# Generic Smoothness Type of Results in Positive Characteristic

Let $f:X\to Y$ be a surjective morphism between two projective varieties over a field of characteristic $p>0$. Also assume that $f_*\mathcal{O}_X=\mathcal{O}_Y$, and $X$ is smooth.

We know that the general fiber of $f$ is not smooth in general. But can we say that the general fiber is an integral scheme?

Maybe this is too much to ask for an arbitrary morphism $f$; are there any known classes of morphsims for which this property holds? What if $f$ is given by an Iitaka fibration of a nef line bundle, does it make any difference?

$\textbf{Note}:$ It is known that if the $\textit{generic}$ fiber of $f$ is $\textit{geometrically integral}$, then the general fiber is an integral scheme. So my question basically reduces to asking what are some known classes of morphisms with geometrically integral generic fibers?

• If $X$ is smooth, then over a dense Zariski open subset of $Y$ the geometric fibers are LCI, hence Cohen-Macaulay. Also, if the natural map $\mathcal{O}_Y\to f_*\mathcal{O}_X$ is generically an isomorphism, then the geometric generic fiber is generically reduced. Hence the geometric generic fiber is (everywhere) reduced. Feb 12, 2016 at 10:11
• Also, because the geometric generic fiber is $S2$, this is really about the codimension $1$ part of the singular locus of the morphism. So, via Bertini, etc., you may as well assume that $X$ is a surface and $Y$ is a curve. Feb 12, 2016 at 10:51
• Actually, as updated below, there are examples where the geometric generic fiber of $f$ is everywhere nonreduced. Feb 12, 2016 at 19:20

Correction. I just realized that there are examples where the geometric generic fiber is NOT generically reduced. In all of my comments and the answer below, I was assuming that the geometric generic fiber is generically reduced. When that is true, then the geometric generic fiber is integral. However, without the hypothesis (or some other hypothesis that implies this hypothesis), the argument below only proves that the geometric generic fiber is an irreducible, LCI scheme. It may be everywhere nonreduced. There is a numerical condition that will guarantee this. For any ample divisor class $A$ on $X$, and for any ample divisor class $B$ on $Y$, if the intersection number $A^{\text{dim}(X)-\text{dim}(Y)}.(f^*B)^{\text{dim}(Y)}$ is prime to $p$, then the geometric generic fiber of $f$ is generically reduced. I noticed the counterexamples below by looking for examples where $p$ divides this integer.

Original post. For a normal, integral scheme $X$, for an integral scheme $Y$, for a proper, locally finitely presented morphism $f: X\to Y$, if the natural map $f^\# : \mathcal{O}_Y\to f_*\mathcal{O}_X$ is an isomorphism, then the geometric generic fiber of $f$ is integral and LCI (thus Cohen-Macaulay), but quite possibly not $R1$, i.e., not normal.

Denote by $K(Y)$ the function field of $Y$. Denote by $X_\eta$ the fiber product of $f$ and the natural morphism $i:\text{Spec}\ K(Y) \to Y$. Since $X$ is a normal scheme and $X_\eta$ is obtained by localization, also $X_\eta$ is a normal $K(Y)$-scheme. Also the natural map $K(Y)\to H^0(X_\eta,\mathcal{O}_{X_\eta})$ is an isomorphism since it is the generic fiber of the isomorphism $f^\#$.

Denote by $K(Y)^{\text{sep}}/K(Y)$ the separable closure of $K(Y)$. Denote by $X_{\eta^s}$ the base change of $X_\eta$ to $K(Y)^{\text{sep}}$. Since $K(Y)^{\text{sep}}/K(Y)$ is a limit of etale extensions, the same holds for $X_{\eta^s}\to X_\eta$. In particular, $X_{\eta^s}$ is a normal scheme. Since the base change is flat, also $K(Y)^{\text{sep}} \to H^0(X_{\eta^s},\mathcal{O}_{X_{\eta^s}})$ is still an isomorphism. Thus, $X_{\eta^s}$ is still a normal, integral $K(Y)^{\text{sep}}$-scheme.

Finally, the field extension to the algebraic closure $\overline{K(Y)}/K(Y)^{\text{sep}}$ is purely inseparable. Thus for the corresponding base change $X_{\overline{\eta}}$ of $X_{\eta^s}$, the morphism $X_{\overline{\eta}} \to X_{\eta^s}$ is also purely inseparable. In particular, it is a homeomorphism (with respect to the Zariski topologies). Since $X_{\eta^s}$ is irreducible, also $X_{\overline{\eta}}$ is irreducible. As explained in my comments above, $X_{\overline{\eta}}$ is LCI and generically reduced, hence (everywhere) reduced. Thus, the geometric generic fiber $X_{\overline{\eta}}$ is integral and LCI. However, there are examples (like quasi-elliptic fibrations) where $X_{\overline{\eta}}$ is not $R1$.

Edit. The OP did not ask this, but I (and my advisees) have occasionally found it useful. In the setting above, since the divisorial part of the closed subscheme $X_{\eta}^\text{sing} \subset X_{\eta}$ (cut out by the Fitting ideal of the sheaf of relative differentials, for example) must be inseparable over $K(Y)$, this forces certain numerical invariants to be divisible by the characteristic $p$. So if you study varieties with those numerical invariants equal to a specified (nonzero) integer, if you exclude "small characteristics" that divide that integer, then $X_{\overline{\eta}}$ will be normal.

Counterexample to Generic Reducedness. Let $k$ be a field of characteristic $p$. Let $Y$ be $\mathbb{P}^2_k = \text{Proj}\ k[s_0,s_1,s_2]$. Let $T$ be $\mathbb{P}^2_k = \text{Proj}\ k[t_0,t_1,t_2]$. Let $X\subset T\times_{\text{Spec}\ k} Y$ be the closed subscheme $\text{Zero}(t_0^ps_0 + t_1^ps_1 + t_2^ps_2)$. The projection $\text{pr}_T:X\to T$ is a smooth, projective morphism; it is Zariski locally a $\mathbb{P}^1$-bundle. Define $f:X\to Y$ to be the other projection. This morphism is nowhere smooth. The geometric generic fiber is everywhere nonreduced.
The argument above does prove that the geometric generic fiber is irreducible. Moreover, if the geometric generic fiber is generically reduced, then it is everywhere reduced. However, there are examples where the geometric generic fiber is everywhere nonreduced.

• Dear Prof. Jason Starr, this is an excellent and concrete answer; Thank you so much! It will be enormously useful in my current research. Feb 12, 2016 at 15:10
• Ok, general fiber is being irreducible is still quite a good property to have. I hope to be able to use it in our advantage. Feb 13, 2016 at 6:03