Algebraic Geometry by Adam Boocher

By Adam Boocher

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Plugging in our two points yields the system A+C +E =0 B + C + F = 0. The solution to this will have dimension 4 as it will have 4 free variables. Note that we could have also written the system in matrix form   A  B     C  1 0 1 0 1 0   = 0. ·  0 1 1 0 0 1  D   E  F 33 Since the matrix has rank 2, we expect the solution space to be 6-2=4 dimensional. We next do an example where equality does not hold in Proposition 24. Consider the set Z = {[1, 0, 1], [0, 0, 1], [−1, 0, 1]}. Then these three points lie in a line.

The dimension of Rd in k[x0 , . . , xn ] is d+n n. Proof. Clearly a basis for Rd is the set S = {xa0 0 · · · xann | a0 + . . + an = d, ai ≥ 0} . The trick will be in counting how many elements it has. This can be done in the following way. Consider n + d bowls in a line. Fill any n with water. From this we construct an element of S as follows. Let a0 be the number of bowls to the left of the first filled bowl. Let a1 be the number of bowls between the first and second filled bowls, etc. Since there are exactly n dry boxes, a0 + · · · + an = d.

One should not abandon hope, however. Although it may be very difficult to compute the entire Hilbert Function, as you may have noticed in the homework computations, eventually the function levels off and remains constant. In fact it is constantly equal to the number of points! Proposition 28. Let Z be a set of r points in P2 . Then HZ (d) = r for all d ≥ r − 1. Furthermore, this is the best lower bound we can achieve. 39 Figure 10: Two cubics intersecting in 9 points Proof. We note that Z imposes independent conditions on forms of degree r − 1 since removing any one point p, we can just draw lines through the remaining r − 1 point, carefully avoiding p.

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