# when is the normal cone a linear scheme?

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 $$Y$$.

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