Let $K$ be a field of characteristic zero and $F_n$ be a free group of rank $n$. What is known about the group of units $K[F_n]^\times$?
In the case of $n=1$, there are only trivial units: $K[F_1]^\times\cong K^\times\times F_1$. I couldn't find any literatures on Kaplansky's unit conjecture dealing with free groups.
Thank you.
A group $G$ is locally indicable if every nontrivial finitely generated subgroup surjects onto the infinite cyclic group $\mathbb{Z}$. The following theorem was proven by Higman in 1940 [1]:
Theorem. If $G$ is locally indicable and $R$ a ring with no zero-divisors, all the units of the group ring $RG$ are trivial.
The proof is short and pleasant. Higman stated a slightly weaker form of the theorem, about groups all of whose subgroups surject onto $\mathbb{Z}$ ("indicable throughout"), but his proof only requires indicability of finitely generated subgroups.
Higman pointed out that his result applies to both free groups and free abelian groups (as their subgroups are themselves free in the appropriate setting). Thus, groups rings of free groups with a field of characteristic zero have only trivial units. This answers the question in the OP.
Lots of groups are locally indicable. In particular, Brodskii [2] and Howie [3] independently proved the following result:
Theorem. Let $A$ and $B$ be locally indicable groups. If $W$ is a cyclically reduced word in the free product $A\ast B$ which is not a proper power (i.e. if $U^n=W$ then $n=\pm1$), then the corresponding one-relator product $(A\ast B)/\langle\langle W\rangle\rangle$ is locally indicable.
It follows that all torsion-free one-relator groups $\langle \mathbf{x}\mid W\rangle$ are locally indicable, and hence surface groups are locally indicable. Combining with Higman's result, this answers the question in the comments.
Lets now shift up a dimension and mention $3$-manifold groups. Locally indicable groups are left orderable, and it is known that there are $3$-manifold groups which are not left orderable. Thus, the above framework is not applicable. Indeed, Gardam's counter-example to the unit conjecture is a $3$-manifold group [4, 5]. On the other hand, Kielak and Linton proved that Kaplansky's zero divisor conjecture holds here when the field is $\mathbb{C}$ [6]. Combining, we get the following for complex group rings:
Theorem [Gardam, Kielak-Linton]. Let $G$ be the fundamental group of a $3$-manifold. Suppose that $G$ is torsion-free. Then $\mathbb{C}G$ can have non-trivial units but cannot have non-trivial zero divisors.
Kielak and Linton's proof is quite different to the "locally indicable" proofs. It is known that if a torsion-free group $G$ satisfies the Strong Atiyah Conjecture over $\mathbb{C}$ then $\mathbb{C}G$ has no non-trivial zero divisors. They prove that $3$-manifold groups satisfy this conjecture by proving that they have bounded torsion, and that they lie in the smallest class $\mathcal{C}$ of groups containing all free groups that is closed under directed unions and extensions by elementary amenable groups. The result then follows immediately from a result of Linnell.
[1] Graham Higman, The units of group-rings. Proc. London Math. Soc. (2) 46 (1940), 231–248. doi
[2] S. D. Brodskiĭ, Equations over groups, and groups with one defining relation. (Russian) Sibirsk. Mat. Zh. (25) 2 (1984), 84–103. doi
[3] James Howie, On locally indicable groups. Math. Z. (180) 4 (1982), 445–461. doi eduml
[4] Giles Gardam A counterexample to the unit conjecture for group rings. Ann. of Math. 194 (2021) 967-979 arxiv doi
[5] Giles Gardam Non-trivial units of complex group rings. arxiv
[6] Dawid Kielak and Marco Linton Group rings of three-manifold groups. Proc. Amer. Math. Soc. (152) 5 (2024), 1939-1946 arxiv doi
Following ADL's answer for completeness here's an elementary proof of Higman's result that
(a) for every locally indicable group $G$ and field $K$, every unit of $KG$ is "trivial", i.e. a nonzero scalar multiple of a basis element $\delta_g$.
I'll use that
(b) under the same assumptions, $KG$ has no nonzero zero divisors.
For $u=\sum_{g\in G}u(g)\delta_g\in KG$, we write $\mathrm{supp}(u)=\{g\in G:u(g)\neq 0\}$.
(b) Suppose that $uv=0$ in $KG$ with $u,v\neq 0$. Write $u=\sum_{g\in G}u(g)\delta_g$. We can suppose that $1$ belongs to the support of $u$ and $v$, and we can suppose that $|\mathrm{supp}(u)|+|\mathrm{supp}(v)|$ is minimal. Let $H$ be the subgroup generated by $\mathrm{supp}(u)\cup\mathrm{supp}(v)$. It is necessarily nontrivial (otherwise $u,v$ would be scalar multiples of $\delta_1$), so there exists $p:H\to\mathbf{Z}$ a surjective homomorphism. Write $u_n=\sum_{g\in p^{-1}(\{n\})}u(g)\delta_g$, so that $u=\sum_{n\in\mathbf{Z}}u_n$ (finitely supported sum), and similarly for $v$. Then $u_0,v_0\neq 0$, and $u\neq u_0$. Let $k,\ell$ be the max of $\mathrm{supp}(u)$ and $\mathrm{supp}(v)$. So $k,\ell\ge 0$. Up to change $p$ into $-p$, $k>0$. So $u_kv_\ell=0$, and by minimality, we get a contradiction.
(a) Let now $u$ be a unit in $KG$, say $uv=1$. We can suppose that $u(1)\neq 0$. By contradiction, suppose that $\mathrm{supp}(u)\neq\{1\}$. Let $H$ the subgroup generated by $\mathrm{supp}(u)\cup\mathrm{supp}(v)$. So, $H\neq\{1\}$.
Let $k<k'$ be the max and min of the support of $u$, and $\ell\le\ell'$ the max and min of the support of $v$. Then $k'+\ell'<k+\ell$, hence at lease one of the two is nonzero. So at least $k+\ell$ one of $u_kv_\ell$ and $u_{k'}v_{\ell'}$ is zero. We get a contradiction with (b).
