# smooth group scheme genrically abelien

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?

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If $S$ is any scheme and $j:U \rightarrow S$ is a quasi-compact open immersion (so $j_{\ast}$ applied to quasi-coherent sheaves commutes with flat base change) such that $j$ is schematically dense (i.e., $O_S \rightarrow j_{\ast}(O_U)$ is injective) then $S$-maps $f, g:X \rightrightarrows Y$ are equal if $X$ is $S$-flat, $Y$ is $S$-separated, and $f_U = g_U$. This is a good exercise. It follows that a flat separated $S$-group $G$ is commutative if $G_U$ is commutative (via Sawin's argument with the commutator morphism, upgrading "set" to "scheme" suitably). – user27056 Oct 20 '12 at 6:36

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.
This argument assumes via the consideration of the identity section as a closed immersion that $J$ is $S$-separated (and that $U$ is non-empty, which the OP clearly meant to assume). Without separatedness it is false over $S = {\rm{Spec}}(R)$ for any discrete valuation ring $R$. Indeed, choose a non-commutative finite group $G$ and let $J$ be the gluing along the open generic point $U$ of copies of $S$ indexed by $G$, equipped with the obvious $S$-group structure. This is quasi-finite etale over $S$ with $J_U$ trivial, but the special fiber is the non-commutative finite constant group $G$. – user27056 Oct 20 '12 at 6:02
Dear @Will, I'm not sure what $S$ is supposed to be smooth over in the question, but in any case, why is a scheme smooth over some base which is also connected necessarily irreducible? A quasi-compact, connected, regular scheme is definitely irreducible, but I believe there are smooth morphisms with the source (and necessarily the target) not regular (I could be wrong). And without some quasi-compactness hypothesis, having all local rings domains and being connected does not imply irreducibility. – – Keenan Kidwell Oct 21 '12 at 4:08
I assumed that $S$ was smooth over the spec of a field, since a base wasn't mentioned. In that case the implication is standard - if there are two different irreducible components, since it's connected they must meet at a point, and that point must have zero divisors in its local ring and thus me singular. If there is a base, and it's reducible, clearly the argument doesn't work. If it's irreducible I think the implication that $S$ is reducible still goes through. – Will Sawin Oct 21 '12 at 5:20
@Will: Did you mean to say "irreducible" at the end of your preceding comment? A smooth connected scheme over an irreducible base need not be irreducible. For example, the projective planar nodal cubic $y^2 = x^2(x-1)$ over an alg. closed field is irreducible but it admits reducible connected finite etale covers of every degree $n > 1$ (namely, the so-called $n$-gon of projective lines). Basically, the problem is that whereas normality is local for the etale topology, irreducibility is not (in the absence of normality). – user27056 Oct 21 '12 at 6:18