Constructibility of the locus of points where the fiber is an isomorphism modulo nilpotents Let $f: X \rightarrow S$ be a finitely presented morphism of schemes and let $$E = \{s \in S \mid \text{ $X_s$ is a point with residue field $\kappa(s)$ }  \}$$
Is $E$ a constructible set?
The basic motivation for my asking this is in to check, for dominant $f$, whether there exists an affine monomorphism $S' \rightarrow X \rightarrow S$ such that $S'$ is $0$-dimensional and dominates $X$.  If $E$ is constructible, then it suffices to look at the generic points of $X$ (or any other subset whose closure contains the generic points).
If we remove the condition on the residue field, then the resulting set certainly is constructible, since e.g. both "$0$-dimensional" and "geometrically irreducible" are constructible properties of fibers.
My reference for this sort of thing is usually appendix $E$ in Görtz and Wedhorn's Algebraic Geometry I.  I don't know of any constructible properties that control residue fields in this way, but the property in question does seem to me like it should be a decent "descent" property.
Linking this relevant MO question for convenience.
 A: Let $k$ be a field of characteristic $p>0$, $X=S=\operatorname{Spec}k[t]$ the affine line, and $f:X\to X$ the $p$-th power map. Then $E$ is dense but does not contain the generic point, so it is not constructible.
[Added August 29] On the other hand, if $S$ is a noetherian $\mathbb{Q}$-scheme, then $E$ is constructible. Indeed, it suffices to see that if $S$ is integral then either $E$ or $S\smallsetminus E$contains a nonempty open set. We may assume $X$ reduced (passing to $X_\mathrm{red}$ does not change $E$, and $X_\mathrm{red}\to S$ is still finitely presented). If $E$ contains the generic point $\eta$, then $f$ is birational and we are done. If not, then the fiber $X_\eta$ is either empty, or positive-dimensional, or finite (reduced) over $\kappa(\eta)$ of degree $>1$. Each of these properties extends to a neighborhood of $\eta$, whence the claim. (The characteristic zero assumption is used via the fact that "reduced"="geometrically reduced".)
Whether one can extend this to arbitrary $\mathbb{Q}$-schemes is unclear to me.
