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By Adam Boocher

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Proof for r = m is left to the reader) Then U ⊥ is at least a two dimensional vector space. Since (v1 , . . , vr ) is independent, vr is not in U so we can choose w in U ⊥ not perpendicular to vr . Then by construction, w is perpendicular to v1 , . . , vr−1 as required. Conversely, suppose that we can remove any point p from Z and there exists a form of degree d vanishing on all points of Z but not p. Then using the notation above, this means for each i there exists a vector vi such that vi ⊥ b1 , b2 , bi−1 , bi+1 , .

If every point takes away one degree of freedom, then we will lose r dimensions. Of course, the conditions imposed by the points might very well be redundant, so we don’t necessarily have equality. We will now do several examples to familiarize you with the abstraction above so that you feel more at ease. Suppose we are looking for polynomials that vanish on the set Z = {[1, 0, 1], [0, 1, 1]}. Clearly no nonzero form of degree 0 can vanish on both of these points, so dim(IZ )0 = 0. ) Next we try to tackle forms of degree 1.

The computation of s(2) is much more interesting. Let C be any conic vanishing on Z. Then by B´ezout’s Theorem we know it must contain the line L through the 4 points as a factor. To get the other factor, we can multiply L by any line passing through the 5th point. Lines are of the form AX + BY + CZ and the one point will clearly impose one condition so we still have two degrees of freedom. Thus, s(2) = 2. Finally, we know that Z imposes independent conditions on forms of degree 3 and higher, so s(d) = d+2 − 5 for 2 all d ≥ 3.

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