# Example of locally complete intersection varieties which are not smooth and not complete intersection

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.

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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.

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Thanks Serge. How does one now that $\pi_p(X)$ is not smooth. Answer to this even in char $0$ is fine. –  Tony Puthenpurakal Feb 25 '13 at 12:39
Zariski tangent space of $X$ at thr point $y$ equals $T_yY\cap T_yQ$; since $Q$ is tangent to $Y$ at $y$, one has $T_yQ\supset T_yY$, whence $\dim T_yX>\dim X$. In char $p$, one can always embed a surface so that there is a quadric tangent to it at exactly one point, so this argument extends to finite characteristic as well. –  Serge Lvovski Feb 25 '13 at 13:24

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.

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Great! That inspired me to find a minimalist reduced example in $\mathbb A^2$: just suppose that $Z\subset\mathbb A^2$ is a triple of non-collinear points. It's an l.c.i. (and even a complete intersection in $\mathbb A^2$, as in your example), but not a c.i. in $\mathbb P^2$. –  Serge Lvovski Feb 25 '13 at 11:03
@Serge: Yes, but it is smooth. And the original problem was about a nonsmooth example! –  Sasha Feb 25 '13 at 11:08
Oops! OK, at least my positive-dimensional example is not smooth;) –  Serge Lvovski Feb 25 '13 at 11:12

At the best of my knowledge what is know about complete intersection is the follwing:

Varieties in an affine space.

• in characteristic zero any lci $X\subseteq\mathbb{A}^n$ is a set-theoretic complete intersection.
• In characteristic $p$ any lci curve $C\subseteq\mathbb{A}^n$ is a set-theoretic complete intersection.

Projective varieties

Let $X\subseteq\mathbb{P}^n$ be a smooth non-degenerate degree $p$ (a prime number) variety of codimension $c$. Then $X$ is not a scheme-theoretic complete intersection. Indeed, if $X = H_1\cap H_2\cap...\cap H_c$, then $deg(H_2)=...=deg(H_c) = 1$ and $deg(H_1) = p$ by Bezout's theorem because $p$ is prime. Therefore $X$ would be degenerate. An example is again the twisted cubic $C\subset\mathbb{P}^3$. However $C$ is a set-theoretic complete intersection. There exist a quadric surface $Q$ and a cubic surface $S$ such that $Q\cap S = 2C$ (i.e. $Q$ and $S$ are tangent along $C$).

Hartshorne Conjecture: If $X\subseteq\mathbb{P}^N$ is a smooth variety of dimension $n$, codimesnion $c$ and $c\geq 2n+1$ then $X$ is a scheme-theoretic complete intersection.

Hartshorne Conjecture has been proven for Fano varieties of codimension two and quadratic varieties (i.e. varieties that can be defined just by quadratic polynomials).

Thanks to Barth’s result: Barth, W.: ”Transplanting cohomology classes in complex-projective space”, Amer. J. Math., 92, 951-967 (1970), and since no indecomposable rank two vector bundle on $\mathbb{P}^N$, $N\geq 5$, is known, it is generally believed that any smooth, codimension two subvariety of $\mathbb{P}^N$, $N \geq 6$, is a complete intersection. The main results for codimension two subvarieties can be summarized as follows: let $\omega_X\cong \mathcal{O}_X(e)$, $d$ the degree of $X$ and $s$ the minimal degree of an hypersurface containing $X$. if $e \leq N + 1$ or if $d < (N − 1)(N + 5)$ or if $s \leq N − 2$, then $X$ is a complet intersection. For $N = 5,6$ we can something more: let $X \subset \mathbb{P}^6$ be a smooth, codimension two subvariety, if $s\leq 5$ or if $d \leq 73$, then $X$ is a complete intersection. Let $X \subset \mathbb{P}^5$ be a smooth, subcanonical threefold. If $s \leq 4$, then $X$ is a complete intersection. This is Theorem 1.1 of http://arxiv.org/abs/math/9909137.

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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 $n+1$.

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Thanks Prof. Mohan Tony Puthenpurakal –  Tony Puthenpurakal Feb 28 '13 at 11:51
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 $d_i\geq 1$.