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What is the most general version of the connectedness principle in algebraic geometry? In particular, I'm interested in cases where there is no field available (eg, $Y$ below is the spectrum of something like $\mathbb{Z}_p$), or if it is then it isn't algebraically closed.

In his algebraic geometry book, Hartshorne gives the following version (ex III.11.4):

Let $k = \bar{k}$ and $\{X_t \}$ be a flat family of closed subschemes of $P^n_k$ parametrized by an irreducible curve $T$ of finite type over $k$. Suppose there is a nonempty open set $U \subset T$ such that for all closed points $t \in U$, $X_t$ is connected. Then $X_t$ is connected for all $t\in T$.

You can also restate this in terms DVR's

Let $X\rightarrow Y$ be a proper faithfully flat map with $Y$ the spectrum of a DVR. Let $y$ be the generic point of $Y$ and assume that $\dim_{k(y)} H^0(X_y, \mathcal{O}_{X_y}) = 1$. In particular, this condition implies that the generic fiber is connected. Then the special fiber is connected.

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EGA III$_1$, 4.3.4 (note the geometric connectedness). –  user30379 Jan 8 '13 at 20:25
    
Your second condition does not appear to require a field, nor an algebraically closed one. What sort of generalization do you want? –  Will Sawin Jan 8 '13 at 20:44
    
The geometric connectedness is of course what the weird dimension condition is achieving. –  Will Sawin Jan 8 '13 at 20:46
    
Will, I don't have a specific application in mind. –  LMN Jan 8 '13 at 21:20
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@LMN: Yes, it is the right reference. The point is that if you can prove a map is its own Stein factorization (say by using normality conditions together with some knowledge of a generic fiber) then all fibers are geometrically connected. I assure you from experience that this is the manifestation of Zariski's connectedness principle that is most widely used in algebraic geometry. –  user30379 Jan 9 '13 at 3:07
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Here are two relevant counterexamples:

Let $R$ be a ring and $I$ an ideal, then the morphism $\operatorname{Spec} R[x]/(x-x^2,Ix) \to \operatorname{Spec} R$ has connected fibers outside $V(I)$ but disconnected fibers inside $V(I)$. This suggests that some kind of flatness condition is unavoidable.

Let $R$ be a non-Henselian local ring and let $f$ be an irreducible polynomial that factors into coprime polynomials modulo the maximal ideal. Then $\operatorname{Spec} R[x]/f(x) \to \operatorname{Spec} R$ has connected generic fiber but disconnected special fiber. This suggests that some sort of geometric connectedness condition fro the generic fiber is unavoidable.

But we do not always need the generic fiber to be "completely" geometrically connected. Let $Y=\operatorname{Spec} k[[x]]$, then if the generic fiber of a flat proper morphism $f: X \to Y$ is connected, then the special fiber is connected. This is because $f_* \mathcal O_X$ is a finite flat algebra. After inverting $x$, and ignoring nilpotents, it injects into a finite field extension of $k((x))$, which must be the field of fractions of a complete local ring, so it itself is a complete local ring.

Presumably we should formulate the most general version locally, with $Y$ the spectrum of a local ring $R$, for simplicity. A sufficient flatness condition is that the morphism is flat, or that $f_* \mathcal O_X$ is flat. A sufficient connectedness condition is that the fiber over the "Henselian generic point", the spectrum of the field of fractions of the Henselization or completion of $R$, is connected.

We can see this by reducing the connectedness of $X$ to the connectedness of the second half of Stein factorization via Zariski's connectedness theorem. We reduce to the case where $R$ itself is Henselian by completing. Then $f_* \mathcal O_X$ is an $R$-algebra which injects into a connected algebra over the field of fractions of $R$, so modulo nilpotents, which we can safely ignore, it injects into a finite field extension of $R$.

Assume that the special fiber is disconnected, that is, $f_* \mathcal O_X/m$ contains an idempotent mod $m$. Lift that idempotent to $f_* \mathcal O_X$, and compute its minimal polynomial. This is irreducible, so by Hensel's Lemma it has a unique root mod $m$, which must be either $0$ or $1$ since $x^2-x$ divides the minimal polynomial mod $m$, so the idempotent is equal to either $0$ or $1$, a contradiction.

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Will: the beef in your argument remains Stein factorization and the geometric connectedness of its fibers (all non-empty). So to say we "do not always need the generic fiber to be `completely' geometrically connected" is somewhat misleading: one is (as you indicated) really disentangling the entire connectedness analysis into two steps, namely the Stein factorization step and the "finite covering" step. It is a waste of time to try to formulate some "most general" theorem along these lines. The result for Stein factorization is the real crux, and the rest is trivial variations on it. –  user30379 Jan 9 '13 at 3:11
    
I agree that Stein factorization is the only hard part. My goal was to create a stronger version of LMN's second statement, that he might be happy with, since he did not seem to feel like Stein factorization was an answer to this question. But you can certainly recast this as an explanation of how to actually apply Stein factorization to a very general case, thus demonstrating the importance of Stein factorization. I am not an expert in these matters and do not know what is most important, just that this argument is in fact accurate. –  Will Sawin Jan 9 '13 at 4:26
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A good, general presentation of Stein factorization and related topics covering even the non-noetherian case can be found in the Stacks Project (stacks.math.columbia.edu). To me the most remarkable fact is, that the finite morphism appearing in the Stein factorization is the normalization of the base scheme $Y$ in $X$. Thus the obstructions to getting geometrically connected fibres lie in the behavior of the normalization. This is particularly useful if for example $X$ is itself normal and $Y$ is affine. –  Hagen Jan 9 '13 at 8:38
    
@Hagen: What definition of "normalization of the base scheme $Y$ in $X$" are you using when $Y$ is reduced and $X$ is non-reduced? Or are you assuming both are reduced? –  user30379 Jan 9 '13 at 15:37
    
@ pravnak: the normalization is defined in a general setting in definition 24.48.3 of the Stacks Project ( stacks.math.columbia.edu/tag/035E ). As you see no reducedness assumption is necessary. –  Hagen Jan 10 '13 at 7:54
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