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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 $X$?

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For something like what you ask to be true you at least need to have geometrically connected and reduced fibres and probably also need $f$ to be proper. But I don't think that even this is enough. –  ulrich Aug 18 '11 at 8:26
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@ulrich: it isn't, take a non-trivial infinitesimal deformation: $X=k[T]/(T^2)$, $Y=Y_0\times X$ where $Y_0/k$ nice having $H^1(Y_0,\mathcal O_{Y_0})\ne0$ –  user2035 Aug 18 '11 at 8:41
    
@a-fortiori: Thanks for the example. (I was implicitly assuming the base is not too bad, say reduced or integral.) –  ulrich Aug 18 '11 at 10:40
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up vote 10 down vote accepted

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 $y$.

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I guess you are assuming that all fibres are reduced (and maybe also characteristic zero)? –  ulrich Aug 18 '11 at 10:44
    
@Ulrich: I think you are right about the reduced fibers, thank you very much. Does the characteristic really matter here? –  rita Aug 18 '11 at 11:10
    
I don't have a counterexample in nonzero characteristics but I also don't see why $f*_M$ is a line bundle without some additional assumptions. –  ulrich Aug 18 '11 at 11:56
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It is a line bundle if $h^0(X_y, M)=1$ for all $y$ by Corollary 12.9 in Hartshorne, `Algebraic geometry'. –  rita Aug 18 '11 at 12:10
    
Thanks for the reference. Given counterexamples with non-reduced fibres or base what you say seems to be the best possible. –  ulrich Aug 18 '11 at 12:40
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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.

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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.

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very nice . –  Qfwfq Aug 18 '11 at 14:04
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