# Is the pushforward of a line bundle on the smooth locus of a terminal singularity again a line bundle?

In algebraic geometry, it is a sad fact of life that pushforward doesn't preserve being a coherent sheaf; for example, the pushforward by the complement of a divisor of the structure sheaf (or more generally a line bundle) has essentially no hope of being again coherent. On the other hand, on a smooth variety, if I pull out a closed subset of codimension $\geq 2$, then the pushforward of the structure sheaf will be be the structure sheaf of the whole thing, essentially by Hartog's theorem.

Now, this is not true for singular varieties; you can have an affine inclusion of varieties where the complement has codimension greater than 1, like removing the singular point from singular plane quadric.

I'm generally curious about when "not too bad" singularities can avoid this problem; I'm particularly interested in whether this works for removing the singular locus of a terminal variety, but would be interested to hear other results along the same lines.

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You probably know this, but it warrants pointing out. Suppose that $X$ is a normal variety. Set $U = X \setminus \text{Sing X}$ with the natural inclusion $i : U \to X$, and pushforward the structure sheaf, then $i_* \mathcal{O}_U = \mathcal{O}_X$. It just doesn't work for arbitrary line bundles.

If your variety is normal and factorial, then it does follow that the pushforward of any line bundle from the smooth locus will still be a line bundle. Actually, being factorial should be equivalent to the property you want.

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On a normal variety, pushing forward a line bundle from the non-singular locus gives you a reflexive sheaf which is essentially the same as taking a Weil divisor representative of your original line bundle and take the Zariski closure of all the components keeping the same coefficients. So, what you get is a Weil divisor.

In other words, only assuming that your variety is normal, you get that the push forward of a line bundle (i.e., the sheaf version of a Cartier divisor) is a reflexive sheaf of rank 1 (i.e., the sheaf version of a Weil divisor (on a normal variety at least)).

The condition you want/need is that every Weil divisor be Cartier. As Karl said, this is exactly the condition of being factorial.

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Thanks, this is really helpful. –  Ben Webster Apr 24 '11 at 17:13
For terminal singularities this will not work already if we apply pushforward to the canonical bundle. Indeed, not all terminal singularities is Gorenstein. A simple contre-example is $\mathbb C^3/ \pm 1$.