2)$ is trivial, and there aren't any homomorphisms $\mathbb{R} \to C_2$, so we get $\text{Aut}(\mathbb{R})$, the group of automorphisms of an uncountable dimensional $\mathbb{Q}$-vector space. Assuming AC this is some huge group $GL_I(\mathbb{Q})$ of matrices over $\mathbb{Q}$.

2. Write $\mathbb{R} \cong \mathbb{Q} \oplus \mathbb{R}/\mathbb{Q}$ (both are just $\mathbb{Q}$-vector spaces). Applying the exponential $\exp (ix)$ gives $S^1 \cong \mathbb{Q}/\mathbb{Z} \oplus \mathbb{R}/\mathbb{Q}$, so $\mathbb{C}^{\times}$ is $\mathbb{Q}/\mathbb{Z}$ plus another uncountable dimensional $\mathbb{Q}$-vector space which I will pretend is isomorphic to $\mathbb{R}$ for ease of notation, although I need AC for this. $\text{Aut}(\mathbb{Q}/\mathbb{Z})$ turns out to be the group of units of the profinite integers $\widehat{\mathbb{Z}}^{\times} \cong \prod_p \mathbb{Z}_p^{\times}$, and there are lots of homomorphisms from a $\mathbb{Q}$-vector space to $\mathbb{Q}/\mathbb{Z}$. Altogether we get

$$\left[ \begin{array}{cc} \widehat{\mathbb{Z}}^{\times} & \prod_I \text{Hom}(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \\ 0 & GL_I(\mathbb{Q}) \end{array} \right].$$

1. Now in addition to the issues that arose in 2 we also need to figure out the automorphism group of $\mathbb{R}$ plus uncountably many copies of $\mathbb{Z}$. There are no nontrivial homomorphisms from $\mathbb{R}$ to any sum of copies of $\mathbb{Z}$, so this again has an upper triangular form, involving $\text{Aut}(\mathbb{R})$ (big matrices over $\mathbb{Q}$), $\text{Aut}(\bigoplus_{\mathbb{C}} \mathbb{Z})$ (big matrices over $\mathbb{Z})$, and homomorphisms from $\bigoplus_{\mathbb{C}} \mathbb{Z}$ to $\mathbb{R}$, of which there are again a lot. Altogether we get

$$\left[ \begin{array}{cc} \widehat{\mathbb{Z}}^{\times} & \prod_I \text{Hom}(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) & \prod_{\mathbb{C}} \mathbb{Q}/\mathbb{Z} \\ 0 & GL_I(\mathbb{Q}) & \prod_{\mathbb{C}} \mathbb{R} \\ 0 & 0 & GL_{\mathbb{C}}(\mathbb{Z}) \end{array} \right]$$

1. Now we need to figure out the automorphism group of $\mathbb{Z}_p \oplus \mathbb{Z}$. At this point I have to confess: I just don't know what this is. I know what the automorphisms of $\mathbb{Z}_p$ are as a profinite group, but not as an abstract group. And I don't know what the homomorphisms $\mathbb{Z}_p \to \mathbb{Z}$, again as abstract groups, are.
• Ack, okay, so above I previously assumed that $\text{Hom}(\mathbb{Q}, \mathbb{Q}/\mathbb{Z}) \cong \mathbb{Q}/\mathbb{Z}$ but in fact this group is bigger and weirder: it's the Pontryagin dual of $\mathbb{Q}$ (math.stackexchange.com/questions/48170/…). – Qiaochu Yuan Mar 26 '16 at 4:58