3 New answer to edited question.

[This is now an answer to the edited question(s), with some details added.My answer to the original question is kept at the very end.]

Firstly: The question is a good one, and it is not easy to find references onthis. I had spent too much time pondering about the failure of the double dualargument (see below) before I finally heard the arguement given in the lastsection below, indirectly from Fantechi, via Faber.

Assume $X$ is smooth projective.

Definition: An $S$-valued point in $M_I(X)$ is an $S$-flat coherent sheaf on$S\times X$, with stable fibres of rank one, and with determinant line bundleisomorphic to $\mathcal{O}_{S\times X}$, modulo isomorphism.

(I do not know if this is what Bridgeland meant, but to me this is resonablystandard.)

Comment: Stability for rank one means torsion free.

Existence: Let $M(X)$ be the (Simpson) moduli space for stable rank onesheaves. Then $M_I(X)$ is the fibre over $\mathcal{O}_X$ for the determinantmap $M(X) \to \mathrm{Pic}(X)$. This map sends a sheaf $I$ (stable rank onefibres) on $S\times X$ to the determinant line bundle $\det(I)$, and it istrivial as a point in $\mathrm{Pic}(X)$ if it is of the form $p^*L$ with$L\in\mathrm{Pic}(S)$. Then $I\otimes p^*L^{-1}$ is equivalent to $I$in $M(X)(S)$, and it has trivial determinant. This shows that $M_I(X)$indeed is a fibre of the determinant map.

Of course the determinant of an ideal $I_Y\subset \mathcal{O}_X$ is nontrivialif $Y$ is a non principal divisor, so you cannot map such ideals to $M_I(X)$.In any case, the ideal of a divisor, without the embedding, would onlyremember the linear equivalence class.

For brevity, let $\mathrm{Hilb}(X)$ be the part of the Hilbert schemeparametrizing subschemes $Y\subset X$ of codimension at least $2$.Then there is a natural map $F: \mathrm{Hilb}(X) \to M(X)$ that sends an ideal$I_Y\subset\mathcal{O}_{S\times X}$ to $I_Y$, forgetting the embedding. Since $Y$ isflat, so is $I_Y$, and its fibres are torsion free (by flatness again) of rankone. By the codimension assumption, the determinant of $I$ is trivial.

Theorem: $F$ is an isomorphism.

Comment: In the literature one sometimes finds the argument that if $I$ is arank one torsion free sheaf with trivial determinant, then $I$ embeds into itsdouble dual, which coincides with its determinant $\mathcal{O}_X$. Thisestablishes bijectivity on points. (For Hilbert schemes of points on surfacesthis is enough to conclude, since you can check independently that both$\mathrm{Hilb}(X)$ and $M_I(X)$ are smooth, and that the induced map on tangentspaces is an isomorphism.) I do not know how to make sense of this argument infamilies.

Sketch proof of theorem: The essential point is to show that every $I$ in$M_I(X)(S)$ has a canonical embedding into $\mathcal{O}_{S\times X}$ such that thequotient is $S$-flat.

Let $U\subset S\times X$ be the open subset where $I$ is locally free. Itscomplement has codimension at least $2$ in all fibres. By the trivialdeterminant assumption, the restriction of $I$ to $U$ is trivial. By codimension $2$,the trivialization extends to a map $I\to \mathcal{O}_{S\times X}$. This map is injective,in fact injective in all fibres: The restriction to eachfibre ${s}\times X$ is nonzero (as $U$ intersectsall fibres) and hence an embedding ($I$ is torsion free in fibres). Itfollows that the quotient is flat. There are some details to check, butthis is the main point, I think.

[End of new answer, here is the original one:]

(I have not studied Bridgeland's paper, so I do not know the intended meaning there.)

2 Forgot: $I_Z$ is $S$-flat

If we attempt to define $M_I(X)(S)$ as the set of $S$-flat ideals $I_Z$ in $\mathcal{O}_{S\times X}$, then that would not be functorial in $S$, as the inclusion $I_Z \subset \mathcal{O}_{S\times X}$ may not continue to be injective after base change (in the counter example in the other answer, restriction to the problematic fibre gives the zero map). We could impose "universal injectivity", but that is just another way of requiring the quotient $\mathcal{O}_Z$ to be $S$-flat, so then we have (re)defined the Hilbert scheme.

Another common way of defining moduli of ideals is as the moduli space for rank one stable sheaves (i.e. torsion free) with trivial determinant line bundle. The resulting moduli space is isomorphic to the Hilbert scheme of subschemes of codimension at least 2.

(I have not studied Bridgeland's paper, so I do not know the intended meaning there.)

1

If we attempt to define $M_I(X)(S)$ as the set of ideals $I_Z$ in $\mathcal{O}_{S\times X}$, then that would not be functorial in $S$, as the inclusion $I_Z \subset \mathcal{O}_{S\times X}$ may not continue to be injective after base change (in the counter example in the other answer, restriction to the problematic fibre gives the zero map). We could impose "universal injectivity", but that is just another way of requiring the quotient $\mathcal{O}_Z$ to be $S$-flat, so then we have (re)defined the Hilbert scheme.

Another common way of defining moduli of ideals is as the moduli space for rank one stable sheaves (i.e. torsion free) with trivial determinant line bundle. The resulting moduli space is isomorphic to the Hilbert scheme of subschemes of codimension at least 2.

(I have not studied Bridgeland's paper, so I do not know the intended meaning there.)