Question

Hi Let $V$ be an affine or projective variety. Recall that $V$ is a local complete intersection (l.c.i) if all its local rings are complete intersection. Also recall that $V$ is a complete intersection (c.i) if $I(V)$ is generated by a regular sequence of length $codim(V)$.

I do not know a single example of a l.c.i variety which is not c.i and NOT smooth. Examples will be greatly appreciated.

Answer 1

If $X\subset \mathbb P^N$ is an l.c.i projective subvariety that linearly spans $\mathbb P^N$, and if $p\in\mathbb P^n\setminus X$ is a point s.t. the projection of $X$ from $p$ into $\mathbb P^{N-1}$ is an isomorphism onto its image $X'$, then $X'$ will not be a complete intersection (since the linear system of its hyperplane section is not complete).

To obtain such an example, suppose that $Y\subset\mathbb P^N$ is a smooth surface not lying in a hyperplane and that $N$ is big enough ($N\ge6$ will suffice). If characteristic is zero, there exists a quadric $Q\subset\mathbb P^N$ that is tangent to $Y$ at exactly one point $y\in Y$. Now if $X=Y\cap Q$ and $P$ is a generic point of $\mathbb P^N$ (in particular, $p$ should not lie in the tangent space $T_yY$), then the projection $\pi_p\colon X\to \mathbb P^{N_1}$ is an isomorphism onto ints image, and $\pi_p(X)\subset\mathbb P^{N-1}$ is an l.c.i that is not a c.i.

Answer 2

Take $Z = Spec k[x,y]/(y-x^2,x^3) \subset A^2 \subset P^2$. This is a nonreduced artinian subscheme of length 3. By definition it is l.c.i. Let us show that it is not a complete intersection. Indeed, since the length is 3, the only possibility would be to have it as a complete intersection of a line and a cubic. But it is easy to see that $Z$ is not contained in a line.

Answer 3

I give an example, which can be generalized in many ways, but I write a specific one for clarity. Let $C\subset\mathbb{P}^2$ be a plane cubic curve (possibly singular). Embed $\mathbb{P}^2$ in $\mathbb{P}^5$ by the Veronese embedding and let $D$ (to avoid confusion) be the image of $C$. Then $\deg D=6$ and $\omega_D=\mathcal{O}_D$. If $D$ is the complete intersection (notice that $D$ is a local complete intersection) of type $d_1,d_2,d_3,d_4$, then we have $6=\deg D=d_1d_2d_3d_4$ and since $\omega_D=\mathcal{O}_D=\mathcal{O}_D(d_1+d_2+d_3+d_4-6)$ and so $d_1+d_2+d_3+d_4=6$. Easy to see that these two have no solutions with all $d_i\geq 1$.

If you are more interested in the affine case, let me know.

Answer 4

Here is a somewhat general example in the affine case. First, it is elementary, but not obvious, the following intrinsic condition. Let $X=\mathrm{Spec} A$ be an affine variety (reduced, irreducible). Then $X$ is a complete intersection in some embedding in affine space if and only if $\Omega_X^1$ has a free resolution of length one over $A$. This implies that if $X,Y$ are affine varieties with $Y$ a complete intersection (in some embedding), then $X\times Y$ is a complete intersection (again in some embedding) if and only if $X$ is. Thus to construct an example as you need, suffices to find a smooth example, since then I can take its product with a singular hypersurface to get a singular example. The simplest one such would be to take $X$ to be the complement of an irreducible hypersurface of degree $d$ in a projective space of dimension $n$ with $d$ not dividing $n+1$. If such an $X$ was a complete intersection in some affine space, then as mentioned, $\Omega_X^1$ will have a free resolution of length 1 over $A$, in particular since $\Omega^1_X$ is $A$-projective, it will be stably free. But stably free modules have determinant trivial, while in the above case determinant is $\mathcal{O}_X(-n-1)$ which is not trivial since $\mathrm{Pic} X=\mathbb{Z}\mathcal{O}_X(1)/d$ and $d$ does not divide $n+1$.