# Does the image of a section lie in the regular part

Say $f:X\to C$ is a family of curves. More precisely, $C$ is a smooth projective irreducible curve over a field, $f$ is a flat morphism of schemes and $X$ is a normal projective irreducible surface.

Say I take a section $P:C\to X$. Does the image of $P$ lie in the nonsingular part of $X$?

What conditions (weaker than regularity) on $X$ do I need? Does it help if every Weil divisor on $X$ is $\mathbf{Q}$-Cartier?

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Of course, as S\'andor points out, there are plenty of counterexamples where your conclusion fails. On the other hand, there are plenty of cases where your conclusion holds. For instance, if there exists a resolution of singularities of X such that the multiplicity of every exceptional divisor is strictly larger than $1$, then your condition does hold. This comes up quite naturally in studying sections of degenerations of Del Pezzo surfaces. So I suggest you might look at Alessio Corti's paper on this topic. –  Jason Starr Jan 10 '12 at 16:09

Example: Let $Y$ be the projective cone over a conic, so for instance, let $X$ be defined by $xz=y^2$ in the projective 3-space with coordinates $[x:y:z:w]$ and consider the projection to the $[x:w]$-axis: $$g:Y\dashrightarrow \mathbb P^1$$ $$[x:y:z:w]\mapsto [x:w]\quad\qquad$$ This is defined everywhere except at $[0:0:1:0]$, so let $X$ be the surface one obtains by resolving the indeterminacies of $g$ and $f:X\to \mathbb P^1$ the induced morphism.

This $X$ is a normal, projective, irreducible surface and $f$ is a dominant morphism onto $C=\mathbb P^1$. It follows that $f$ is flat. $X$ is also non-singular everywhere except at $[0:0:0:1]$ where it has an $A_1$-singularity which is one of the simplest surface singularities. In particular, it is $\mathbb Q$-factorial, i.e., every Weil divisor is $\mathbb Q$-Cartier.

There is also a section that goes through the singular point: $\sigma: \mathbb P^1\to X$ defined by $[x:w]\mapsto [xw^2:x^2w:x^3:w^3]$ which originally maps to $Y$, but lifts to $X$ by taking the strict transform of the image. Notice that the image of this section is $\mathbb Q$-Cartier.

Analysis: The above example shows that unfortunately what you want does not necessarily hold, even under pretty strict conditions. The only condition short of non-singularity I could imagine is to assume that $X$ is factorial, that is, every Weil divisor (or at least the section you are considering) is Cartier. This is of course very easy: the section is non-singular since it is isomorphic to $C$ and if a variety contains a smooth Cartier divisor, then it is smooth along that divisor.

One could prove this last statement in a little more complicated way that shows why this fails in the example above:
The intersection of the section and any fiber has to be $1$, because the projection is an isomorphism. If the section is Cartier, this means that the fiber cannot be singular at the intersection point, so not only $X$ is non-singular, but even $f$ is smooth along the section. The problem with this argument in the case when the section is only $\mathbb Q$-Cartier is that the intersection number of the section with a fiber could be $1$ and the fiber still be singular.
In the example, the section is a smooth curve, but it is not Cartier (this actually follows from the above, but it is usually proved by computing the self-intersection, which is $\frac 12$), but the fiber through the singular point of $X$ is singular and the section goes through that point. Twice the section is Cartier and its intersection with the singular fiber is $2$, so the intersection of the section with the fiber is $1$ and we get no contradiction.

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So if the section $P$ I consider is $\mathbf{Q}$-Cartier, then $nP$ is Cartier for some positive integer $n$. The support of $P$ is the support of $nP$, and we know that $nP$ is disjoint from the singular locus. Therefore, $P$ is disjoint from the singular locus. Here we use that if a variety contains a smooth Cartier divisor, then it is smooth along that divisor. Does that hold in more generality? That is, suppose that D is a Cartier divisor on a normal integral surface $X$ over a Dedekind scheme $S$. Assume that $D$ is smooth over $S$. Then $X$ is smooth over $S$ along $D$? –  Ali Jan 10 '12 at 7:15
No,no,no,no,no! $nP$ is not non-singular! It is not even reduced! It does not have a single non-singular point. Compare with the example at the beginning. On the other hand, if a Cartier divisor is non-singular, then the ambient variety is non-singular regardless of anything else. The proof of this is that if a local ring mod by a single non-zero divisor non-unit is regular, then the original local ring had to be regular by a simple dimension count. –  Sándor Kovács Jan 10 '12 at 7:45
My God. What a mistake!! Thank you for this! –  Ali Jan 10 '12 at 19:41