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A useful criterion for triviality of a line bundle $\mathscr{L}$ on an integral curve $C$ is that the trivial line bundle is the unique line bundle of degree $0$ which admits a global section. This is useful e.g. in relative settings in order to characterize the locus in the base where a line bundle is fiber-wise trivial.

Can something similar be said in higher dimensions? In particular, let $\mathscr{L}$ be a line bundle on an integral variety $X$. If we assume that it admits a global section - so we can think of it as an effective divisor $D \subseteq X$ - can we show that if it is also "degree $0$", then it is trivial?

Here, the first natural generalization of "degree $0$" to higher dimension which comes to mind is the criterion of numerical triviality, but I would also be happy to use homological or algebraic triviality if it turns out that torsion is the issue. (This question came from a discussion about varieties over $\mathbb{C}$, but I would be happy to see a solution or counterexample in any characteristic!).

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    $\begingroup$ As long as $X$ is projective, this should be fine (even if $D$ is merely a Weil divisor), since the correct power of an ample line bundle should meet $D$ in a positive number of points. But if $X$ is proper with $\operatorname{Pic}(X) = 0$, then every irreducible codimension $1$ subscheme gives an effective divisor that is numerically trivial (in fact it's rationally equivalent to $0$). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 15, 2017 at 4:19
  • $\begingroup$ @R.vanDobbendeBruyn: That was my first thought to give a counter-example in the non-projective case, but if $\mathrm{Pic} X=0$, then every divisor is trivial, so the statement is actually true. $\endgroup$
    – Sándor Kovács
    Commented Feb 15, 2017 at 21:56
  • $\begingroup$ @SándorKovács But every variety has codimension $1$ points by dimension theory, so wouldn't those give nonzero elements of $\operatorname{Div}(X)$? But I agree that the class in $\operatorname{Pic}(X)$ is zero, which is how the question is formulated. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 16, 2017 at 7:28
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    $\begingroup$ @R.vanDobbendeBruyn, I think you answered your question. Obviously a non-zero effective divisor is not zero as a divisor (i.e., as an element of $\mathrm{Div} X$), but the question seemed to be about the triviality of the associated line bundle. If we interpret the question to ask whether the divisor is trivial, then the answer is that clearly it fails on any variety, including projective ones: take a non-zero principal divisor. It is numerically trivial and has degree $0$ according to any sane definition of those notions, yet not necessarily trivial as a divisor. $\endgroup$
    – Sándor Kovács
    Commented Feb 16, 2017 at 16:52
  • $\begingroup$ @SándorKovács I think I agree with your interpretation of the question. But a nonzero principal divisor on a projective variety cannot be effective (again: intersect with an ample divisor). So $\operatorname{Pic}(X) = 0$ really gives a weird phenomenon that we are not used to. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 16, 2017 at 19:09

2 Answers 2

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If $X$ is a smooth projective variety, then the line bundle is trivial if and only if $D$ is the zero divisor. That's because $L$ is trivial only if $H^{n-1} \cdot D =0$, where $H$ is an ample divisor and $n=\dim X$. For effective divisor $D$, the latter condition implies that $D$ is the zero divisor. This argument works for case that $X$ has some mild singularities.

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The first question is what one should mean by "degree $0$". As you put in the title, there is actually a notion of a Cartier divisor being numerically trivial, which means that the intersection number with any integral curve is zero. A way to define "degree" of irreducible subschemes of a projective variety is to choose an ample Cartier divisor, say $H$, and intersect the subscheme with the appropriate power of $H$ (the appropriate power is the co-dimension of the subscheme).

This actually gives you what you want: the degree you get this way is essentially what you get if you take a high enough power of the ample so it gives an embedding to projective space and then this degree is a positive constant multiple of the degree of the subscheme in the ambient projective space. Since that is never $0$ for a non-empty projective scheme, you have your statement. (And notice that this does not need the original variety to be smooth, only projective).

If $X$ is not projective, then this could get tricky. It is already not clear how to define "degree". I suppose you would need $X$ to be at least proper so this would be even reasonable to ask, but if it's not projective then there are absolutely no ample line bundles (Cartier divisors), so if you try to define degree using some other Cartier divisor, then you are bound to run into "degree $0$", but not trivial divisors.

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