TITLE: Does a semistable curve descend to a regular base?
QUESTION [9 upvotes]: Let $f\colon X \rightarrow S$ be a semistable curve of genus $g \ge 0$. Being a semistable curve means that $f$ is a morphism of schemes such that

$f$ is proper, flat, and of finite presentation;
The geometric fibers of $f$ are reduced, connected, one-dimensional, of arithmetic genus $g$, and smooth away from finitely many points at which the singularities are required to be ordinary double points.

Does $f$, Zariski locally on $S$, come as a base change of a semistable curve $f' \colon X' \rightarrow S'$ with $S'$ regular? 
A lemma of this sort seems to be used in the proof of 9.4/1 of "Neron models," i.e., in the proof of the representability of $\mathrm{Pic}^0_{X/S}$ by a scheme. The authors cite the paper of Deligne and Mumford for this reduction to a regular base. I can't see how the citation justifies the claim because Deligne and Mumford seem to deal with a narrower class of curves, namely, they impose an additional condition:

If a geometric fiber of $f$ has an irreducible component isomorphic to $\mathbb{P}^1$, then that component meets the other components in $\ge 3$ points.

EDIT. The real question is: for a semistable curve $f$ as above, is $\mathrm{Pic}^0_{X/S}$ a scheme, as claimed in 9.4/1 of "Neron models"? The proof given there seems incomplete (see the comments below for some discussion). Any ideas are welcome.

REPLY [2 votes]: The answer to the original problem (not the one in the edit) is: No.
Let $S$ be a nodal curve over the complex numbers with normalization $T$. Say $t, t' \in T$ are the inverse images of the node. Let $E$ be an elliptic curve. Take $E \times T$ and glue the fibres $E \times t$ and $E \times t'$ by translating with a nontorsion point to get $X \to S$.
This is the standard example of a smooth genus $1$ fibration $X \to S$ which is not locally projective.
Such a thing can never come from $X' \to S'$ with $S'$ regular. If it did then you would be able to Zariski shrink $S'$ and assume that $X' \to S'$ is smooth and proper. Then you would take a divisor on the generic fibre and take the closure $D'$ in $X'$. A small argument then shows $D'$ is relatively ample hence the pullback to $X$ is relatively ample over $S$. Contradiction.