Smooth diagonalizable groups over algebraically closed fields have only principal cocycles into smooth commutative unipotent groups

Fix $n>1$ in ${\mathcal{N}}_k$ and $x\in G_n(k)$, and sum the identity $f(x)=f(xy)-x\cdot f(y)$ over all $y\in G_n(k)$: since the action of $x$ is a group automorphism, ${\sum }_{y\in G_n(k)}f(xy)={\sum }_{y^{\mathrm{\prime }}\in G_n(k)}f(y^{\mathrm{\prime }})=s$ and ${\sum }_{y}x\cdot f(y)=x\cdot s$, so $e_{n}f(x)=s-x\cdot s$ where $e_n=\mathrm{\mid }{G}_n(k)\mathrm{\mid }$ divides a power of $n$, hence $e_n\cdot 1_k\mathrm{\ne }0$. By [F2] $e_n$ is invertible on $M(k)$, so $f(x)=x\cdot m_n-m_n$ with $m_n=-e_n^{-1}s$ for all $x\in G_n(k)$: the restriction of $f$ to $G_n$ is principal.

For each $n\in {\mathcal{N}}_k$ let $M(n)=\{m\in M(k):f(x)=x\cdot m-m\text{ for all }x\in G_n(k)\}$. Each $M(n)$ is nonempty by [step 1.1] (for $n>1$; for $n=1$ the condition is vacuous and $M(1)=M(k)$) and is the set of $k$-points of the closed subscheme of $M$ defined by the finitely many equations $f(x)=x\cdot m-m$ for $x$ running over a generating set of $G_n$; the family is directed downwards under divisibility: if $n\mid n^{\mathrm{\prime }}$, then $G_n\subseteq G_{n^{\mathrm{\prime }}}$ and $M(n^{\mathrm{\prime }})\subseteq M(n)$.

Choose a divisibility-increasing cofinal sequence $n_1\mid n_2\mid \cdots$ in ${\mathcal{N}}_k$ (take $n_j$ to be the least common multiple of the integers at most $j$ belonging to ${\mathcal{N}}_k$). The sets $M(n_1)\supseteq M(n_2)\supseteq \cdots$ form a descending chain of nonempty closed subsets of $M(k)$, which is Noetherian as a subspace of the Noetherian finite-type scheme $M$ (Chain dimension and the empty-space convention), so it stabilizes at some index $j_0$; choose $m\in M(n_{j_0})$, which is nonempty. Then $f(x)=x\cdot m-m$ for all $x$ in ${\bigcup }_j{G}_{n_j}(k)$.

Consider the morphisms $G\to M$, $x\mapsto f(x)$ and $x\mapsto x\cdot m-m$; their equalizer $Z$ is a closed subscheme of $G$ (closed immersions and equalizers into the separated $M$) with $Z(k)\supseteq {\bigcup }_{n\in {\mathcal{N}}_k}{G}_n(k)$ by [step 3.1], since every $n\in {\mathcal{N}}_k$ divides some $n_j$ and hence has its $G_n$ contained in some $G_{n_j}$. As $G$ is a diagonalizable group variety, it is smooth of multiplicative type over the algebraically closed field $k$, [F3] gives $Z=G$; therefore $f(x)=x\cdot m-m$ for all points $x$, and $f$ is principal.