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I know that the following holds in much more generality, but lets say everything happens in the toric case over $\mathbb{C}$.

Setting: Given a smooth variety $X$ then there is an isomorphism between the group of Cartier divisors of $X$ and the group of Weil divisors of $X$. I know and understand how to come from a Weil divisor $D$ to its Cartier divisor and back.

Question: What fails in the singular case? I think of a Cartier divisor as a locally principal Weil divisor. Given a Weil Divisor $D$ on $X$ and let $U \subset X$ be the singular locus of $X$. My guess is, that $D|_U$ fails to be principal. But what does this exactly mean?

Johannes
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  • Would you have a reference for the isomorphism between the group of Cartier divisors and the group of Weil divisors? Thank you very much! – Watson Apr 20 '17 at 17:41
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    @Watson: Because every Cartier divisor is also a Weil divisor by definition the crucial point is Weil -> Cartier. In the toric setting this can be found in: Toric Varieties, Cox, Little, Schenk Proposition 4.2.6. – Johannes Apr 21 '17 at 07:23

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Imagine the doubly-infinite (singular) cone $x^2+y^2=z^2$ in 3-space. If we intersect with the plane $x=z$, we get a single line $x=z, y=0$, a Weil divisor. But it cannot be Cartier, since it is not locally principal at the origin; any polynomial vanishing on this line will vanish with multiplicity, or else vanish on more than one component.

Algebraically speaking, the corresponding ideal of the local ring is not principal (though its square is).

Andrew Dudzik
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  • Thnx for that answer. That's what I expected when I thought about this question after I put it here. I see the algebraic point, but why has a polynomial vanishing on the line to vanish with multiplicity? – Johannes Jan 27 '14 at 19:48
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    @Johannes This is just the same statement as the statement that the ideal is not principal at the origin. It's not hard to show this algebraically, but an easy way to see it intuitively is that a surface meeting the cone only at this line must be tangent to the cone as well. – Andrew Dudzik Jan 28 '14 at 02:02