Let J a smooth group scheme over a smooth connected base S.


I assume, that over an open subset U of S, J is a torus, do I have that J is abelian?



smooth group scheme genrically abelien
Yes. Consider the map: $J \times\_S J \to J$ that sends $(a,b)$ to $aba^{-1}b^{-1}$. The pullback of the zero section along this map is a closed set. It is exactly the set of commuting pairs. It includes the inverse image of $U$, which is nonempty open.


Since $S$ is smooth and connected it is irreducible, so since $J$ is a torus over $U$ it has only one irreducible component over $U$, and any other irreducible component of $J$ would have to live entirely over $U^c$ which is impossible since it's flat, so $J$ is irreducible, so the inverse image of $U$ is dense, so the set of commuting pairs is the whole set, so it's commutative.


Edit: Per xbnv's comment, yes only if it's separated.


