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? 



Why does the divisor $Z$ homologous to $0$ in projective mainfold satisfy that every irreducible hypersurface appears in $Z$ with multiplicity $1$?
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


1. every irreducible component of $Z\_1'$ and $Z\_2'$ appears with multiplicity $1$;
2. 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.


