# Question about Exercise 11.2 in Atiyah-Macdonald

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.

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.