Let $Y$ be a nonsingular variety and $X\subset Y$ a closed subscheme. A linear scheme over $X$ is a scheme of the form $\textbf{Spec}\, ( Sym _{\mathcal{O}_X} F) $, where $F$ is a coherent sheaf on $X$.

If the embedding $X\subset Y$ is regular, i.e. if every point of $Y$ has a neighborhood over which the ideal $I$ defining the above embedding is generated by a regular sequence, then it is well known that the normal cone $C_{X/Y}= \textbf{Spec}\, (\oplus _{i\geq 0} I^i/I^{i+1})$ is isomorphic to the linear scheme $ \textbf{Spec}\, ( Sym _{\mathcal{O}_X} I/I^2) $- the total space of the conormal sheaf.

Question: Is the converse true? That is, suppose that $X$ is a closed subscheme of a nonsingular variety $Y$ and $C_{X/Y}$ is isomorphic to a linear scheme. Is it true that $X\subset Y$ is regular embedding?


I believe the answer is negative. Take $Y= \mathbb{A}^3= \textbf{Spec}\, k[x,y,z]$ and $X= V(xz,yz)$, so $I=(xz,yz)$. Note that $X$ is the union of the plane $z=0$ and the line $x=y=0$. Then $X\subset Y$ is not regular: any neighborhood of the origin contains a point in the plane and a point in the line, so their local rings have different dimensions.

Now, let $A=k[x,y,z]$, $\overline{A}=\dfrac{k[x,y,z]}{(xz,yz)}$. The surjection of graded rings $$A[S,T]\to \oplus _{i\geq0} I^i$$ which sends S to $xz$ and $T$ to $yz$ has kernel $(yS-xT)$ so it induces a surjection $$\overline{A}[S,T]\to \oplus _{i\geq0} \dfrac{I^i}{I^{i+1}} $$ with kernel $R=(\bar{y}S-\bar{x}T)$. Since $R$ is defined by an equation in degree one, and the above surjection factors through $Sym I/I^2$, we actually have

$$\frac{\overline{A}[S,T]}{R}\cong\oplus _{i\geq0} \dfrac{I^i}{I^{i+1}}\cong Sym I/I^2 $$

so the normal cone $C_{X/Y}$ is the linear scheme associated to the conormal sheaf of $X$ in $Y$.

  • $\begingroup$ I guess the embedding $X \subset Y$ is regular (according to the definition given in the question). So this is not a counterexample. $\endgroup$ – Sasha Oct 4 '19 at 13:32
  • $\begingroup$ If it were regular, $X$ would be smooth. $\endgroup$ – sky223 Oct 4 '19 at 18:34
  • $\begingroup$ Did you check the definition? Why do you think the regularity of the embedding implies the smoothness of the source? $\endgroup$ – Sasha Oct 5 '19 at 6:54
  • $\begingroup$ You are right- I changed the answer. $\endgroup$ – sky223 Oct 5 '19 at 12:33
  • $\begingroup$ The new example is good! $\endgroup$ – Sasha Oct 5 '19 at 13:00

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