Fibers of the normal cone

For any embedding of smooth varieties $$X\subset Y$$ given by an ideal sheaf $$I\subset \mathcal {O}_Y$$, it is well known that the normal cone $$C _{X/Y}= \textbf{Spec}(\oplus_{\geq 0} I^i/I^{i+1})$$ is isomorphic to the normal bundle of $$X$$ to $$Y$$. Therefore, its fiber over each point $$x \in X$$ is the affine space associated to the quotient $$T_xY/T_xX$$ of the tangent space of $$Y$$ by the tangent space of $$X$$ at $$x$$.

Now let $$X$$ be an arbitrary subscheme of a smooth variety $$Y$$. Suppose further that over any point $$x\in X$$ the fiber $$(C _{X/Y})_x$$ of the canonical morphism $$C _{X/Y}\to X$$ over $$x$$ is isomorphic to the affine space $$T_xY/T_xX$$.

Question: Is it true that X must be smooth?

For example, if, in addition to the above assumptions, both $$X$$ and $$C _{X/Y}$$ are irreducible varieties, then the answer is positive. Indeed, $$\begin{gather}dim (C _{X/Y})_x\geq dim C _{X/Y} -dim X \\= dim Y - dim X\geq dim T_xY -dim T_xX \end{gather}$$ so all inequalities must be equalities, hence $$dim T_xX =dim X$$ for all $$x$$, i.e. $$X$$ is smooth.

Yes, $$X$$ must be smooth. The question is local at $$x \in X$$. Write $$R$$ for the local ring $$\mathcal{O}_{Y,x}$$, and $$\mathfrak{m}$$ for the maximal ideal of $$R$$. Write $$I$$ for the ideal defining $$X$$. Then $$(C_{X/Y})_x = \mathrm{Spec}(\oplus I^n/mI^n)$$. Its Krull dimension is called the analytic spread of $$I$$; it is at least $$\mathrm{height}(I)$$. (See the chapter on "Minimal reductions" in the book by Swanson and Huneke on integral closure.) On the other hand, $$\dim T_xX = \dim_{R/\frak{m}}(\frak{m}/(I + \frak{m}^2)) \geq \dim R/I$$. Since $$\dim R = \dim R/I + \mathrm{height}(I)$$, the given hypothesis implies that $$\dim T_xX = \dim R/I$$, i.e., $$R/I$$ is a regular local ring.