Let $F$ be a field and let $G$ be a smooth, commutative and connected $F$-group scheme of finite type. Let $K/F$ be a finite Galois extension of fields. Then there exists a canonical norm (or *trace*, if $G$ is written additively) smooth and surjective morphism of $F$-group schemes
$$
R\_{K/F}(G\_{K})\to G,
$$
where $R\_{K/F}$ denotes Weil restriction of scalars. When $G$ is an $F$-torus, the kernel of the above morphism is an $F$-torus which is denoted by $R\_{K/F}^{(1)}(G\_{K})$ and called the *norm 1 torus*. This case is well-known. I need information about the other cases, e.g., when $G$ is an abelian variety, which should be "well-known" (= "treated somewhere in the literature at the required level of generality"). I searched the web for the term "trace 0 abelian variety" and I got tons of references to cryptography but no general reference work from which to get the information that I need, e.g., is that kernel always connected, e.g., when G is a semiabelian variety?



Norm 1 (or trace 0) commutative algebraic groups
You can do the same construction when $K$ is any étale $K$-algebra, not necessarily a field. If $K$ is the trivial $K$-algebra $F^n$, then $R\_{K/F}G\_K\cong G^n$ and the norm map is just the multiplication $G^n\to G$, hence $R^{(1)}\_{K/F}G\_K\cong G^{n-1}$.


Now observe that the construction commutes with ground field extension. If you extend the ground field to an algebraic closure $\overline{F}$, then $K$ is replaced by $\overline{F}\otimes\_F K\cong \overline{F}^n$ (with $n=[K:F]$), which proves that $R^{(1)}\_{K/F}G\_K$ is a form of $G^{n-1}$, hence connected if $G$ is.


