By Adam Boocher

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**Example text**

The first set imposes independent conditions on forms of degree 1. To see this, remove any point. Then there is a line (a form of degree 1) passing through the other two but not the point you removed. Thus the first set imposes independent conditions on forms of degree one. The second set does not impose independent conditions on forms of degree 1 because to draw a line through two points, the line will also contain the third. We make the note that both sets impose independent conditions on forms of degree 2.

To do this, we will need more algebraic machinery, however. In the meantime, here’s another interesting fact about Hilbert functions. Proposition 29. If Z is a set of points in P2 then for any t, hz (t) ≤ hz (t + 1).

Proposition 26. Let Z be a set of points in P2 that imposes independent conditions on forms of degree d. Then Z also imposes independent conditions on forms of degree d + 1. Remark: This proposition along with induction shows that once a set starts 37 imposing independent conditions, it does so thereafter. Thus once we get independent conditions, we know how to compute dim(IZ )d thereafter. Proof. To show that Z imposes independent conditions on forms of degree d+1, pick any point p ∈ Z. Since Z imposes independent conditions on forms of degree d, we know there exists a form of degree d (call it C) that passes through all of Z except the point P .

### Algebraic Geometry by Adam Boocher

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