Question. Let $B \twoheadrightarrow C$ be a fully faithful homomorphism of finite connected commutative group schemes over a perfect field $k$. Let $T = k[x^{1/p^\infty}]/(x) = \varinjlim k[t]/(t^p)$. Is the map $B(T) \rightarrow C(T)$ surjective?
For $m < n$, the underlying space of natural quotients $\mu_{p^n} \twoheadrightarrow \mu_{p^m}$ and $\alpha_{p^n} \twoheadrightarrow \alpha_{p^m}$ are induced by the natural ring monomorphism $k[t]/(t^{p^m}) \hookrightarrow k[t]/(t^{p^n})$. Since every element in $T$ has a unique $p$-th root of unity, any ring homomorphism $k[t]/(t^{p^m}) \rightarrow T$ naturally lifts to $k[t]/(t^{p^n}) \rightarrow T$.
I know that that the underlying space of a connected finite group scheme is a product of $\mathrm{spec}\,k[t]/(t^{p^n})$ for various $n$. However, I do not know whether the morphisms between the underlying spaces of $B \rightarrow C$ are given by a product of $\mathrm{spec}\,k[t]/(t^{p^n}) \twoheadrightarrow \mathrm{spec}\,k[t]/(t^{p^m})$. If this is true, the earlier argument implies that $B(T) \rightarrow C(T)$ is surjective.
