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.

  • 2
    $\begingroup$ You can find such $W_i$ if $X$ admits an ample invertible sheaf. You can also find such $W_i$ if $X$ is everywhere regular. The easiest counterexample I know begins with a "nice" $X'$ that fibers over a hyperelliptic curve $C$ and such that every divisor class on $X'$ is the pullback of a divisor class on $C$. Now glue a section of the fibration to itself via the hyperelliptic involution. Remove a general point $t$ on the resulting $\mathbb{P}^1$. Let $Y$ be the fiber over one point of $C$ mapping to $t$, let $Z_1$ be the fiber over the other point of $C$. $\endgroup$ May 24, 2016 at 12:50
  • $\begingroup$ @JasonStarr: Thanks a lot for your comment. Could you please hint at how X admitting an ample invertible sheaf helps to prove the result? $\endgroup$ May 24, 2016 at 13:33
  • $\begingroup$ If you want to see a really baby version of this I think Ex. IV.1.9 in Hartshorne is good. Of course, you said you already knew how to do curves and there are many more assumptions there, but still. $\endgroup$
    – Hoot
    May 24, 2016 at 15:00

1 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.


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