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Extra resources for Algebraic Geometry, Edition: version 29 Jan 2007

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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 , . . , br , and vi · bi = 0. Then we claim that b1 , . . , br are linearly independent. Suppose that c1 b1 + . . cr br = 0. Then dotting both sides with vi shows that ci = 0, so 35 the bi are linearly independent and thus the matrix has rank r meaning that the dimension of (IZ )d is m − r as required.

Note that this is not a ring for many reasons (find at least two for homework). Rd ∪ {0} is a vector space over R, however. Indeed, adding 30 two forms of degree d is another form of degree d, and similarly for scalar multiplication. Now that we have a vector space, we recall that every vector space has a basis. So what is a basis of R0 ? A form of degree 0 is just a constant, so {1} is a basis. A basis for forms of degree 1 is {x0 , . . , xn }, so R1 has dimension n + 1. To make the notation easier for higher degrees, we simplify to three variables, R(x, y, z).

Homework 36. Compute the Hilbert function of d collinear points in P2 . In general, computing the Hilbert function of a set of points is highly nontrivial. The function is extremely dependent on the geometry of the set of points. For instance, if the points are all collinear, it is easy to compute the function. In general, this is not the case. To illustrate why this is not the case consider the following. Suppose you had 9 points in P2 and you were wondering if they imposed independent conditions on forms of degree 3.

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