When \(M=0\) we take \(n=0\) and there is nothing to prove. When \(M\neq 0\) we have \(\operatorname{Ass}(M)\neq \varnothing \) and we can take \({\mathfrak {p}}_1\) be any element in \(\operatorname{Ass}(M)\) and we obtain \(M_1:=A/{\mathfrak {p}}_1\hookrightarrow M\). If \(M_1\neq M\), do the same for \(M/M_1\) to get \(M_2/M_1\cong A/{\mathfrak {p}}_2\hookrightarrow M/M_1\). Since \(M\) is a Noetherian \(A\)-module, the chain \(M_1\subset M_2\subset \cdots \) must stop after a finite number of steps.

If \({\mathfrak {p}}\) is a minimal prime over-ideal of \({\mathfrak {a}}\), then \({\mathfrak {a}}A_{\mathfrak {p}}\subset {\mathfrak {p}}A_{\mathfrak {p}}\), and \({\mathfrak {p}}(A_{\mathfrak {p}}/{\mathfrak {a}}A_{\mathfrak {p}})\), which is the only maximal ideal of \(A_{\mathfrak {p}}/{\mathfrak {a}}A_{\mathfrak {p}}\), is also a minimal prime ideal. Therefore \(A_{\mathfrak {p}}/{\mathfrak {a}}A_{\mathfrak {p}}\) is Artinian, and so \({\mathfrak {a}}A_{\mathfrak {p}}\) is an open ideal. It follows that \({\mathrm{ht}}({\mathfrak {p}})=\dim (A_{\mathfrak {p}})\leq n\).

Conversely, if \({\mathfrak {p}}\) is a prime ideal of height \(\leq n\), then \(\dim (A_{\mathfrak {p}})\leq n\), so there exists an ideal of definition \(IA_{\mathfrak {p}}\) of \(A_{\mathfrak {p}}\) (i.e. an open ideal with respect to \({\mathfrak {p}}A_{\mathfrak {p}}\)-adic topology), generated by \(n\) elements. Clearing their denominators we may assume that the generators take the form \(\frac{x_1}s,\cdots ,\frac{x_n}s\) for some \(x_1,\cdots ,x_n\in I\) and \(s\in A\setminus {\mathfrak {p}}\). Since \({\mathfrak {p}}^NA_{\mathfrak {p}}\subset IA_{\mathfrak {p}}\subset {\mathfrak {p}}A_{\mathfrak {p}}\) for some integer \(N\geq 1\), we obtain \({\mathfrak {p}}^N\subset I\subset {\mathfrak {p}}\), and \(IA_{\mathfrak {p}}=(x_1,\cdots ,x_n)A_{\mathfrak {p}}\) (hence we may replace \(I\) by \((x_1,\cdots ,x_n)\)), and that \({\mathfrak {p}}\) is a minimal prime over-ideal of \(I\) (since if \(I\subset {\mathfrak {p}}'\subset {\mathfrak {p}}\) for some prime ideal \({\mathfrak {p}}'\), then \({\mathfrak {p}}^N\subset {\mathfrak {p}}'\), hence \({\mathfrak {p}}\subset {\mathfrak {p}}'\), so \({\mathfrak {p}}={\mathfrak {p}}'\)).

“\(\Rightarrow \)”: Let \({\mathfrak {p}}\) be a prime ideal of height \(1\), and let \(a\in {\mathfrak {p}}\) be a non-zero element. Factor \(a\) into a product of irreducible elements (= prime elements) \(a=p_1\cdots p_r\). Then by \(a\in {\mathfrak {p}}\) we obtain \(p_i\in {\mathfrak {p}}\) for some \(i\), since \(({\mathfrak {p}}_i)\neq 0\) is a prime ideal contained in \({\mathfrak {p}}\), we have \({\mathfrak {p}}=(p_i)\).

“\(\Leftarrow \)”: It suffices to show that every irreducible element is prime. Let \(p\in A\) be irreducible, and \({\mathfrak {p}}\) be a prime ideal of \(A\) minimal containing \(p\). Then \({\mathfrak {p}}\) is of height \(1\) by Krull’s Hauptidealsatz (Theorem 2.2), so \({\mathfrak {p}}=(\pi )\) is principal. Since \(p\in {\mathfrak {p}}=(\pi )\), \(\pi \mid p\), and since \(p\) is irreducible, \(p=u\pi \) for some unit \(u\), therefore \(p\) is prime (which generates \({\mathfrak {p}}\)).

Let \(Y:=\Lambda x_1+\cdots +\Lambda x_t\) be the \(\Lambda \)-submodule of \(X\) generated by \(x_1,\cdots ,x_t\). Then by the image of \(Y\) in \(X/TX\) is \(X/TX\), we know that \(Y+TX=X\). Note that \(Y\) is compact, so it is closed in \(X\), so \(Z:=X/Y\) is a pro-\(p\) abelian group, and the image of \(TX\) in \(Z\) is \(TZ\). On the other hand, by \(Y+TX=X\), the image of \(TX\) in \(Z\) is also \(Z\), so \(TZ=Z\). Since \(T\) is topologically nilpotent in \(\Lambda \), for any open subgroup \(U\) of \(Z\), there exists \(N\geq 0\) such that for any \(n\geq N\), \(T^nZ\subset U\). But by \(TZ=Z\) we know that \(T^nZ=Z\) for any \(n\geq 0\). Therefore \(Z\) must be zero.

