Postmodern Algebra

Since \(A\) is Noetherian, a has a minimal primary decomposition \(\mathfrak{a}=\bigcap_{i=1}^{n} \mathfrak{q}_{i}\) by (7.13), where each \(\mathfrak{q}_{1}\) is say \(\mathfrak{p}_{1}\)-primary. Since \(\operatorname{dim} A=1\) and \(A\) is an integral domain, each non-zero prime ideal of \(A\) is maximal, hence the \(\mathfrak{p}_{i}\) are distinct maximal ideals (since \(\mathfrak{p}_{\mathfrak{i}} \supseteq \mathfrak{q}_{\mathfrak{i}} \supseteq \mathfrak{a} \neq 0\) ), and are therefore pairwise coprime. Hence by (1.16) the \(\mathfrak{q}_{1}\) are pairwise coprime and therefore by (1.10) we have \(\prod \mathfrak{q}_{\mathfrak{i}}=\bigcap \mathfrak{q}_{\mathfrak{i}}\text{.}\) Hence \(\mathfrak{a}=\prod \mathfrak{q}_{\mathfrak{i}}\text{.}\)

Before we start going the rounds, we make two remarks: (A) If \(\mathfrak{a}\) is an ideal \(\neq 0\text{,}\) (1), then \(\mathfrak{a}\) is \(\mathfrak{m}\)-primary and \(\mathfrak{a} \supseteq \mathfrak{m}^{n}\) for some \(n\text{.}\) For \(r(\mathfrak{a})=\mathfrak{m}\text{,}\) since \(\mathfrak{m}\) is the only non-zero prime ideal; now use (7.16). (B) \(\mathfrak{m}^{n} \neq \mathfrak{m}^{n+1}\) for all \(n \geqslant 0\text{.}\) This follows from (8.6). i) \(\Rightarrow\) ii) by (5.18). ii) \(\Rightarrow\) iii). Let \(a \in \mathfrak{m}\) and \(a \neq 0\text{.}\) By remark (A) there exists an integer \(n\) such that \(\mathfrak{m}^{n} \subseteq(a), \mathfrak{m}^{n-1} \ddagger(a)\text{.}\) Choose \(b \in \mathfrak{m}^{n-1}\) and \(b \notin(a)\text{,}\) and let \(x=\) \(a / b \in K\text{,}\) the field of fractions of \(A\text{.}\) We have \(x^{-1} \notin A\) (since \(b \notin(a)\) ), hence \(x^{-1}\) is not integral over \(A\text{,}\) and therefore by (5.1) we have \(x^{-1} \mathfrak{m} \ddagger \mathfrak{m}\) (for if \(x^{-1} \mathfrak{m} \subseteq \mathfrak{m}\text{,}\) m would be a faithful \(A\left[x^{-1}\right]\)-module, finitely generated as an \(A\)-module). But \(x^{-1} \mathfrak{m} \subseteq A\) by construction of \(x\text{,}\) hence \(x^{-1} \mathfrak{m}=A\) and therefore \(\mathfrak{m}=A x=(x)\text{.}\) iii) \(\Rightarrow\) iv). By (2.8) we have \(\operatorname{dim}_{k}\left(\mathfrak{m} / \mathfrak{m}^{2}\right) \leqslant 1\text{,}\) and by remark (B) \(\mathfrak{m} / \mathfrak{m}^{2} \neq 0\text{.}\) iv) \(\Rightarrow v\) ). Let \(\mathfrak{a}\) be an ideal \(\neq(0)\text{,}\) (1). By remark (A) we have \(\mathfrak{a} \supseteq \mathfrak{m}^{n}\) for some \(n\text{;}\) from (8.8) (applied to \(A / \mathfrak{m}^{n}\) ) it follows that \(\mathfrak{a}\) is a power of \(\mathfrak{m}\text{.}\) v) \(\Rightarrow\) vi). By remark (B), \(\mathfrak{m} \neq \mathfrak{m}^{2}\text{,}\) hence there exists \(x \in \mathfrak{m}, x \notin \mathfrak{m}^{2}\text{.}\) But \((x)=\mathfrak{m}^{r}\) by hypothesis, hence \(r=1,(x)=\mathfrak{m},\left(x^{k}\right)=\mathfrak{m}^{k}\text{.}\) vi) \(\Rightarrow\) i). Clearly \((x)=\mathfrak{m}\text{,}\) hence \(\left(x^{k}\right) \neq\left(x^{k+1}\right)\) by remark (B). Hence if \(a\) is any non-zero element of \(A\text{,}\) we have \((a)=\left(x^{k}\right)\) for exactly one value of \(k\text{.}\) Define \(v(a)=k\) and extend \(v\) to \(K^{*}\) by defining \(v\left(a b^{-1}\right)=v(a)-v(b)\text{.}\) Check that \(v\) is well-defined and is a discrete valuation, and that \(A\) is the valuation ring of \(v\text{.}\)

i) \(\Leftrightarrow\) iii) by (9.2) and (5.13). ii) \(\Leftrightarrow\) iii). Use (9.2) and the fact that primary ideals and powers of ideals behave well under localization: (4.8), (3.11).

(9.1) and (9.3).

