Extending a prime divisor to a principal divisor

Question

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.

Answer 1

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.