# Use the degree of Hilbert polynomial to define the dimension of a ring

Let $R$ be a local ring , $\mathfrak{m}$ the maximal ideal, $q$ is $\mathfrak m$-primary . Then we can prove that there exists a polynomial $F_q(t)\in \mathbb{Q}(t)$ such that $F_q(n)=\mathcal{l}(R/\mathfrak q^n)$ for $n>>0$ . Let $d_q(R)$ be the degree of $F_q(t)$ . Then we can prove that the value of $d_q(R)$ don't depend on the choice of $q$ .

Now we want to show that $d(R/(a))\leqq d(R)-1$ where $a$ is not a zero-divisor. Then the degree of $\mathcal l(\bar R/\bar {\mathfrak m}^n)$ where $\bar R=R/(a)$ and $\bar {\mathfrak m}=\mathfrak m/(a)$ is less than $d(R)-1$ .

My question is how to get the conclusion we want by this?

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See Theorem 13.3 in Matsumura's Commutative Ring Theory and apply to exact sequence $0 \to R \overset{a}{\to} R \to \bar{R} \to 0$. Otherwise see page 120 of Atiyah-Macdonald. –  Andrea Sep 1 '11 at 10:42
Thank you . In fact , it is the definition of $d(R/(a))$ which I didn't recognise . Thank you all the same . –  molan Sep 3 '11 at 2:09