# Why is the number of irreducible components upper semicontinuous in nice situations?

Suppose $f: X \rightarrow Y$ is a flat projective morphism of finite type schemes over an algebraically closed field so that the fibers over the closed points of Y are (geometrically) reduced. Why is it true (or do I need some additional assumptions?) that the number of irreducible components of the geometric fibers is upper semicontinuous on Y?

I have asked Joe Harris and Allen Knutson about this statement, and they both believed it was true, but I have not been able to find a reference or a proof.

• The robust constructibility techniques in EGA IV$_3$ are what you are missing. For proper flat $f$ between noetherian schemes (fibral geometric reducedness is irrelevant, but see 12.2.4(v)), by constructibility considerations (9.7.8) and the generization criterion for a constructible set to be open we may assume $Y={\rm{Spec}}(R)$ for a discrete valuation ring $R$. The Zariski closure of a closed subset $Z$ of $X_{\eta}$ is now $R$-flat (!), so it has equidimensional non-empty (!) special fiber of dimension $d$ if $Z$ is so (14.3.10). The rest is now clear. (The result is not in EGA, AFAIK.) Dec 8 '15 at 7:45
• You might also check Jouanolou, "Th'eor`emes de Bertini". Although this is not a primary source, if there is no precise reference in EGA (yet the result does follow from EGA), you might prefer a precise reference in Jouanolou's book. Dec 8 '15 at 11:45
• @nfdc23 You need to use (geometric) reducedness somewhere -- otherwise, consider $\mathrm{Spec} \mathbb{C}[x^2] \to \mathrm{Spec} \mathbb{C}[x]$. The fibers over $x \neq 0$ have $2$ points, the fiber over $x=0$ has one point. Dec 8 '15 at 16:13
• Count Dracula's answer identifies the geometric gap in my remark above. Via references in that remark, it suffices to consider a dvr $R$ and proper flat $R$-scheme with reduced geometric fibers. There are no embedded points in geometric fibers (reducedness!), so the number of irreducible components in a geometric fiber is the sum of the "total multiplicity" at the generic points of the actual fiber in the sense of EGA IV$_2$ 4.7.12 (also see 4.7.4ff). Hence, IV$_3$ 12.2.4(ix) [real content is 12.2.1(xi)!] gives the result without projective methods (i.e., no Hilbert polynomials). Dec 9 '15 at 1:52
• @JasonStarr: This result is definitely not in the Bertini book by Jouanolou. As for the rigid-analytic case, that is an interesting question. I don't see offhand how to attack it using Raynaud's formal models (which interact well with conditions such as proper, flat, & geometrically reduced fibers). I suppose one could check if the EGA proof can be adapted; if that is not feasible then I recommend you email Ducros with this question, as it is right up his alley (he recently wrote a book all about making analogues of these EGA IV$_3$ techniques in non-archimedean geometry). Dec 9 '15 at 20:01

Here is an alternative to Count Dracula's (correct) argument that emphasizes instead the constancy of the Hilbert polynomial for a flat family of projective schemes.

As above, assume that $Y$ is a DVR. For one fixed irreducible component $Z_{\eta}$ of $X_\eta$ of minimal dimension $d$, denote by $Z$ the Zariski closure of $Z_{\eta}$ in $X$ together with its closed immersion $u:Z\to X$. Denote by $W_{\eta}$ the union of all other irreducible components of $X_{\eta}$, and denote by $v:W\to X$ the closure of $W_\eta$ in $X$. By construction, both $Z$ and $W$ are flat over $Y$, since every associated point is a generic point of $Z_\eta$, resp. $W_\eta$.

The closed immersion $u$ and $v$ determine an associated morphism of $\mathcal{O}_X$-modules, $$(u^\#, v^\#):\mathcal{O}_X \to u_*\mathcal{O}_Z \oplus v_*\mathcal{O}_W.$$ The restriction of $(u^\#,v^\#)$ on $X_\eta$ is injective. Thus the kernel of $(u^\#,v^\#)$ is a subsheaf of $\mathcal{O}_X$ that is torsion for $\mathcal{O}_Y$. Since $\mathcal{O}_X$ is flat over $\mathcal{O}_Y$, the kernel of $(u^\#,v^\#)$ is the zero sheaf.

Denote the quotient of $(u^\#,v^\#)$ by $\mathcal{Q}$. The restriction of $\mathcal{Q}$ on $X_\eta$ has support whose dimension is strictly smaller than the dimension of any irreducible component of $X_\eta$. In particular, the Hilbert polynomial of $\mathcal{O}_{X_\eta}$ agrees with the Hilbert polynomial of $\mathcal{O}_{Z_\eta}\oplus \mathcal{O}_{W_\eta}$ modulo the subspace of numerical polynomials of degree strictly less than $d = \text{dim}(Z_\eta)$.

