Q:
How to show that on the smooth projective variety $X$, any divisor $D$ can be represented by the difference of two effective divisors, i.e. $D=D_1-D_2$.
If $X$ is just any complex manifold, does above also hold?
Divisor is the difference of effective divisor
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algebraic-geometry
divisors-algebraic-geometry
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1Isn't this true by definition of a divisor? A divisor is just a formal $\mathbf{Z}$-linear combination of codimension one subvarieties of a variety $X$. So you just express $D$ as the part of the linear combination with positive coefficients, minus the part with negative coefficients. Depending on what you mean by a divisor on a complex manifold, the same argument works. – 2017-02-12
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1A more subtle question is whether you can choose $D_1,D_2$ to be *smooth* divisors. See [this other question](http://math.stackexchange.com/a/1732888/116766) for how you can ensure this version, at least on a smooth projective variety over an infinite field (note it is written for surfaces, so just replace "curve" with "divisor" to make the proof work in arbitrary dimension). – 2017-02-12