Let $R$ be a DVR with fraction field $K$ and a perfect residue field $k$. Consider a commutative connected group scheme $G$ over $R$, say smooth and of finite type over $R$. So we may use Chevalley's Theorem on algebraic groups and obtain a unique exact sequence $$ 0 \to H_K \to G_K \to A_K \to 0, $$ where $H_K$ is a connected affine group scheme and $A_K$ is an abelian variety. $H_K$ contains a maximal torus $T_K$, more exactly $H_K \cong T_K \times U_K$, where $U_K$ is unipotent and both are unique. This is the argumentation in [2] on p. 4. The same argumentation can be applied to the special fiber $G_k$ of $G$. Denote by $U_k$ the unique unipotent subgroup of $G_k$ obtained by the above method.

Now the authors claim, that $U_K = 0$ iff $U_k = 0$.

Question: Why is that true?


This is false as you state (but I don't think this is what the authors said).

Consider the Néron model $G$ of an elliptic curve $G_K$ with additive reduction of type II (say). Then $H_K=0$ (hence $U_K=0$), but $H_k=U_k=G_k=\mathbb G_{a,k}$.

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    $\begingroup$ Oh, I see now. First, they take $K$ as $Quot(R)$ and in the next line as a different field $K$, a perfect extension of $k$. Thanks for the counterexample. $\endgroup$ – boxdot Oct 30 '12 at 7:38

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