Looking for a generalization of Wedderburn's Little theorem with weaker hypothesis

Question

Let $R$ be a division ring such that $(R^* , .)$ is a finitely generated group ($R^*:=R \setminus \{0\}$), then is it true that $R$ is a field? If not, then is there any such related fact in general of any weakening of the hypothesis of Wedderburn's Little theorem which would make a division ring a field?

Any result or reference would be very helpful. Thanks in advance.

Answer 1

[This is only a partial answer, due to a mistake pointed out in the comments]

A division ring with finitely-generated unit group must be finite, and therefore a field.

Since $R$ is a ring, it is an abelian group (additively) so $R$ is a $\mathbb{Z}$-module. But every nonzero element is invertible ($R$ being a division ring), so $R$ must in fact be a vector space over $\mathbb{Q}$ (if $\text{char } R = 0$) or over $\mathbb{F}_p$ (if $\text{char }R = p$). Here we identify the scalar field with the field of fractions of the elements $1_R + ... + 1_R$ where $1_R$ is the ring's identity element (these elements always commute with every element of $R$).

If the unit group of $R$ is finitely-generated, then $R$ must be finite-dimensional because otherwise $R$ would be uncountable, while $R\backslash\{0\}$ would be countable (being a finitely-generated group).

Thus $R$ is a finite-dimensional vector space over $\mathbb{Q}$ or $\mathbb{F}_p$ with dimension $n$. [As pointed out in the comments, this was wrong, we can have countable infinite-dimensional vector spaces. I was hoping for a convenient way of excluding the infinite-dimensional case.].

Also the [finite-dimensional] case over $\mathbb{Q}$ is impossible. Let's have a set of generators $\{r_1, ..., r_k\}$ of $R^*$. Let $R_i \in GL_n\left(\mathbb{Q}\right)$ be the matrix associated to the linear transformation of multiplication by $r_i$. Then every element generated by the $r_i$ has components (over $\mathbb{Q}$) generated by the entries of the matrices $R_i$ and $R_i^{-1}$, which are finitely many. Thus the set of components of these elements is contained in a finitely-generated subring of $\mathbb{Q}$. But $\mathbb{Q}$ itself is not finitely-generated as a ring so this is a proper subset and $R^* = \mathbb{Q}^n \backslash \{0\}$ is not finitely-generated as a group.

So $R$ is a finite-dimensional vector space over $\mathbb{F}_p$, and is therefore finite.