Postmodern Algebra

Let \(\mathfrak{p}\) be a prime ideal of \(A\text{.}\) Then \(B=A / \mathfrak{p}\) is an artinian integral domain. Let \(x \in B, x \neq 0\text{.}\) By the d.c.c. we have \(\left(x^{n}\right)=\left(x^{n+1}\right)\) for some \(n\text{,}\) hence \(x^{n}=x^{n+1} y\) for some \(y \in B\text{.}\) Since \(B\) is an integral domain and \(x \neq 0\text{,}\) it follows that we may cancel \(x^{n}\text{,}\) hence \(x y=1\text{.}\) Hence \(x\) has an inverse in \(B\text{,}\) and therefore \(B\) is a field, so that \(\mathfrak{p}\) is a maximal ideal.

Consider the set of all finite intersections \(\mathfrak{m}_{1} \cap \ldots \cap \mathfrak{m}_{r}\text{,}\) where the \(\mathfrak{m}_{i}\) are maximal ideals. This set has a minimal element, say \(\mathfrak{m}_{1} \cap \cdots \cap \mathfrak{m}_{n}\text{;}\) hence for any maximal ideal \(\mathfrak{m}\) we have \(\mathfrak{m} \cap \mathfrak{m}_{1} \cap \cdots \cap \mathfrak{m}_{n}=\mathfrak{m}_{1} \cap \cdots \cap \mathfrak{m}_{n}\text{,}\) and therefore \(\mathfrak{m} \supseteq \mathfrak{m}_{1} \cap \cdots \cap \mathfrak{m}_{n}\text{.}\) By (1.11) \(\mathfrak{m} \supseteq \mathfrak{m}_{i}\) for some \(i\text{,}\) hence \(\mathfrak{m}=\mathfrak{m}_{\mathfrak{i}}\) since \(\mathfrak{m}_{i}\) is maximal.

By d.c.c. we have \(\mathfrak{R}^{k}=\mathfrak{R}^{k+1}=\cdots=\mathfrak{a}\) say, for some \(k>0\text{.}\) Suppose \(\mathfrak{a} \neq 0\text{,}\) and let \(\Sigma\) denote the set of all ideals \(\mathfrak{b}\) such that \(\mathfrak{a} \mathfrak{b} \neq 0\text{.}\) Then \(\Sigma\) is not empty, since \(a \in \Sigma\text{.}\) Let \(\mathrm{c}\) be a minimal element of \(\Sigma\text{;}\) then there exists \(x \in \mathrm{c}\) such that \(x \mathfrak{a} \neq 0\text{;}\) we have \((x) \subseteq \mathfrak{c}\text{,}\) hence \((x)=\mathfrak{c}\) by the minimality of \(\mathfrak{c}\text{.}\) But \((x \mathfrak{a}) \mathfrak{a}=x \mathfrak{a}^{2}=x \mathfrak{a} \neq 0\text{,}\) and \(x \mathfrak{a} \subseteq(x)\text{,}\) hence \(x \mathfrak{a}=(x)\) (again by minimality). Hence \(x=x y\) for some \(y \in \mathfrak{a}\text{,}\) and therefore \(x=x y=x y^{2}=\cdots=x y^{n}=\cdots\text{.}\) But \(y \in \mathfrak{a}=\mathfrak{N}^{k} \supseteq \mathfrak{N}\text{,}\) hence \(y\) is nilpotent and therefore \(x=x y^{n}=0\text{.}\) This contradicts the choice of \(x\text{,}\) therefore \(\mathfrak{a}=0\text{.}\)

\(\Rightarrow\text{:}\) By (8.1) we have \(\operatorname{dim} A=0\text{.}\) Let \(\mathrm{m}_{i}(1 \leqslant i \leqslant n)\) be the distinct maximal ideals of \(A(8.3)\text{.}\) Then \(\prod_{i=1}^{n} \mathfrak{m}_{i}^{k} \subseteq\left(\bigcap_{i=1}^{n} \mathfrak{m}_{i}\right)^{k}=\mathfrak{R}^{k}=0\text{.}\) Hence by (6.11) \(A\) is Noetherian.