Now consider the restriction $(u_0^\#,v_0^\#)$ of $(u^\#,v^\#)$ to the closed fiber $X_0$. By the flatness hypothesis, the Hilbert polynomials of the domain and target of this homomorphism equal the Hilbert polynomials on the generic fiber. Thus the difference of these Hilbert polynomials on $X_0$ equals the difference of these polynomials on $X_\eta$, and we know that this difference is a polynomial of degree strictly less than $d$. The cokernel of $(u_0^\#,v_0^\#)$ equals the restriction $\mathcal{Q}_0$. If the induced morphism $\mathcal{O}_{Z_0}\to \mathcal{Q}_0$ is nonzero at some generic point of $Z_0$, then the support of $\mathcal{Q}_0$ has an irreducible component of dimension $\geq d$. Thus the Hilbert polynomial of $\mathcal{Q}_0$ has degree $\geq d$. Since the difference polynomial has degree strictly less than $d$, the kernel of $(u_0^\#,v_0^\#)$ has Hilbert polynomial of degree $\geq d$ counterbalancing the Hilbert polynomial of $\mathcal{Q}_0$. In particular, the kernel of $(u_0^\#,v_0^\#)$ is not zero.

By the flatness hypothesis, every associated point of $X$ is contained in $X_\eta$. Thus, every generic point $\xi$ of $X_0$ is the specialization of a generic point that is either in $Z$ or in $W$. Thus the localization $\mathcal{O}_{X_0,\xi}$ either factors through $\mathcal{O}_{Z_0,\xi}$ or factors through $\mathcal{O}_{W,\xi}$. Therefore the kernel of $(u_0^\#,v_0^\#)$ is in the kernel of the localization at every generic point of $X_0$. Since the kernel is nonzero, $X_0$ has embedded associated points, contradicting the hypothesis that $X_0$ is geometrically reduced. Therefore, by way of contradiction, the support of $\mathcal{Q}_0$ does not contain $Z_0$. So $Z_0$ is not contained in $W_0$.

Now we continue by induction on the number of irreducible components, replacing $X$ by $W$.

OK, the comment of nfdc23 reduces the question to the case where the base is a discrete valuation ring. I also agree with what she says about closures, but I think there is a small part missing: why is the closure of an irreducible component of $X_\eta$ not contained in the closure of another irreducible component? For example, why can't it happen that $X_\eta$ is the union of a threefold and a point and $X_0$ just a threefold? (I suggest keeping this example in mind when reading below.)

I'm sure there is an easy solution to this, but I find it fun to deduce this from a result of Hartshorne about connectedness of punctured spectra. Namely, let $A, B \subset X$ be closures of irreducible components of $X_\eta$. (By the way, you can always first make a finite extension of the base dvr to make sure that the irreducible components of the generic fibre are geometrically irreducible.) Assume that $A_0 \subset B_0$ to get a contradiction. Let $x \in A_0$ be a generic point of an irreducible component. Consider the local ring $O_{X, x}$.

Case I. $\dim(O_{X, x}) = 1$. In this case $O_{X, x}$ is a dvr because $X_0$ is reduced. In this case it is clear that there is a unique point of $X_\eta$ specializing to $x$ and we get our desired contradiction.

Case II. $\dim(O_{X, x}) \geq 2$. Because $X_0$ is reduced we see that $O_{X, x}/\pi$ has depth at least $1$ where $\pi$ is the uniformizer of the base dvr $R$. Then Hartshorne's connectedness result shows that the punctured spectrum $U$ of $O_{X, x}$ is connected. But the generic point of $A$ is an isolated point of $U$ which is a contradiction unless the generic point of $B$ is the generic point of $A$ and we win.

• Dracula: Certain the threefold / point situation is ruled out by considering $H^0(X_\eta,\mathcal{O}_{X_\eta})$ and $H^0(X_0,\mathcal{O}_{X_0})$. I imagine that you can rule out similar cases with irreducible components of unequal dimension by taking hyperplane sections. Then for components of the same dimension, you should be able to use the degree to rule this out. Dec 8 '15 at 15:19
• Ah right, sorry for overlooking the issue of geometry of overlaps of irreducible components. My follow-up remark above gives the appropriate theorem in EGA that proves the result without projectivity hypotheses (just properness). Dec 9 '15 at 1:55

An expanded version of nfdc23's answer in the comments to this question can be found at http://arxiv.org/pdf/1601.05840v1.pdf, Proposition 2.9. An even more expanded version can be found at http://arxiv.org/pdf/1605.01117v1.pdf, Proposition 3.2.5.