Given a semi-abelian scheme, is the set of points such that the fibres are abelian varities open?

Let $$\pi:\mathcal{A}\rightarrow C$$ be a semi-abelian scheme, i.e. $$\mathcal{A}$$ is a smooth separated commutative group scheme over $$C$$ via $$\pi$$ with geometrically connected fibres, such that each fibre $$\mathcal{A}_v$$, where $$v\in C$$, is an extension of an abelian variety $$\mathcal{B}_v$$ by a torus $$T_v$$ over the residue field $$\kappa(v)$$, or equivalently it is a semi-abelian variety.

I would like to ask if the set $$\{v\in C:\mathcal{A}_v\textrm{ is an abelian variety}\}$$ is open. In particular where $$C$$ is a smooth, projective, geometrically integral curve over a perfect field of characteristic $$p$$, and the set contains the generic point of $$C$$.

I would like to use the constructibility results from EGA, but all of them relies on the condition that "the morphism is of finite presentation", which I can't show. To be clear, it is actually the property of "quasi-compactness" of $$\pi$$ which I can't show, with quasi-compactness then I can show finite presentation hence the above set is constructible, and it contains the generic point so it contains a dense open subset. On a curve its complement is a finite set of closed points, so itself must be open.

Update: I have found a lemma for it, which is [SGA3, Expose VI B, Corollary 5.5], it says that:

Let $$G$$ be a group scheme over $$S$$, if $$G$$ is locally of finite presentation and universally open over $$S$$ with connected fibres, then $$G$$ is separated and of finite presentation over $$S$$.

The universally openness can be replaced by flatness because "locally of finite presentation" + "flat" imply "universally open", see tag 01UA.

• Welcome new contributor. For every field $k$, for every connected group $k$-scheme $(G,m:G\times_{\text{Spec}\ k} G \to G)$, for every nonempty open affine $U$, the product morphism $m:U\times_{\text{Spec}\ k} U \to G$ is surjective. Thus, for every affine scheme $S$, for every smooth group $S$-scheme $(\pi:G\to S,m:G\times_S G \to G)$ with geometrically connected fibers, for every open affine $U$ of $G$ that surjects to $S$, the morphism $m:U\times_S U \to \pi^{-1}(V)$ is surjective. Thus $\pi^{-1}(V)$ is quasi-compact. Mar 22 '19 at 20:46
• @JasonStarr Thank you! I understand the part until $m: U\times_k U\rightarrow G$ is surjective. But after that I can barely follow. Here is what I think which is probably what you mean, pick $V$ open affine in $C$, $\pi^{-1}(V)$ is a smooth group $V$-scheme with geometrically connected fibers, then find an open affine subset $U$ of $\pi^{-1}(V)$ such that sujects to $V$, then the morphism $m:U\times_V U\rightarrow \pi^{-1}(V)$ is surjective. The first point is how do we find an open affine subset $U$ surjects to $V$, the second is that why is the final morphism surjective?
– Z Wu
Mar 22 '19 at 21:16
• Welcome new contributor. Every point of a scheme is contained in an open affine. Thus, for every point $s$ of $S$, for every point $g$ of $G$ mapping to $s$, there exists an open affine containing $g$. Since $G$ is smooth over $S$, the image in $S$ of this open affine is open. Now choose $V$ to be an open affine neighborhood of $s$ in this open image, and choose $U$ to be the inverse image of $V$ in the original open affine. To see that $G\times_S V$ is quasi-compact, consider the inverse image of any open covering in the affine scheme $U\times_V U$. Mar 22 '19 at 21:27
• I bascially understood but why is $U\times_V U$ affine? We have $V$ is affine and $U$ is the intersection of preimage of $V$ and the original open affine, I don't see how is the fiber product affine.
– Z Wu
Mar 22 '19 at 21:46
• Welcome new contributor. If $V$ is $\text{Spec}(B)$ and if $U$ is $\text{Spec}(A)$, the $U\times_V U$ is $\text{Spec}(A\otimes_B A)$. Mar 22 '19 at 22:04

Let $$k$$ be a field. Let $$(G,m:G\times_{\text{Spec}\ k}G \to G)$$ be a locally finitely presented group scheme over $$\text{Spec}\ k$$. For every open $$U$$, denote by $$m_U$$ the restriction of $$m$$, $$m_U:U\times_{\text{Spec}\ k}U \to G.$$

Lemma 1. If $$G$$ is connected, then for every nonempty open $$U$$, the morphism $$m_U$$ is surjective. In particular, $$G$$ is quasi-compact.

