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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 $X$ in $Y$. 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.

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.

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 $X$ in $Y$. 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.

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 $X$ in $Y$. 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.

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 $X$ in $Y$. 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.

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 $X$ in $Y$. 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.