I don't believe this question has been asked here before, so here we go: I'm having trouble with Exercise 11.2 in Atiyah-Macdonald, which states:
Let $A$ be a Noetherian local ring of dimension $d$ with maximal ideal $\mathfrak{m}$ which is $\mathfrak{m}$-adically complete, let $x_1, \ldots, x_d$ be a system of parameters for $A$, and denote the $\mathfrak{m}$-primary ideal they generate by $\mathfrak{q} = (x_1, \ldots, x_d)$. If $k \subset A$ is a field mapping isomorphically onto $A/\mathfrak{m}$, then the homomorphism $k[[t_1, \ldots, t_d]] \to A$ given by $t_i \mapsto x_i$ ($1 \leq i \leq d$) is injective and $A$ is a finitely-generated module over $k[[x_1, \ldots, x_n]]$.
I have already shown that the homomorphism $k[[t_1, \ldots, t_d]] \to A$ exists (i.e. is well-defined) and is injective, as follows: the corresponding homomorphism $k[t_1, \ldots, t_d] \to A$ maps each ideal $(t_1, \ldots, t_n)^n$ onto $\mathfrak{q}^n$, thereby inducing the desired homomorphism $k[[t_1, \ldots, t_d]] \to A$, since $A$ is complete. We prove injectivity by induction: let $f(t)$ be a power series in $k[[t_1, \ldots, t_d]]$ lying in the kernel, and let $f_n(t)$ denote the $n$th homogeneous component of $f$. It is clear that $f_0(t)$ is the zero polynomial, since otherwise $f$ would be a unit and $k[[t_1, \ldots, t_d]] \to A$ would be the zero map, a contradiction. Now suppose $n > 0$ and that $f_k(t) = 0$ for all $k < n$. Evaluating $f(t) – f_n(t)$ at $x_1, \ldots, x_d$ then implies $f(x_1, \ldots, x_d) – f_n(x_1, \ldots, x_d) \in \mathfrak{q}^{n+1}$. Since $f(x_1, \ldots, x_d) = 0$, it follows (by the independence property of a system of parameters) that the coefficients of $f_n$ lie in $\mathfrak{m}$. But the coefficients are also in $k$; since $k \cap \mathfrak{m} = 0$, this implies that $f_n(t) = 0$. Hence $f = 0$ and $k[[t_1, \ldots, t_d]] \to A$ is injective.
I'm struggling to prove the last part, that $A$ is a finitely-generated $k[[x_1, \ldots, x_n]]$-module; following the book's hint, I've reduced it to showing that $G_\mathfrak{q}(A)$ is a finitely-generated graded module over the graded ring $G_{(t_1, \ldots, t_n)} k[[t_1, \ldots, t_n]]$, which is isomorphic to $G_{(t_1, \ldots, t_n)} k[t_1, \ldots, t_n] \cong k[t_1, \ldots, t_n]$. It is clear by construction that $k[t_1, \ldots, t_n]$ maps surjectively onto the generators (as $A/\mathfrak{q}$-modules) of each homogeneous component $\mathfrak{q}^n/\mathfrak{q}^{n+1}$ in $G_\mathfrak{q}(A)$ of degree $n \geq 0$, so it would suffice to show that $A/\mathfrak{q}$ is finitely-generated as a $k$-module. Why should this be true? I suspect that I'm supposed to use the fact that $A/\mathfrak{q}$ is an Artin local ring, but I'm not too sure how to go about it.
Edit: Immediately after posting this question, I found a solution by proving the following lemma: If $A$ is an Artin local ring, $\mathfrak{m}$ its maximal ideal, and $k \subset A$ a field mapping isomorphically onto $A/\mathfrak{m}$, then $A$ is a finite-dimensional $k$-vector space.
Proof: Since $A$ is an Artin local ring, there exists an integer $n$ for which $\mathfrak{m}^{n+1} = 0$. Each $\mathfrak{m}^k/\mathfrak{m}^{k+1}$ is an Artin $A$-module, hence of finite length; naturally regarded as an $A/\mathfrak{m} \cong k$-module, this implies that each $\mathfrak{m}^k/\mathfrak{m}^{k+1}$ has finite dimension as a $k$-vector space. Thus, $\dim_k A = \dim_k (A/\mathfrak{m}^{n+1}) = \sum_{k=0}^n \dim_k(\mathfrak{m}^k/\mathfrak{m}^{k+1})$ is finite.
Best Answer
(I am self-answering for the sake of producing an accepted answer.) As noted above, the exercise boils down to proving the following lemma:
If $A$ is an Artin local ring, $\mathfrak{m}$ its maximal ideal, and $k\subset A$ a field mapping isomorphically onto $A/\mathfrak{m}$, then $A$ is a finite-dimensional $k$-vector space.
Proof: Since $A$ is an Artin local ring, there exists an integer $n$ for which $\mathfrak{m}^{n+1}=0$. Each $\mathfrak{m}^k/\mathfrak{m}^{k+1}$ is an Artin $A$-module, hence of finite length; naturally regarded as an $A/\mathfrak{m}≅k$-module, this implies that each $\mathfrak{m}^k/\mathfrak{m}^{k+1}$ has finite dimension as a $k$-vector space. Thus, $\dim_kA=\dim_k(A/\mathfrak{m}^{n+1})=\sum_{k=0}^n \dim_k(\mathfrak{m}^k/\mathfrak{m}^{k+1})$ is finite.