Why is the Hilbert polynomial of projective varieties an invariant

algebraic-geometrycommutative-algebra

Suppose $S=k[x_0,…,x_n]$ and $M$ is the homogeneous coordinate ring of some projective variety in $P^n$. Then $M$ is a graded $S$ module. By a theorem of Hilbert-Serre, Hilbert polynomial is defined to be the unique polynomial, $P_M$, such that $P_M(l)=\dim_k (M_l)$.

Is the Hilbert polynomial an invariant under birational map or isomorphism? Why?

Best Answer

No, the Hilbert polynomial is defined as a function of the specific projective embedding. Its degree is an invariant (always the dimension of the projective variety), but the general polynomial isn’t.

For instance, the Hilbert polynomial of the identity inclusion $\mathbb{P}^1 \subset \mathbb{P}^1$ is $H(T)=T+1$. But the Hilbert polynomial of the Veronese embedding $\mathbb{P}^1 \longmapsto \mathbb{P}^n$ is $H(T)=nT+1$ (I think).

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