Complete intersection

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 n − m homogeneous polynomials

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

Geometrically, each F_i 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 n − m ≥ 2. When n − m = 1 then V is automatically a hypersurface and there is nothing to prove.

A classic example is the twisted cubic in ${\displaystyle \mathbb {P} ^{3}}$: 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 ${\displaystyle 2\times 2=4,}$ and the twisted cubic has degree 3, so the only way to get degree 4 is to add a line.

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