Proof of Proposition 11.20 of Atiyah-Macdonald I struggle with verifying the pole order inequality asserted in the proof of proposition 11.20. (Full statement and proof of the proposition can be found here: Atiyah-Macdonald 11.20 and 11.21)
My question is: how to prove this inequality?
I find several online resources covering various issues with the book, but I find nothing on this particular problem. I think it would be beneficial that some reference to this is made available as well, as an insightful answer might be helpful to anybody trying to learn the subject from this book.
In case it is of interest, I've based my own efforts on the following additional assumptions:

*

*$d((A/\mathfrak{q})[t_1,...,t_d]/(\bar{f}))$ must refer to the pole order as the other $d$ (the degree of the characteristic polynomial) is defined for local rings only.

*The graded structure of this ring is $\bigoplus A_n/\bar{f}A_{n-s}$, where $\bigoplus A_n$ is the standard grading of $(A/\mathfrak{q})[t_1,...,t_d]$.

EDIT: I guess the problem is not clear enough unless one is quite deep in the book, so I will provide a short summary of the relevant results found in Chapter 11 up to (11.20): For a Noetherian graded ring $A$ generated as an $A_0$-algebra by $s$ homogeneous elements of degree 1, Theorem (11.1) states that the Poincaré series $P(M,t) = \sum^\infty_{n=0}\lambda(M_n)t^n$ of any finitely-generated graded $A$-module $M$ has a pole of order $d(M)\leq s$ at $t=1$. This gives an upper bound for $d(A)$ when taking $M=A$. The inequality in (11.20), however, introduces a lower bound for $d((A/\mathfrak{q})[t_1,...,t_d]/(\bar{f}))$. A lower bound of the pole order occurs earlier in the text only in the form of an equality, namely in the very specialized case that the graded ring is the associated graded ring $G_\mathfrak{q}(A)$ of a Noetherian local ring $A$ wrt. an $\mathfrak{m}$-primary ideal $\mathfrak{q}$ [the pole order of $G_\mathfrak{q}(A)$ is in this case equal to dim $A$]. Hence the difficulty lies in the lack of results for determining lower bounds of the pole order.
 A: Let $\bigoplus A_n$ be the standard grading of $(A/\mathfrak{q})[t_1,\dots,t_d]$. The homomorphism of graded rings $\bigoplus A_n \to \bigoplus A_n/\bar{f}A_{n-s}$ is surjective and has kernel $(\bar{f})$, hence $\bigoplus A_n/\bar{f}A_{n-s}$ is a grading of $(A/\mathfrak{q})[t_1,\dots,t_d]/(\bar{f})$. $\alpha$ induces a map $\bigoplus A_n/\bar{f}A_{n-s} \to \bigoplus \mathfrak{q}^n/\mathfrak{q}^{n+1}$ since $(\bar{f}) \subseteq \textrm{Ker}(\alpha)$, and so we obtain the following surjective homomorphisms of graded rings:
$$ \bigoplus A_n \to \bigoplus A_n/\bar{f}A_{n-s} \to \bigoplus \mathfrak{q}^n/\mathfrak{q}^{n+1}. $$
Note that $A_n/\bar{f}A_{n-s}$ and $\mathfrak{q}^n/\mathfrak{q}^{n+1}$ are $A/\mathfrak{q}$-modules for all $n$ (assuming $s > 0$), and must therefore have finite length since $A/\mathfrak{q}$ is Artin. Since $\mathfrak{q}^n/\mathfrak{q}^{n+1}$ is the homomorphic image of $A_n/\bar{f}A_{n-s}$, we also have that $l(\mathfrak{q}^n/\mathfrak{q}^{n+1}) \leq l(A_n/\bar{f}A_{n-s})$. Finally observe that since $\bigoplus A_n$ is generated as an $A/\mathfrak{q}$-algebra by $t_1,\dots,t_d$, the two other rings are generated by the respective images of these. As these images are all homogeneous of degree 1, we get from (11.2) that for all large $n$, $l(\mathfrak{q}^n/\mathfrak{q}^{n+1})$ is a polynomial $g(n)$ of degree $d(\bigoplus \mathfrak{q}^n/\mathfrak{q}^{n+1}) - 1$ and $l(A_n/\bar{f}A_{n-s})$ is a polynomial $h(n)$ of degree $d(\bigoplus A_n/\bar{f}A_{n-s}) - 1$. Now since
$$g(n) = l(\mathfrak{q}^n/\mathfrak{q}^{n+1}) \leq l(A_n/\bar{f}A_{n-s}) = h(n)$$
for all large $n$, we must have that $\deg g(n) \leq \deg h(n)$, thus
$$ d(\bigoplus \mathfrak{q}^n/\mathfrak{q}^{n+1}) = \deg g(n) + 1 \leq \deg h(n) + 1 = d(\bigoplus A_n/\bar{f}A_{n-s}) $$
which proves the inequality.
