# Hilbert Polynomial of zero dimensional module

Let $M$ be a finitely generated graded $A$-module with $\dim M=0.$ Let $M$ be $\mathbb Z$-graded and $A$ be $\mathbb N$-graded Noetherian ring. Then how can we say that the Hilbert function of $M$ is polynomial equivalent to a constant polynomial ?

What I know that under these condition $l_A(M)< \infty.$

Help me. Thanks.

• i was reading a theorem from Bruns and Herzogs book on commutative algebra (thm 4.1.3) which uses some induction on dimension of M, whose base of induction is my qsn. – user371231 Dec 1 '16 at 20:16
• In fact, the Hilbert polynomial is zero in this case. Since $M$ is artinian we have $M_i=0$ for $i$ large, otherwise we get a strictly descending chain of submodules. Suppose there is an infinite sequence $i_1<i_2<\cdots$ such that $M_{i_k}\ne0$ for all $k$. Then $\oplus_{j\ge i_1}M_j\supsetneq \oplus_{j\ge i_2}M_j\supsetneq\cdots$ is a strictly descending chain of submodules, a contradiction. – user26857 Dec 1 '16 at 20:47
• Thank you.I proved that M is Artinian module but didn't note that $M_i=0$ for large i. – user371231 Dec 1 '16 at 20:50