\(\Leftarrow\text{:}\) Since the zero ideal has a primary decomposition (7.13), \(A\) has only a finite number of minimal prime ideals, and these are all maximal since \(\operatorname{din} A=0\text{.}\) Hence \(\Re=\bigcap_{i=1}^{n} m_{i}\) say; we have \(\mathfrak{R}^{k}=0\) by (7.15), hence \(\prod_{i=1}^{n} \mathfrak{m}_{l}^{k}=0\) as in the previous part of the proof. Hence by (6.11) \(A\) is an Artin ring.

Suppose \(\mathfrak{m}^{n}=\mathfrak{m}^{n+1}\) for some \(n\text{.}\) By Nakayama’s lemma (2.6) we have \(\mathfrak{m}^{n}=0\text{.}\) Let \(\mathfrak{p}\) be any prime ideal of \(A\text{.}\) Then \(\mathfrak{m}^{n} \subseteq \mathfrak{p}\text{,}\) hence (taking radicals) \(\mathfrak{m}=\mathfrak{p}\text{.}\) Hence \(\mathfrak{m}\) is the only prime ideal of \(A\) and therefore \(A\) is Artinian.

Let \(\mathfrak{m}_{i}(1 \leqslant i \leqslant n)\) be the distinct maximal ideals of \(A\text{.}\) From the proof of (8.5) we have \(\prod_{i=1}^{n} \mathfrak{m}_{i}^{k}=0\) for some \(k>0\text{.}\) By (1.16) the ideals \(\mathfrak{m}_{i}^{k}\) are coprime in pairs, hence \(\cap \mathfrak{m}_{i}^{k}=\prod \mathfrak{m}_{i}^{k}\) by (1.10). Consequently by (1.10) again the natural mapping \(A \rightarrow \prod_{i=1}^{n}\left(A / \mathfrak{m}_{i}^{k}\right)\) is an isomorphism. Each \(A / \mathfrak{m}_{i}^{k}\) is an Artin local ring, hence \(A\) is a direct product of Artin local rings.

Conversely, suppose \(A \cong \prod_{i=1}^{m} A_{i}\text{,}\) where the \(A_{i}\) are Artin local rings. Then for each \(i\) we have a natural surjective homomorphism (projection on the \(i\) th factor) \(\phi_{i}: A \rightarrow A_{i}\text{.}\) Let \(\mathfrak{a}_{i}=\operatorname{Ker}\left(\phi_{i}\right)\text{.}\) By (1.10) the \(\mathfrak{a}_{i}\) are pairwise coprime, and \(\cap a_{i}=0\text{.}\) Let \(\mathfrak{q}_{i}\) be the unique prime ideal of \(A_{i}\text{,}\) and let \(\mathfrak{p}_{i}\) be its contraction \(\phi_{i}^{-1}\left(\mathfrak{q}_{t}\right)\text{.}\) The ideal \(\mathfrak{p}_{t}\) is prime and therefore maximal by (8.1). Since \(\mathfrak{q}_{i}\) is nilpotent it follows that \(a_{i}\) is \(\mathfrak{p}_{i}\)-primary, and hence \(\cap a_{i}=(0)\) is a primary decomposition of the zero ideal in \(A\text{.}\) Since the \(a_{i}\) are pairwise coprime, so are the \(\mathfrak{p}_{1}\text{,}\) and they are therefore isolated prime ideals of (0). Hence all the primary components \(a_{i}\) are isolated, and therefore uniquely determined by \(A\text{,}\) by the 2nd uniqueness theorem (4.11). Hence the rings \(A_{i} \cong A / \mathfrak{a}_{i}\) are uniquely determined by \(A\text{.}\)