Proof. The identity component of every group scheme is geometrically irreducible. Since $$G$$ is connected, it is geometrically irreducible. Thus, for an algebraically closed field extension $$L/k$$, for every $$L$$-point $$g$$ of $$G$$, the base change open $$U_L\subset G_L$$, its inverse $$U_L^{-1}$$, and its translate $$g\cdot U_L^{-1}$$ are each dense opens in $$G_L$$. Therefore, the intersection of $$g\cdot U_L^{-1}$$ and $$U_L$$ is nonempty. So there exists $$g\cdot h^{-1}\in g\cdot U_L$$ and $$h'\in U_L$$ that are equal, i.e., $$g=h\cdot h'$$ for $$(h,h')\in U_L \times_{\text{Spec}\ L}U_L.$$ QED

Lemma 2. For every flat, locally finitely presented morphism of schemes, every point of the domain has an open affine neighborhood that is fppf over an open affine neighborhood of the image point in the target.

Proof. Without loss of generality, assume that the target and the domain are affine. Thus the morphism corresponds to ring homomorphism, $$\phi:B\to A.$$ Since the morphism is flat and locally finitely presented, the image is open in $$\text{Spec}(B)$$. A basis for the topology on $$\text{Spec}(B)$$ consists of distinguished opens $$\text{Spec}(B_f)$$. Thus, the image open equals the union of all distinguished opens $$\text{Spec}(B_f)$$ that it contains. For each such, the following ring homomorphism gives an fppf morphism of schemes, $$\phi_f:B_f \to A_f.$$ QED

Let $$S$$ be a scheme. Let $$(\pi_S:G_S\to S, m_S:G_S\times_S G_S \to G_S, i_S:G_S\to G_S,e_S:S\to G_S),$$ be a flat, locally finitely presented group scheme over $$S$$.

Proposition 3. If the fibers of $$\pi_S$$ are connected, then $$\pi_S$$ is finitely presented.

Proof. It suffices to prove this locally on the target. For every point $$s$$ of $$S$$, for the image point $$e_S(s)$$, by Lemma 2 there exists an open affine neighborhood $$V$$ of $$s$$ in $$S$$ and an open affine neighborhood $$U$$ of $$e_S(s)$$ in $$G_S$$ such that the morphism $$U\to V$$ is fppf.

By Lemma 1, the following morphism is surjective on fibers over geometric points of $$V$$, $$m_U: U\times_V U \to V \times_S G_S.$$ Since $$U$$ and $$V$$ are affine, also $$U\times_V U$$ is affine. Thus, $$U\times_V U$$ is quasi-compact. Since $$V\times_S G_S$$ is the image of a surjective morphism from a quasi-compact scheme, also $$V\times_S G_S$$ is a quasi-compact scheme. QED

• In proposition 3, it seem one didn't need to "prove this locally on the target" since you didn't use this condition. Also, why does the surjectivity on fibers over geometrical points of $V$ implies the surjectivity of $m_U$.
– Z Wu
Mar 23 '19 at 12:15
• @LanceWu. Quasi-compactness of a morphism can be checked after replacing the target by each open in an open affine covering of the target (and replacing the domain by the inverse image open). Surjectivity of a morphism can be checked on geometric points. Every geometric point of $G$ maps to a geometric point of $S$. Thus, it suffices to check surjectivity after restricting to the fiber in $G_S$ over every geometric point of $S$. Mar 23 '19 at 14:00
• In lemma 1, can you eleborate more on the part $h\in U_L^{-1}$, I think it is a bit handwaving there, I know they are suppose to be $L$-rational points considering the morphism $G_L\times_L G_L\rightarrow G_L$ induces the mulitplication map $G_L(L)\times G_L(L)\rightarrow G_L(L)$. Can we show that if $g\cdot {U_L}^{-1}\cap U_L$ is non-empty then it contains a $L$-rational point? I guess this is where algebraic closure comes in. Also the lemma seems only prove the surjectivity of geometrical points.
– Z Wu
Mar 23 '19 at 14:43
• @LanceWu. Surjectivity of a morphism of schemes can be checked on geometric points. Mar 23 '19 at 14:47