Let $S$ be a scheme and $A$ an abelian $S$-scheme, i.e., $A \rightarrow S$ is a proper smooth $S$-group scheme whose fibers are $g$-dimensional abelian varieties. Suppose that one has a fiberwise dense (and perhaps $S$-quasi-compact) open subscheme $U \subset A$ and an effective relative (to $S$) Cartier divisor $D \subset U$. One takes the schematic image $D'$ of $D \rightarrow A$. Is the closed subscheme $D' \subset A$ a relative (to $S$) effective Cartier divisor? I.e., is $D'$ flat over $S$ and locally on $A$ cut out by a single nonzero divisor?


Something like this comes up in the construction of a $\Theta$-divisor for a proper smooth $S$-curve of genus $g \ge 2$. Namely, my question is inspired by the desire to understand the sentence "Furthermore, $W^{g - 1}$ is an effective relative Cartier divisor on $P$, usually denoted by $\Theta\_\sigma$." on p. 261 of "Neron models."



Schematic image of a relative Cartier divisor of a fiberwise dense open
Certainly it does not suffice to take the schematic closure if $S$ is nonreduced. For instance, let $S$ be $\text{Spec}\ k[x,y]/\langle x^2, xy \rangle$. Let $A$ be $E \times\_{\text{Spec} k} S$, where $E$ is an elliptic curve over $k$ with specified zero point $z$. Let $p \in S$ be the closed point with maximal ideal $\langle x,y\rangle$. Let $U$ be the open complement in $A$ of the closed point $(z,p)$. Let $\zeta:S\to A$ be the zero section with image $\{z\}\times S$. Let $D$ be the intersection of $U$ with the image Cartier divisor $\zeta(S)$. Then $D'$ is the underlying reduced scheme of $\zeta(S)$, and this is not a Cartier divisor in $A$, nor is it flat over $S$.


