Is there a Rosenlicht's lemma for semi-abelian varieties?

Question

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 $$\mathrm{ker}(\mathrm{Div}_{\bar{D}}^0(\bar{C}) \rightarrow \mathrm{Pic}_{\bar{D}}^0(\bar{C}))?$$

Answer 1

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?