# Is this a map $X \to \text{Sym}^n Y$?

Let $X$ and $Y$ be smooth projective curves over an algebraically closed field $k$ and let $D$ be a divisor on $X \times Y$ meeting every vertical fiber ($X$ is the horizontal axis). Let's assume that $D$ contains no vertical components. We get a map of sets $X(k) \to \text{Sym}^nY(k)$ by the following recipe:

For each $P \in X(k)$, map $P$ to the unordered $n$-tuple of points on the fiber $F_P = Y$ of $X \times Y$ over $P$, where each point $Q$ in this intersection is counted $n$ times, where $n$ is the intersection multiplicity $(D.F_P)_Q$.

Is this set-map algebraic? More generally, if I drop the assumption on vertical components, is it a rational map (which therefore actually extends to a morphism)? I have no idea how I would prove this, as the universal property of the symmetric product is for maps out, not for maps in.

I am particularly interested in the characteristic $p$ case. More generally, is there an analogous map if $X$ and $Y$ are just one-dimensional schemes, or arbitrary schemes?

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The answer to my question is yes, and it's proved in Milne's online notes on Abelian Varieties, p. 98. –  user29743 May 26 '12 at 22:06
For an answer in a more general situation (no base field), see Bosch, Lütkebohmert, Raynaud "Néron models", p. 252-254. –  user18119 May 27 '12 at 17:50
thanks for the reference –  user29743 May 28 '12 at 4:41