triviality of determinant sheaf

On a smooth algebraic variety X, every coherent sheaf F has a finite resolution by locally free sheaves. Using such resolution, we can define the determinant of F, det F, which is a line bundle on X.

My question is :

why if the support of F is of codimension greater or equal to 2 is the determinant of F trivial ?

It is mentionned without proof on the book "The geometry of moduli spaces of sheaves", D. Huybrechts, M. Lehn. I have verified this result on some explicit examples for which I know some explicit locally free resolutions but I don't see how to do the general case.

• The best reference for these kinds of questions is the following article. MR0437541 (55 #10465) Reviewed Knudsen, Finn Faye; Mumford, David The projectivity of the moduli space of stable curves. I. Preliminaries on "det'' and "Div''. Math. Scand. 39 (1976), no. 1, 19–55. 14H10 (14F05 14C05) Aug 7, 2012 at 19:22

Outside the support of $F$, the resolution is an exact sequence, so the alternating tensor product of the determinants is trivial. On a smooth scheme, a line bundle trivial outside a codimension $2$ subset is trivial.
Just an idea, using the first Chern class which should live in the cohomology with support in Supp($F$), you should then get that $c_1(F) = 0$ which makes $F$ trivial since it's a line bundle (perhaps modulo linear equivalence).
• $c_1(F) = 0$ does not imply that $F$ is trivial: Consider an elliptic curve $E$. Set $X = E \times E$. Let $g : (x,y) \mapsto (x+1/2,-y)$ be an involution w/o fixed points on $X$. Then the canonical bundle of the quotient $Z = X / \langle g \rangle$ has zero first Chern class, but is not trivial since it has no non-zero sections. Aug 7, 2012 at 18:46