MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer
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
My wording was weird so I should clarify. I mean to ask: why is a connected scheme which is smooth over some base necessarily irreduicble? – Keenan Kidwell Oct 21 '12 at 4:13
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.