Let $S$ be a base-scheme, for our purposes we can assume it is the spectrum of a field $k$. Given a smooth, proper and connected $S$-scheme $X$ and an integer $n\geq 1$, is then the fibre-product $X\times_{S}\mathbb{P}^{n}_{S}$ still connected? If not, are their any assumptions on $X$ respectively the base scheme $S$ that would guarantee connectedness?
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3$\begingroup$ The fibered product is just P^n_X, so it should be connected if X is. $\endgroup$– Mattia TalpoCommented Aug 25, 2010 at 13:49
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4 Answers
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This question is taking ridiculously many attempts to get a clean, correct answer.
First: As inkspot points out, a scheme is connected iff it has no idempotent global sections other than zero and one. Thus, whether a scheme is connected is determined purely by its ring of global sections.
Second: As Mattia Talpo and Charles Siegel both observe, $X \times_S \mathbb{P}^n_S$ is naturally isomorphic to $\mathbb{P}^n_X$.
Third: The global section ring of $\mathbb{P}^n_X$ is isomorphic to the global section ring of $X$. To see this, first observe that if $f \colon \mathbb{P}^n_X \to X$ is the obvious morphism, then $$\Gamma(\mathbb{P}^n_X, \mathcal{O}_{\mathbb{P}^n_X}) \cong \Gamma(X, f_{*}\mathcal{O}_{\mathbb{P}^n_X}).$$ This follows from the definition of $f_{*}$. Thus, it suffices to show that $f_* \mathcal{O}_{\mathbb{P}^n_X} = \mathcal{O}_X$. For this, we can work locally on $X$. If $U \subset X$ is an open affine, then our desired statement $f_* \mathcal{O}_{\mathbb{P}^n_U} = \mathcal{O}_U$ is true by standard facts about projective space over an affine scheme.
Conclusion: $X \times_S \mathbb{P}^n_S$ is connected iff $X$ is. This statement requires no hypotheses on either $X$ or $S$ (not even local Noetherianness).
answered Aug 25, 2010 at 18:39
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$\begingroup$ BCnrd's answer was perfectly clean and correct. $\endgroup$– AngeloCommented Aug 26, 2010 at 4:09
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$\begingroup$ I think I wrote that in large part out of a sense of frustration that all of the actual answers so far have required corrections in the comments. $\endgroup$ Commented Aug 26, 2010 at 15:43
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A scheme $Z$ is connected if and only if the only idempotent $f\in H^0(Z,\mathcal O_Z)$ are $f=0,1$. Since $H^0(S,\mathcal O_S)=H^0(S\times Y,\mathcal O_{S\times Y})$ for any proper $k$-variety $Y$ such that $k$ is algebraically closed in $H^0(Y,\mathcal O_Y)$, it follows that $S$ is connected if and only if $S\times Y$ is so.
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$\begingroup$ Your statement $H^0(S, \mathcal{O}_S) = H^0(S \times Y, \mathcal{O}_{S \times Y})$ for any proper $k$-variety $Y$ is not true if $k$ is not algebraically closed. For instance, there are proper (even projective) varieties over $\mathbb{R}$ having global sections $\mathbb{C}$. However, the statement should hold if $Y$ is, specifically, a projective space. $\endgroup$ Commented Aug 25, 2010 at 18:22
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$\begingroup$ Charles, you're quite right, thank you. I've added an appropriate hypothesis. $\endgroup$– inkspotCommented Aug 25, 2010 at 18:29
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$\begingroup$ Charles, this is a good illustration of why "$k$-variety" should never be used with clarifying the intended meaning: no one knows if (in addition to "finite type and separated") it means "irreducible and reduced", or "geometrically integral over $k$", etc. If one takes the definition in Hartshorne, namely to require geometric integrality (with any $k$) then what inkspot originally wrote is correct. Inkspot's new condition that $k$ be algebraically closed in the $k$-finite domain ${\rm{H}}^0(Y,O_Y)$ is the same as the more concrete statement ${\rm{H}}^0(Y,O_Y) = k$ (via the canonical map). $\endgroup$– BCnrdCommented Aug 26, 2010 at 1:22
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$\begingroup$ BCnrd: Your comment took me a bit to parse, so let me see if I understand it. Let's define a variety over an algebraically closed field to be a separated integral scheme of finite type over $k$. I have never actually seen a scheme-theoretic definition of variety over a non-algebraically closed field, but in my comment, I was assuming this still means "separated integral scheme of finite type over $k$." You point out that if one instead requires that the pullback to $\bar{k}$ be a variety over $\bar{k}$, then inkspot's original statement was correct. $\endgroup$ Commented Aug 26, 2010 at 15:55
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The product is just $\mathbb{P}^n_X$ (because it is $X\times_S (S\times\mathbb{P}^n_\mathbb{Z})$. The fibers then of $\mathbb{P}^n_X\to X$ are projective spaces, and thus connected, so all you need is $X$ is connected.
answered Aug 25, 2010 at 13:54
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7$\begingroup$ Not every continuous surjective map with connected fibers onto a connected base has connected source; must use that the map to the base is either open (since flat and locally of finite presentation; concretely in this case it is Zariski-locally an affine space, for which life is easy) or closed (because of properness). $\endgroup$– BCnrdCommented Aug 25, 2010 at 14:24
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1$\begingroup$ A natural place where this happens is for instance the case on the smooth stratification of the union $X$ of the two axes in the affine plane. Obviously the scheme $X$ is connected; on the other hand, we can look at the stratification of $X$ into smooth strata, and in this case we find the complement of the origin in $X$ and the origin. Clearly there is a surjective morphism from the disjoint union of the two strata to $X$, each fiber is connected (and consists of a single point), and the target is connected. Clearly again, the whole space is not connected! $\endgroup$– damianoCommented Aug 25, 2010 at 14:50
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1$\begingroup$ Wouldn't noting that the map $\mathbb{P}^n_X \to X$ admits a section suffice? $\endgroup$ Commented Aug 25, 2010 at 19:45
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To second BCnrd's remark about the need to have assumptions on the surjection, this is a purely topological issue that can occur whether you're looking at schemes or not. For instance, the projection of $$ [0,1) \times \{0\}\ \cup\ [1,2]\times \{1\} $$ on the first factor has both a connected base and connected source (but is not a proper map).
answered Aug 25, 2010 at 14:49