Let $X$ be an irreducible projective variety over $\mathbb{C}$ (note that I do not assume $X$ smooth) and let $ p : X \longrightarrow S$ be a projective surjective morphism. For any open $U \subset S$, I consider the natural map:


$$ \pi : \mathrm{Hilb}^2\_{U}(p^{-1}(U)) \longrightarrow S^2(p^{-1}(U)/U)),$$
where $\mathrm{Hilb}^2\_U(p^{-1}(U))$ is the punctual Hilbert scheme parametrizing relative subschemes of $p^{-1}(U)$ of length $2$ and $S^2(p^{-1}(U)/U))$ is the relative symmetric square of $p^{-1}(U)$ over $U$. Let me finally denote by $\mathcal{H}\_U$ the closure in $\mathrm{Hilb}^2\_U(p^{-1}(U))$ of :
$$ \pi^{-1}\left(S^{2}(p^{-1}(U)/U) \backslash \Delta\_{p^{-1}(U)/U} \right), $$
where $\Delta\_{p^{-1}(U)/U)}$ is the relative diagonal.


**Question** : I would like to know if there exists a non-empty $U$ such that $\mathcal{H}\_U$ is irreducible?


The comment below [this question](https://mathoverflow.net/questions/212571/hilbert-scheme-of-relative-subschemes-of-lenght-2) seems to suggest (by generic smoothness) that it could be true if $X$ is smooth. However I am interested in the general case. I would also be interested in a reference (or a short proof) if the answer to the question happens to be positive.


**Edit** : In a former (naive) version of the question, I asked if the whole relative Hilbert scheme could be irreducible over some non-empty open $U$. The answer is trivially "no", as observed by Jason Starr in the comment below.



irreducibility punctual Hilbert scheme of relative subschemes of length $2$
If I understand the question correctly, the subset $\mathcal{H}\_U$ is the closure of the locus parametrizing two distinct points. Over this locus, the Hilbert-Chow morphism $\pi$ is an isomorphism so irreducibility of $\mathcal{H}\_U$ is equivalent to irreducibility of $S^2(p^{-1}(U)/U \setminus \Delta\_{p^{-1}(U)/U})$ which it turn would follow from irreducibility of $S^2(p^{-1}(U)/U)$.




---


**Lemma:** Suppose $p : X \to S$ is a flat and proper surjective morphism of varieties such that $S$ is irreducible and the generic fiber of $p$ is geometrically irreducible. Then the symmetric powers $S^n(X/S)$ are irreducible.


**Proof:** Since flatness is stable under base change and composition, the $n$-fold fiber powers $X^n\_S := X \times\_S \ldots \times\_S X$ are flat over $S$. Moreover, the generic fiber of $X^n\_S \to S$ is the $n$-fold product of the generic fiber of $p$ and thus geometrically irreducible. By [Tag 0559](https://stacks.math.columbia.edu/tag/0559), there exists a non-empty open subvariety $U \subset S$ such that the pullback $X^n\_U \to U$ has irreducible fibers. By flatness, $X^n\_S$ is the closure of $X^n\_U$ and by [this answer](https://math.stackexchange.com/a/584919/111541), $X^n\_U$ is irreducible. Thus $X^n\_S$ is irreducible and we conclude that $S^n(X/S) = X^n\_S/\mathfrak{S}\_n$ is irreducible.


**Corollary:** Suppose $p : X \to S$ is a proper surjection of varieties with geometrically irreducible generic fiber. Then there exists a nonempty open subset $U \subset S$ such that $S^n(p^{-1}(U)/U)$ is irreducible.


**Proof** By generic flatness, there exists a nonempty open subset $U \subset S$ over which $p$ is flat and this open set does the job by the Lemma.




---


If we don't assume that the generic fiber of $p$ is geometrically irreducible, then the result is false. For example we can take $p : X \to S$ to be the map $\mathbb{G}\_m \to \mathbb{G}\_m$ given by $z \mapsto z^5$. Then I believe that $S^2(X/S)$ has 3 irreducible components which all map $5$-to-$1$ onto $S$. One of them is the diagonal which we remove but then whats left is still not irreducible and doesn't become irreducible if we shrink $S$.


