# Algebraic Geometry by Adam Boocher

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Sample text

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 , . . , br , and vi · bi = 0. Then we claim that b1 , . . , br are linearly independent.

We can actually do this for each point pi so we in fact get a system of equations c11 a1 c21 a1 cr1 a1 + ... + ... + 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.

An cr1 · · · crm Call the rows of this matrix b1 , . . , br . Since the points impose independent conditions, the dimension of the nullspace of this matrix is m − r, so it follows that the bi are all linearly independent. Let our “coefficient vector” be v = (a1 , . . , am ). Then the matrix equation tells us that bi · v = 0 for each i. In other words, each row is perpendicular to v. We now pause to collect our thoughts and think about what we are trying to prove. Goal: Pick any point p ∈ Z.