\(K\) is a separable extension of \(\mathbf{Q}\) (because the characteristic is zero), hence by (5.17) there is a basis \(v_{1}, \ldots, v_{n}\) of \(K\) over \(\mathbf{Q}\) such that \(A \subseteq \Sigma \mathbf{Z}^{\prime}\text{,}\) Hence \(A\) is finitely generated as a \(\mathbf{Z}\)-module and therefore Noetherian. Also \(A\) is integrally closed by (5.5). To complete the proof we must show that every nonzero prime ideal \(\mathfrak{p}\) of \(A\) is maximal, and this follows from (5.8) and (5.9): (5.9) shows that \(\mathfrak{p} \cap \mathbf{Z} \neq 0\text{,}\) hence \(\mathfrak{p} \cap \mathbf{Z}\) is a maximal ideal of \(\mathbf{Z}\) and therefore \(\mathfrak{p}\) is maximal in \(A\) by (5.8).

i) \(\Rightarrow\) ii): \(A_{\mathfrak{p}}=(M \cdot(A: M))_{\mathfrak{p}}=M_{\mathfrak{p}} \cdot\left(A_{\mathfrak{p}}: M_{\mathfrak{p}}\right)\) by (3.11) and (3.15) (for \(M\) is finitely generated, because invertible). ii) \(\Rightarrow\) iii) as usual. iii \(\Rightarrow \mathrm{i})\) : Let \(\mathfrak{a}=M \cdot(A: M)\text{,}\) which is an integral ideal. For each maximal ideal \(\mathfrak{m}\) we have \(\mathfrak{a}_{\mathfrak{m}}=M_{\mathfrak{m}} \cdot\left(A_{\mathfrak{m}}: M_{\mathfrak{m}}\right)\) (by (3.11) and (3.15)) \(=A_{\mathfrak{n}}\) because \(M_{\mathfrak{m}}\) is invertible. Hence \(\mathfrak{a} \neq \mathfrak{m}\text{.}\) Consequently \(\mathfrak{a}=A\) and therefore \(M\) is invertible.

\(\Rightarrow\text{.}\) Let \(x\) be a generator of the maximal ideal \(\mathfrak{m}\) of \(A\text{,}\) and let \(M \neq 0\) be a fractional ideal. Then there exists \(y \in A\) such that \(y M \subseteq A\) : thus \(y M\) is an integral ideal, say \(\left(x^{r}\right)\text{,}\) and therefore \(M=\left(x^{r-s}\right)\) where \(s=v(y)\text{.}\)

\(\Leftarrow\) : Every non-zero integral ideal is invertible and therefore finitely generated; so that \(A\) is Noetherian. It is therefore enough to prove that every nonzero integral ideal is a power of \(m\text{.}\) Suppose this is false; let \(\Sigma\) be the set of nonzero ideals which are not powers of \(\mathfrak{m}\text{,}\) and let \(\mathfrak{a}\) be a maximal element of \(\Sigma\text{.}\) Then \(\mathfrak{a} \neq \mathfrak{m}\text{,}\) hence \(\mathfrak{a} \subset \mathfrak{m}\text{;}\) hence \(\mathfrak{m}^{-1} \mathfrak{a} \subset \mathfrak{m}^{-1} \mathfrak{m}=A\) is a proper (integral) ideal, and \(\mathfrak{m}^{-1} \mathfrak{a} \supseteq \mathfrak{a}\text{.}\) If \(\mathfrak{m}^{-1} \mathfrak{a}=\mathfrak{a}\text{,}\) then \(\mathfrak{a}=\mathfrak{m} \mathfrak{a}\) and therefore \(\mathfrak{a}=0\) by Nakayama’s lemma (2.6); hence \(\mathfrak{m}^{-1} \mathfrak{a} \supset \mathfrak{a}\) and hence \(\mathfrak{m}^{-1} \mathfrak{a}\) is a power of \(\mathfrak{m}\) (by the maximality of \(\mathfrak{a}\) ). Hence \(\mathfrak{a}\) is a power of \(\mathfrak{m}\) : contradiction.

\(\Rightarrow\) : Let \(M \neq 0\) be a fractional ideal. Since \(A\) is Noetherian, \(M\) is finitely generated. For each prime ideal \(\mathfrak{p} \neq 0, M_{\mathfrak{p}}\) is a fractional ideal \(\neq 0\) of the discrete valuation ring \(A_{\mathfrak{p}}\text{,}\) hence is invertible by (9.7). Hence \(M\) is invertible, by (9.6).

\(\Leftarrow\text{:}\) Every non-zero integral ideal is invertible, hence finitely generated, hence \(A\) is Noetherian. We shall show that each \(A_{\mathfrak{p}}(\mathfrak{p} \neq 0)\) is a discrete valuation ring. For this it is enough to show that each integral ideal \(\neq 0\) in \(A_{\mathfrak{p}}\) is invertible, and then use (9.7). Let \(\mathfrak{b} \neq 0\) be an (integral) ideal in \(A_{\mathfrak{p}}\text{,}\) and let \(\mathfrak{a}=\mathfrak{b}^{\mathfrak{c}}=\mathfrak{b} \cap A\text{.}\) Then \(\mathfrak{a}\) is invertible, hence \(\mathfrak{b}=\mathfrak{a}_{\mathfrak{p}}\) is invertible by (9.7).