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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.

C1m am + c2m am ... + . . + c1m am = = 0 0 = 0. Remember, our goal is to find the constants a1 , . . , am so this is now just a question of linear algebra. We have m unknowns and r equations. A nice intuitive way to think about this is that to begin with, your homogeneous polynomial is free to be whatever it wishes if it has no vanishing restrictions. But as soon as the function is required to vanish at a certain point, that places one restriction on its behavior. It “loses a degree of freedom” if you will.

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 , .

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Algebraic Geometry by Adam Boocher

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