Extending a prime divisor to a principal divisor Let $X$ be a noetherian,integral,separated scheme which is nonsingular in codimension $1$. Let $Z_1, \ldots, Z_k$ be a fixed set of prime divisors. Now given a prime divisor $Y$, does there exist prime divisors $W_1, \ldots, W_m$, none of them equaling any of the $Z_j$s, such that $Y +\sum n_i W_i = 0$ in $\text{Cl}(X)$?
Further, when can we do it with $n_i \ge 0$? For example in $\mathbb P^n_k$ one can not.
I can easily do the first part for curves over algebraically closed fields by using a Chinese remainder type argument. But I don't see it in the general case or know if its true. Basically I am interested to find out how much I can modify the support of a Weil divisor. If true, my statement would inductively imply that we can always remove any chosen set of prime divisors from the support and still maintain the same linear equivalence.
 A: It is easier to write this as an answer than as a comment.  A good resource for some questions of this type is "Ample subvarieties of algebraic varieties" by Robin Hartshorne.
Let $A$ be an ample divisor class.  Then there exists an integer $n_0$ such that for every integer $n\geq n_0$ and for every $j=1,\dots,k$,
both the sheaf $\mathcal{E}_j=\mathcal{O}(nA-(\sum_{i\neq j} Z_i))$ is globally generated and the sheaf $\mathcal{F}_j=\mathcal{O}(nA-Y-(\sum_{i\neq j} Z_i))$ is globally generated.  Thus, there exists a section $s_j$ of $\mathcal{E}_j$, resp. $t_j$ of $\mathcal{F}_j$ that is nonzero at some point $p_j$ of $Z_j$.  Now consider every $s_j$, resp. every $t_j$, as a section of $\mathcal{O}(nA)$, resp. $\mathcal{O}(nA-Y)$.  The sum $s=\sum_j s_j$, resp. $t=\sum_j t_j$ is nonzero at every point $p_j$.  The zero divisor $W_-$ of $s$, resp. $W_+$ of $t$, is an effective Cartier divisor whose prime Weil divisors are distinct from every $Z_j$.  Of course $W_-$ is linearly equivalent to $Y+W_+$.  Thus $Y+W_+-W_-$ is linearly equivalent to zero.
For the counterexample, begin with a fiber variety $F$ such that the only Cartier divisor on $F$ is the zero Cartier divisor, e.g., 
let $F$ be the glueing of a line to a disjoint conic in $\mathbb{P}^3$.  Note that $F$ is quite singular, but it is regular in codimension $1$.  Now let $C$ be a hyperelliptic curve, and let $f:C\to B$ be the quotient by the hyperelliptic involution.  Let $X'$ be $C\times F$.  Let $y_0\in F$ be a general closed point of $F$, i.e., a smooth point of $F$.  Form $\overline{X}$ as the scheme obtained as the coproduct of $C\times\{y_0\} \to X'$ and $C\times\{y_0\} \to B\times\{y_0\}$.  Now let $t\in B$ be a point that is not any of the branch points of $f$.  Let $X\subset \overline{X}$ be the open subset obtained by removing the closed point $(t,y_0)$.  
Every Cartier divisor on $C\times F$ is of the form $D\times F$ for a Cartier divisor $D$ on $C$.  Since $C\times F$ is smooth at the two points lying over $(t,y_0)$, the pullback to $C\times F$ of every Cartier divisor on $X$ extends to a Cartier divisor on all of $C\times F$.  Restricting the Cartier divisor on $X$ to the curve $(B\setminus\{t\})\times \{y_0\}$, the Cartier divisor $D$ must be of the form $f^*E + m \underline{p} + n\underline{q}$, where $f^{-1}(\{t\})$ equals $\{p,q\}$.  Thus, if we let $Y$ be $\underline{p}\times F$ and if we let $Z_1$ be $\underline{q}\times F$, then there is no choice of $W_i$ satisfying the conditions.   
