In the literature, there are a number of results called *Rosenlicht's lemma*, but I am talking about the following one:


Let $T$ be a torus over a, in my case, number field $k$. Denote by $\bar{k}[T]$ the ring of regular functions on $T$, and by $\widehat{T} := \mathrm{Hom}\_{\bar{k}}(T\_{\bar{k}},\mathbb{G}\_m)$ the group of characters of $T$, where $T\_{\bar{k}} := T \times\_k \bar{k}$. Rosenlicht's lemma states that one has $$\widehat{T} \cong \bar{k}[T]^\*/\bar{k}^\*.$$


Now let $C$ be a smooth projective curve of genus $\geq 1$ over $k$ and for some $n \geq 1$, identify the closed points $p\_0,...,p\_n$ to obtain a singular curve $C'$ with an $(n+1)$-fold singularity $Q$. Then one has a morphism from the smooth affine curve $X :=C' \setminus \{Q\}$ into the *generalized Jacobian variety* $S$ of $C$ w.r.t. the divisor $D = p\_0+...+p\_n$, sitting in the middle of the exact sequence


$$0 \rightarrow T \rightarrow S \rightarrow A \rightarrow 0,$$


where $T$ is a torus of dimension $n$ and $A$ is the usual Jacobian variety of $C$, assuming of course we have an embedding $C \hookrightarrow J$. The geometric points of $\widehat{T}$ is dual $\mathrm{Div}\_{\bar{D}}^0(\bar{C})$. This is the subgroup of the group of divisors of $\bar{C}$, supported on $\bar{D}$ and algebraically equivalent to zero, i.e., of degree zero. By Rosenlicht's lemma, one has $\bar{k}[T]^\*/\bar{k}^\* \cong \mathrm{Div}\_{\bar{D}}^0(\bar{C})$.


**Question 1.** Does one have $\widehat{S} \cong \bar{k}[S]^\*/\bar{k}^\*$?


I have a feeling this is true, but I cannot explain precisely why, only provide some intuition, sorry if it doesn't make sense. Since $S$ is the extension of $A$ by $T$, for a regular function on $S$ on has to consider it on the abelian variety part and the torus part. For the case of $A$, all regular functions are constant and so $\bar{k}[A]^\*/\bar{k}^\* = 0$ and so we can sort of ignore it. Thus we only need to consider the torus part which Rosenlicht's lemma tells us that $\bar{k}[S]^\*/\bar{k}^\* \cong \widehat{S}$.


Here I am not saying that $\widehat{S} \rightarrow \widehat{T}$ is an isomorphism, but I think this is an inclusion? Intuitively, a regular function on the whole of $S$ can always restrict to a regular function on $T$.


**Question 2.** One has an exact sequence of Galois modules
$$1 \rightarrow \bar{k}^\* \rightarrow \bar{k}[X]^\* \rightarrow \mathrm{Div}\_{\bar{D}}^0(\bar{C}) \rightarrow \mathrm{Pic}\_{\bar{D}}^0(\bar{C}),$$
where $X$ is the affine curve defined above. Could $\widehat{S}$ be the subgroup $$\mathrm{ker}(\mathrm{Div}\_{\bar{D}}^0(\bar{C}) \rightarrow \mathrm{Pic}\_{\bar{D}}^0(\bar{C}))?$$



Is there a Rosenlicht's lemma for semi-abelian varieties?
I am not sure which Lemma of Rosenlicht you mean. In "Toroidal algebraic groups. Proc. Amer. Math. Soc. 12 (1961), 984–988" he proves:


**Theorem 3:**
Let $\phi:\Gamma\to G$ be an everywhere defined rational map from
a connected algebraic group $\Gamma$ into a toroidal algebraic group $G$, with
$\phi(e)=0$. Then $\phi$ is a homomorphism.


Apply this to $G=\mathbf G\_m$. Since a morphism $\phi:\Gamma\to\mathbf G\_m$ is just a unit the theorem implies that $\chi:=\phi(e)^{-1}\phi$ is automatically a character. Thus
$$k[\Gamma]^\times=\operatorname{Hom}(\Gamma,\mathbb G\_m)\times k^\*$$ for any connected algebraic group, even nonlinear and noncommutative ones.


Does this settle **Question 1**?


