Complete intersection

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For complete intersection rings in commutative algebra,, see complete intersection ring.

In mathematics, an algebraic variety V in projective space is a complete intersection if the ideal of V is generated by exactly codim V elements. That is, if V has dimension m and lies in projective space Pn, there should exist nm homogeneous polynomials

Fi(X0, ..., Xn), 1 ≤ inm,

in the homogeneous coordinates Xj, which generate all other homogeneous polynomials that vanish on V.

Geometrically, each Fi defines a hypersurface; the intersection of these hypersurfaces should be V. The intersection of n-m hypersurfaces will always have dimension at least m, assuming that the field of scalars is an algebraically closed field such as the complex numbers. The question is essentially, can we get the dimension down to m, with no extra points in the intersection? This condition is fairly hard to check as soon as the codimension nm ≥ 2. When nm = 1 then V is automatically a hypersurface and there is nothing to prove.

Example of a space curve that is not a complete intersection[edit]

A classic example is the twisted cubic in \mathbb{P}^3: it is a set-theoretic complete intersection, i.e. as a set it can be expressed as the intersection of 2 hypersurfaces, but not an ideal-theoretic (or scheme-theoretic) complete intersection, i.e. its homogeneous ideal cannot be generated by 2 elements.

Its degree is 3, so to be an ideal-theoretic complete intersection it would have to be the intersection of two surfaces of degrees 1 and 3, by the hypersurface Bézout theorem. In other words, it would have to be the intersection of a plane and a cubic surface. But by direct calculation, any four distinct points on the curve are not coplanar, so it cannot lie in a plane, ruling out the only possible case. The twisted cubic lies on many quadrics, but the intersection of any two of these quadrics will always contain the curve plus an extra line, since the intersection of two quadrics has degree 2\times 2 = 4, and the twisted cubic has degree 3, so the only way to get degree 4 is to add a line.

On the other hand, the twisted cubic, as a set, is the intersection of the quadric surface x z - y^2 = 0 and the cubic surface  z (y w - z^2) - w (x w - y z)=0 in \mathbb{P}^3. Formally the degree of that intersection is 6, so in a more refined sense, the intersection is actually the twisted cubic counted with multiplicity two.


A complete intersection has a multidegree, written as the tuple (properly though a multiset) of the degrees of defining hypersurfaces. For example taking quadrics in P3 again, (2,2) is the multidegree of the complete intersection of two of them, which when they are in general position is an elliptic curve. The Hodge numbers of complex smooth complete intersections were worked out by Kunihiko Kodaira.

General position[edit]

For more refined questions, the nature of the intersection has to be addressed more closely. The hypersurfaces may be required to satisfy a transversality condition (like their tangent spaces being in general position at intersection points). The intersection may be scheme-theoretic, in other words here the homogeneous ideal generated by the Fi(X0, ..., Xn) may be required to be the defining ideal of V, and not just have the correct radical. In commutative algebra, the complete intersection condition is translated into regular sequence terms, allowing the definition of local complete intersection, or after some localization an ideal has defining regular sequences.


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