Let $f:X\rightarrow \mathbb{P}^2$ be a fibration, here $X$ is a projective variety of dimension three.

Assume that there exixts a smooth curve $C\subset\mathbb{P}^2$ such that for any $p\in\mathbb{P}^2\setminus C$ the fiber $f^{-1}(p)$ is a smooth curve and when $p\in C$ then $f^{-1}(p) = A_p\cup B_p$ where $A_p, B_p$ are smooth curves intersecting transversally in a point $x_p = A_p\cap B_p$.

In this situation what could be said on the singularities of $X$? For instance deos this imply that $X$ is singular at most in the points of the form $x_p$ and that the locus of these points is a locus of at worst canonical singularities for $X$ ?

  • $\begingroup$ If $f:X\to \mathbb P^2$ is flat, then (under your assumptions) the morphism $X\to \mathbb P^2$ is smooth at all $x\in X$ such that $x\neq x_p$ with $p\in C$. In particular, $X$ is nonsingular at those points. $\endgroup$ Apr 14 '17 at 20:51

This is true. Actually, even better, these singularities will be terminal and Gorenstein, so as mild as it can get. Well, at least if we assume that you are working over an algebraically closed field, but otherwise you would have to be more careful about what exactly do you mean by these assumptions and questions, so I'll assume that that's what you meant.

Your assumptions imply that all the fibers are Gorenstein. Since $\mathbb P^2$ is smooth, this implies that then $X$ is Gorenstein. Furthermore, your assumptions imply that all the fibers are $1$-dimensional, so $f$ is equidimensional, and hence it is flat by "miracle flatness". So, as Ariyan noted, then $f$ is smooth at every $x\in X$ such that $x\neq x_p$ and hence $X$ is smooth at all of those points. This also implies that $X$ is normal (it is $S_2$ since it is Gorenstein and its singular locus has at most codimension $2$, so it is also $R_1$).

Now, looking at one of these $x_p$'s, we still have that $X$ is Gorenstein and we know that a complete intersection curve through this point is a simple node. As an exercise, try to prove directly (say via a direct local computation) that this implies that then $X$ is canonical (of index $1$) at these points.

If you get stuck, then use this argument: Let $C_1$ and $C_2$ be two smooth curves in $\mathbb P^2$ that intersect transversally at $p$ and let $D_i=f^*C_i\subseteq X$ for $i=1,2$. (For the record, the $D_i$ are reduced divisors on $X$.) Now, $D_1\cap D_2 =f^{-1}(p)$, which is a nodal curve, so it has slc singularities. Then by inversion of adjunction (applied twice) $(X,D_1+D_2)$ is log canonical. Now if you scrape away the $D_i$, this actually means that $X$ is terminal at the points of the form $x_p$. (Note that $(D_1, D_1\cap D_2)$ is also log canonical, so $D_1$ is canonical. It is locally isomorphic to a cone over a quadric.)

  • $\begingroup$ Thanks a lot for the answer. In my case $X$ is a complete intersection $3$-fold in $\mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^2$. So $X$ is Gorenstein. I checked that all the fibers of $f$ have dimension $1$ so from what you said $f:X\rightarrow \mathbb{P}^2$ is flat. But I just found a reducible fiber with a non reduced component (of multiplicity 2). How can this happen? In this situation shouldn't all the fibers be reduced? $\endgroup$
    – TopGatLu
    Apr 15 '17 at 17:24
  • $\begingroup$ OK, so your fibers are not exactly as you originally described, but having reduced components in the fibers is not unusual. For instance, if you resolve the indeterminacy of the map $(x,y)\mapsto {y^2}/x$, you get exactly that behavior. $\endgroup$ Apr 16 '17 at 1:26

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