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? 
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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. 



Fibers of the normal cone
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.


