Let \(\ts{X_t}\) be a flat family of closed subschemes of \(\PP_k^n\) parametrized by an irreducible curve \(T\) of finite type over \(k\). Suppose there is a nonempty open set \(U \subseteq T\) such that \(X_t\) is connected for all closed points \(t \in U\). Prove that \(X_t\) is connected for all \(t \in T\).
Solution
Let \[ \pi:\mathcal X\longrightarrow T \] denote the total family, so that \[ X_t=\mathcal X\times_T\Spec\kappa(t). \] Because \(\mathcal X\) is a closed subscheme of \(\PP_k^n\times T\), the morphism \(\pi\) is projective. It is flat by hypothesis and of finite presentation because all schemes involved are of finite type over \(k\).
1. For \(t\in T\), let\[ c(t) \]be the number of connected components of the geometric fibre\[ \mathcal X_{\overline t}. \]Then \(c:T\to\ZZ_{\ge0}\) is lower semicontinuous.
We give the Stein-factor argument in the form needed here. Fix \(t\in T\) and put \[ r=c(t). \] Take the Stein factorization \[ \mathcal X\xrightarrow{h}S\xrightarrow{g}T, \] where \[ S=\Spec_T(\pi_*\mco_{\mathcal X}), \] \(g\) is finite, and \(h\) has geometrically connected fibres.
More explicitly, after replacing \(T\) by an étale neighbourhood of \(t\), the points of the finite fibre of \(S\) corresponding to the \(r\) geometric connected components of \(\mathcal X_{\overline t}\) lie in pairwise disjoint open-and-closed pieces \[ V_1,\ldots,V_r\subseteq S. \] This is the standard local decomposition of a finite morphism around the points of one fibre; see the Stein-factor proof of lower semicontinuity.
Put \[ \mathcal X_i=h^{-1}(V_i). \] Each \(\mathcal X_i\) is open in \(\mathcal X\), hence \[ \mathcal X_i\longrightarrow T \] is flat and of finite presentation. A flat morphism locally of finite presentation is open, so its image \[ W_i=\pi(\mathcal X_i) \] is an open neighbourhood of \(t\). Therefore \[ W=W_1\cap\cdots\cap W_r \] is an open neighbourhood of \(t\).
For every \(u\in W\), each \(\mathcal X_i\) meets the fibre over \(u\). Since the \(V_i\) are pairwise disjoint open-and-closed subsets of \(S\), these intersections lie in distinct connected components of the geometric fibre \(\mathcal X_{\overline u}\). Hence \[ c(u)\ge r=c(t) \] for all \(u\) in an open neighbourhood of \(t\). This is precisely lower semicontinuity.
2. For every closed point \(u\in U\),\[ c(u)=1. \]
Because \(T\) is a finite-type curve over the algebraically closed field \(k\), every closed point has residue field \(k\). Thus \[ \mathcal X_{\overline u}=X_u. \] By hypothesis \(X_u\) is connected, so it has exactly one connected component.
3. No point \(t\in T\) can satisfy\[ c(t)\ge2. \]
Suppose that \(c(t)\ge2\) for some \(t\in T\). By lower semicontinuity from step 1, there is a nonempty open neighbourhood \[ W\subseteq T \] of \(t\) such that \[ c(w)\ge2 \] for every \(w\in W\).
The curve \(T\) is irreducible, so any two nonempty open subsets meet. Therefore \[ W\cap U\ne\varnothing. \] Since \(T\) is of finite type over a field, its closed points are dense; choose a closed point \[ u\in W\cap U. \] Then step 2 gives \[ c(u)=1, \] whereas the defining property of \(W\) gives \[ c(u)\ge2, \] a contradiction.
4. Every fibre \(X_t\) is connected.
Step 3 gives \[ c(t)\le1 \] for every \(t\in T\). The family has nonempty fibres everywhere. Indeed, \(\pi\) is projective, hence its image is closed, and it is flat of finite presentation, hence its image is open. The image contains the nonempty open set \(U\), so irreducibility of \(T\) forces \[ \pi(\mathcal X)=T. \] Thus every geometric fibre is nonempty and consequently \[ c(t)=1 \] for all \(t\).
A scheme whose geometric fibre is connected has connected ordinary fibre. Hence every \(X_t\) is connected.
5. Q.E.D.
Steps <1>1–<1>3 rule out the appearance of a second connected component under specialization, and step 4 supplies nonemptiness and concludes that every fibre has exactly one connected component.