Line bundles on fibrations - MathOverflow

Line bundles on fibrations

2011-08-18T07:42:40Z

Let $f:Y \to X$ be a flat morphism with positive dimensional fibers. Is it always true that line bundles that are trivial along each fiber are of type $f^*L$ for $L$ a line bundle on $X$?

Answer by a-fortiori for Line bundles on fibrations

2011-08-18T08:20:37Z

No. Let $X$ be an elliptic curve, $p\colon Y=X\times\mathbf P^1\to X$ the projection, $g\colon X\to X$ multiplication by $2$, and $f=gp$. Take a line bundle $M$ of degree $1$ on $X$. Then, $p^*M$ is trivial on the fibres of $f$. Suppose $p^*M\cong f^*L$ for some line bundle $L$ on $X$. Since $p^*$ is injective on $\mathrm{Pic}$ (see Hartshorne, Ex. III, 12.5), we have $M\cong g^*L$, but the degree of $g^*L$ is even, contradiction.

Answer by Georges Elencwajg for Line bundles on fibrations

2011-08-18T08:38:21Z

No.

Take $X=Spec(k[x^2,x^3])$, the cusp over the field $k$ , the trivial bundle $Y=X\times_k\mathbb A^1_k $ and the first projection $f=pr_X:Y\to X$ .
Every line bundle$M$ on $Y$ is trivial on the fibers, since said fibers are affine lines over the field $k$. However not every line bundle $M$ on $Y$ can be written $f^*L$ with $L$ a line bundle on $X$.
Here is why:

A ring $R$ is called semi-normal if whenever elements $a,b \in R$ satisfy $a^3=b^2$, you can conclude that there exists $r\in R$ with $a=r^2, b=r^3$ . The ring is then automatically reduced (Costa). This notion is due to Traverso and Swan.
[Strange condition, eh? For example, a normal domain $R$ is semi-normal: take $r=b/a\in Frac(R)$, which by normality must be in $R$ since it satisfies the integrality equation $r^2-a=0$. ]

A theorem of Swan then states that given a ring $R$, the map from $R$ to its polynomial ring $j:R\to R[T]$ induces a surjection $j^*:Pic(R)\to Pic(R[T])$ if and only if the reduced ring $R_{red}=R/Nil(R)$ is semi-normal. This proves the above claim about the cusp (and much more).

Bibliography:
a) Here is a nice, completely self-contained survey by Lombardi and Quitté on semi-normal rings. Its bibliography will lead you to the original articles by Traverso and Swan.
b) And there is another very nice survey by Vitulli.

Answer by rita for Line bundles on fibrations

2011-08-18T09:25:23Z

It is true with some extra assumptions. If $f$ is projective (EDIT: in fact proper is enough) and has connected and (EDIT) reduced fibers and $M$ is a line bundle that is trivial on every fiber, then $h^0(X_y, M)=1$ for every $y\in Y$. If $Y$ is integral, then it follows that $L:=f_*M$ is a line bundle and the natural map $f^*L\to M$ is an isomorphism.

EDIT: the argument works provided $h^0(X_y, {\mathcal O})=1$ for every $y$. So in some cases one can remove the assumption that all the fibers are reduced. For instance if $X$ is a smooth complex surface and $Y$ is a smooth curve, then by Zariski's lemma every fiber $X_y$ is either $1$-connected or $X_y=mD$, where $D$ a $1$-connected divisor and $D|_D$ is torsion of order $m$. Using the fact that $h^0({\mathcal O}_D)=1$ if $D$ is $1$-connected and applying induction, one gets $h^0(X_y, {\mathcal O})=1$ for every $y$.