irreducibility punctual Hilbert scheme of relative subschemes of length $2$

Question

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 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.

Answer 1

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, 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, $X^n_U$ 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 $S$.