Connectedness principle in algebraic geometry - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T12:59:21Z http://mathoverflow.net/feeds/question/118387 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118387/connectedness-principle-in-algebraic-geometry Connectedness principle in algebraic geometry LMN 2013-01-08T19:37:36Z 2013-01-22T23:22:00Z <p>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.</p> <p>In his algebraic geometry book, Hartshorne gives the following version (ex III.11.4):</p> <p>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 $t\in T$.</p> <p>You can also restate this in terms DVR's</p> <p>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.</p> http://mathoverflow.net/questions/118387/connectedness-principle-in-algebraic-geometry/118404#118404 Answer by Will Sawin for Connectedness principle in algebraic geometry Will Sawin 2013-01-08T23:11:16Z 2013-01-08T23:18:37Z <p>Here are two relevant counterexamples:</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p>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 $R$. </p> <p>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.</p>