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.

  • $\begingroup$ Thank you very much. $\endgroup$ – sky223 Oct 15 '19 at 21:54

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