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The answer is no, since first-order deformations can be obstructed, so that they do not give necessarily global embedded deformations of the subscheme.

For instance, Mumford gives an example of a smooth surface $X$ containing a curve $C$ such that $h^0(N_{C/X}) \neq 0$ (and so $C$ can be deformed in $X$ infinitesimally to the first order) but $C$ cannot be moved globally; in other words, no effective cicle different from $C$ arises in $X$ arises from the first-order deformations of $C$.

The reduced subscheme of the Hilbert scheme $\textrm{Hilb}_X^{C}$ in a neighborhood of the point $\xi$ corresponding to $C$ consists only of $\xi$, but the tangent space of $\textrm{Hilb}_X^C$ at $\xi$ is $1$-dimensional. This means that the local ring of $\textrm{Hilb}_X^C$ at $\xi$ contains nilpotents elements, and shows that the appearance of nilpotents is unavoidable also in entirely "classical" questions of algebraic geometry.

For further details, see Shafarevich's book Classical algebraic geometry 2: schemes and complex manifolds, page 111 and Mumford's Lectures on curves on algebraic surfaces, Lecture 22.

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The answer is no, since the infinitesimally first-order deformations can be obstructed, so that they do not give necessarily global embedded deformations of the subscheme.

For instance, Mumford gives an example of a smooth surface $X$ containing a curve $C$ such that $h^0(N_{C/X}) \neq 0$ (and so $C$ can be deformed in $X$ infinitesimally at to the first order) but $C$ cannot be moved globally; in other words, no effective cicle in $X$ arises from the first-order deformations of $C$.

The reduced subscheme of the Hilbert scheme $\textrm{Hilb}_X^C$ \textrm{Hilb}_X^{C}$in a neighborhood of the point$\xi$corresponding to$C$consists only of$\xi$, but the tangent space of$\textrm{Hilb}_X^C$at$\xi$is$1$-dimensional. This means that the local ring of$\textrm{Hilb}_X^C$at$\xi$contain contains nilpotents elements, and shows that the appearance of nilpotents is unavoidable also in entirely "classical" questions of algebraic geometry. For further details, see Shafarevich's book Classical Algebraic Geometry algebraic geometry 2: Schemes schemes and Complex complex manifolds, page 111 and Mumford's Lectures on curves on algebraic surfaces, Lecture 22. 1 The answer is no, since the infinitesimally deformations can be obstructed. Mumford gives an example of a smooth surface$X$containing a curve$C$such that$h^0(N_{C/X}) \neq 0$(and so$C$can be deformed in$X$infinitesimally at the first order) but$C$cannot be moved globally; in other words, no effective cicle in$X$arises from the first-order deformations of$C$. The reduced subscheme of the Hilbert scheme$\textrm{Hilb}_X^C$in a neighborhood of the point$\xi$corresponding to$C$consists only of$\xi$, but the tangent space of$\textrm{Hilb}_X^C$at$\xi$is$1$-dimensional. This means that the local ring of$\textrm{Hilb}_X^C$at$\xi\$ contain nilpotents elements, and shows that the appearance of nilpotents is unavoidable also in entirely "classical" questions of algebraic geometry.

For further details, see Shafarevich's book Classical Algebraic Geometry 2: Schemes and Complex manifolds, page 111.