Omitted.

Suppose \(x\in {\mathrm{Frac}}(A)\) is integral over \(A\). Then it’s also integral over \(A_{\mathfrak {p}}\) for all \({\mathfrak {p}}\in {\mathrm{Spec}}(A)\). Hence \(x\in A_{\mathfrak {p}}\) for all \({\mathfrak {p}}\in S\). So \(x\in \bigcap _{{\mathfrak {p}}\in S}A_{\mathfrak {p}}=A\).

“\(\Rightarrow \)”: Lemma 2.8.

“\(\Leftarrow \)”: Let \({\mathfrak {p}}\) be a height one prime of \(A\). Then \(A_{\mathfrak {p}}\) is integrally closed (isIntegrallyClosed_of_isLocalization). We have \({\mathrm{Spec}}(A_{\mathfrak {p}})=\{ 0,{\mathfrak {p}}A_{\mathfrak {p}}\} \) hence \(A_{\mathfrak {p}}\) is also a Noetherian local domain of dimension one. Now by Proposition 2.6, \(A_{\mathfrak {p}}\) is a DVR.

??? ???

Suppose \(M_{{\mathfrak {m}}_i}\) is generated by \(\{ x_{i,1},\cdots , x_{i,n_i}\} \) for each maximal ideal \({\mathfrak {m}}_i\) of \(A\). Let \(y_{i,k}\) be a numerator of \(x_{i,k}\), which is in \(A\).

We prove that \(\bigcup _{i}\{ y_{i,1} \cdots , y_{i,n_i}\} \) is a finite set (Since \(A\) is semilocal) of generators of \(M\). Let \(N\) be the \(A\)-submodule of \(M\) generated by \(\bigcup _{i}\{ y_{i,1} \cdots , y_{i,n_i}\} \).

By local property (Submodule.eq_top_of_localization_maximal), it suffices to show that \(N_{{\mathfrak {m}}_{i}} = M_{{\mathfrak {m}}_{i}}\) for all maximal ideals \({\mathfrak {m}}_i\) of \(A\). It’s clear that \(N_{{\mathfrak {m}}_{i}} \subseteq M_{{\mathfrak {m}}_{i}}\), so we only need to show that \(M_{{\mathfrak {m}}_{i}} \subseteq N_{{\mathfrak {m}}_{i}}\).

In other words, need to prove that \(M_{{\mathfrak {m}}_{i}} \subseteq \operatorname {span}_{A_{{\mathfrak {m}}_i}} f_i (\bigcup _{i}\{ y_{i,1} \cdots , y_{i,n_i}\} )\) where \(f_i : M \to M_{{\mathfrak {m}}_i}\) is the localization map. It is enough to show the set of generators of \(M_{{\mathfrak {m}}_{i}}\) is contained in right hand side.

Take any \(x\) in the set of generators of \(M_{{\mathfrak {m}}_{i}}\). It remains to show that the image of a numerator of \(x\) under \(f_i\) is in the generators of right hand side by Submodule.mem_of_numerator_image_mem and Submodule.mem_span. i.e. \(f_i(y) \in f_i (\bigcup _{i}\{ y_{i,1} \cdots , y_{i,n_i}\} )\) for some numerator \(y\) of \(x\).

We take \(y\) as some \(y_{i,k}\) as above. By definition of \(y_{i,k}\), we have \(f_i(y_{i,k})\) in right hand side.

It’s known that a semilocal Dedeking domain is a PID (IsPrincipalIdealRing.of_finite_primes). So we only need to show \(A\) is a Dedekind domain.

Let \(I\) be any ideal of \(A\). Apply Lemma 2.10 to \(I\) we know that \(I\) is finitely generated. Hence \(A\) is a Noetherian ring.

Let \({\mathfrak {p}}\neq 0\) be a prime ideal of \(A\). Choose a maximal ideal \({\mathfrak {m}}\) containing \({\mathfrak {p}}\). Then \(0\neq {\mathfrak {p}}A_{\mathfrak {m}}\subset {\mathfrak {m}}A_{\mathfrak {m}}\). Since \(A_{\mathfrak {m}}\) is a PID (hence DVR), we have \({\mathrm{Spec}}(A_{\mathfrak {m}})=\{ 0,{\mathfrak {m}}A_{\mathfrak {m}}\} \), so know that \({\mathfrak {p}}A_{\mathfrak {m}}={\mathfrak {m}}A_{\mathfrak {m}}\), hence \({\mathfrak {p}}={\mathfrak {m}}\) is maximal.

It’s known that if the localizations of a domain at all maximal ideals are integrally closed, then the domain itself is integrally closed (IsIntegrallyClosed.of_localization_maximal). Hence our \(A\) is integrally closed. This completes the proof.