Rational quintic curves.

 

Here we encounter an exceptional fact, namely:

every rational quintic curve in P3 has a 4-secant line

This is clear if the curve is contained in a quadric surface (which is necessarly smooth; the curve is then of type (4, 1)).

But (and this is why the existence of the 4-secant line is exceptional) the general rational quintic is not contained in a quadric surface. For rational quintics on a smooth quadric surface every thing is clear, see the minimal free resolution of a curve on a quadric.).

From now on we assume that C is a rational quintic curve not contained in a quadric surface. Such a curve has a unique 4-secant line. Moreover h1(IC(3)) = 0(otherwise C would be linked to a line by a complete intersection of two cubic surfaces and this would imply that C is projectively normal) and IC is 4-regular (Castelnuovo-Mumford's lemma).

It follows that I(C) is generated by its elements of degree <= 4. <= 4. Clearly there are 4 (="h0(IC(3)))" minimal generators of degree 3.

The problem is to determine how many generators of degree 4 do occur.

The answer is that I(C) is generated by 4 cubics and 1 quartic. Here is a sketch of a  "simple" way to see it:

let X be the union of C and L, where L is 4-secant line to C. The curve X has degree 6 and arithmetic genus, pa(X) = 3. Since h0(IX(2)) = 0, X is a.C.M. (arithmetically Cohen-Macaulay); its minimal free resolution is:

 0 -> 3.O(-4) -> 4.O(-3) -> IX -> 0 (*)

We have I(C)3 = I(X)3 (every cubic containing C intersects L in 4 points hence contains also L, i.e. the cubic contains X; since dimI(C)3 = dimI(X)3 = 4, we conclude).

Consider the natural map:

 mC(3): I(C)3 x V -> I(C)4: F x H -> FH (here x = tensor product)

The kernel of this map is the same as the kernel of the map:

 I(X)3 x V -> I(X)4

From the minimal free resolution (*), this kernel has dimension 3. It follows that dim(Im(mC(3)))=13, hence Im(mC(3)) has codimensione one in I(C)4, i.e. we need to add to it a 1-dimensional supplementary subspace to get the whole of I(C)4: there is one minimal generator in degree 4. This argument also shows that there are 3 relations of degree 1 involving the 4 generators of degree 3.