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I'm studying chapter 11 (Dimension Theory) in Atiyah / Macdonald - Intro to Commutative Algebra. Let $ A $ be a Noetherian local ring with $\mathfrak{m}$-primary ideal $\mathfrak{q}$. The book defines $d (A)$ as the common degree of the characteristic polynomial $\chi_q(n) = l(A/q^n) $ for $n $ large enough. Let $ G_{\mathfrak{q}}(A )$ be associated the graded ring. The book defines $d(G_{\mathfrak{q}}(A ))$ to be the order of the pole $1$ in the Poincare series.

On page 119, the book says $d(G_{\mathfrak{q}}(A )) = d( A)$ in light of corollary 11.2. This corollary says

For all sufficiently large $ n $, $l(M_n) $ is a polynomial in $ n $ of degree $ d - 1 $.

$d $ in the corollary is the order of pole $ 1$. My question: Doesn't this imply that $ d(G_{\mathfrak{q}}(A) ) = d( A) + 1$ instead?

PeterM
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1 Answers1

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From Corollary 11.2 we have that $\lambda(m^n/m^{n+1})$ is a polynomial in $n$ (for $n$ large enough) of degree $d-1$, where $d$ is the order of the pole at $t=1$ of the Poincare series of $G_m(A)$. But $$\lambda(m^n/m^{n+1})=\lambda(A/m^{n+1})-\lambda(A/m^n).\qquad (*)$$ We also know that $\chi_m(n)=\lambda(A/m^n)$ is a polynomial in $n$ (for $n$ large enough) whose degree is $d(A)$. Now one can easily conclude from $(*)$ that $d-1=d(A)-1$, that is, $d=d(A)$.

user26857
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