TITLE: Why does the divisor $Z$ homologous to $0$ in projective mainfold satisfy that every irreducible hypersurface appears in $Z$ with multiplicity $1$?
QUESTION [1 upvotes]: In Voisin's book "Hodge theory and complex algebraic geometry I",
the proof of proposition 12.7 (page 296) says that 
if $X$ is projective, then every divisor $Z$ homologous to $0$ can be written as a sum of divisors with multiplicity $1$.
Why is it true?

REPLY [4 votes]: I guess Voisin means that if $X$ is projective then  every divisor $Z$ homologous to $0$ is linearly equivalent to a divisor $Z'$ which is a sum of divisors with multiplicity $1$.
In fact, the element  $\alpha_Z \in \textrm{Pic}^0(X)$ she wants to define only depends on the linear equivalence class of $Z$.
We start by recalling that if $X$ is projective then every divisor $Z$ is linearly equivalent to $Z_1 -Z_2$, where $Z_i$ is very ample. In fact, choose a very ample divisor $H$ and $m >>0$ such that $mH+Z$ is very ample, and set
$$Z_1:=mH+Z, \quad Z_2:=mH.$$
Now, since $Z_1$ and $Z_2$ are very ample, it is possible to choose $Z_1'$ linearly equivalent to $Z_1$ and $Z_2'$ linearly equivalent to $Z_2$ such that

every irreducible component of $Z_1'$ and $Z_2'$ appears with multiplicity $1$;
no component of $Z_2'$ is contained in $\textrm{Supp}(Z_1')$.

Therefore $Z':=Z_1'-Z_2'$ is a divisor linearly equivalent to $Z$ and with the desired property. 
By the way, I do not think that the assumption "degree $0$" is really important here.

REPLY [1 votes]: This is not an answer, but an explanation. The proof in the book says

Assume for simplicity that $Z$ satisfies the property that every irreducible
  hypersurface appears in $Z$ with multiplicity 1. (Note that if $X$ is projective, then
  every divisor homologous to $0$ can be written as a sum of divisors satisfying
  this property.)

Here, $X$ is a compact Kähler manifold and $Z$ is a cycle of codimension one in $X$. The statement seems to be pretty much true by definition.
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