By Proposition 2.1, we may let \(0=M_0\subset M_1\subset \cdots \subset M_n=M\) be a filtration of \(M\) such that for each \(1\leq i\leq n\), \(M_i/M_{i-1}\cong A/{\mathfrak {p}}_i\) for some prime ideal \({\mathfrak {p}}_i\) of \(A\). Note that if \({\mathfrak {p}},{\mathfrak {q}}\) are prime ideals of \(A\), then \((A/{\mathfrak {p}})_{\mathfrak {q}}\neq 0\) if and only if \({\mathfrak {p}}\subset {\mathfrak {q}}\). Therefore by \(M\) is torsion \(A\)-module, we obtain that \({\mathrm{ht}}({\mathfrak {p}}_i)\geq 1\) for all \(1\leq i\leq n\), and if \({\mathfrak {p}}\) is a height one prime, then \(M_{\mathfrak {p}}\neq 0\) if and only if \({\mathfrak {p}}_i\subset {\mathfrak {p}}\) for some \(i\), by height considerations this means that \({\mathfrak {p}}_i={\mathfrak {p}}\) for some \(i\), hence such \({\mathfrak {p}}\) are only finitely many.

To prove \(\ell _{A_{\mathfrak {p}}}(M_{\mathfrak {p}}){\lt}\infty \), we only need to show that if \({\mathfrak {p}},{\mathfrak {q}}\) are prime ideals of \(A\) such that \({\mathrm{ht}}({\mathfrak {p}})\geq 1\) and \({\mathrm{ht}}({\mathfrak {q}})=1\), then \((A/{\mathfrak {p}})_{\mathfrak {q}}\) is an \(A_{\mathfrak {q}}\)-module of finite length. In fact, by height considerations we know that \((A/{\mathfrak {p}})_{\mathfrak {q}}\neq 0\) if and only if \({\mathfrak {p}}={\mathfrak {q}}\), in this case \((A/{\mathfrak {q}})_{\mathfrak {q}}=A_{\mathfrak {q}}/{\mathfrak {q}}A_{\mathfrak {q}}=k({\mathfrak {q}})\) is the residue field of \({\mathfrak {q}}\), which is an \(A_{\mathfrak {q}}\)-module of length one.

(Another proof without using Proposition 2.1. Note that \(M_{\mathfrak {p}}=0\) for all minimal prime ideals of \(A\), therefore if \({\mathfrak {p}}\) is of height one such that \(M_{\mathfrak {p}}\neq 0\), then \({\mathfrak {p}}\) is a minimal element in \({\mathrm{Supp}}(M)\), hence \({\mathfrak {p}}\in \operatorname{Ass}(M)\) which is a finite set. So there are only finitely many height one primes \({\mathfrak {p}}\) of \(A\) such that \(M_{\mathfrak {p}}\neq 0\).

Suppose \({\mathfrak {p}}\) is a height one prime such that \(M_{\mathfrak {p}}\neq 0\). To prove that \(M_{\mathfrak {p}}\) is an \(A_{\mathfrak {p}}\)-module of finite length, we only need to prove that the ring \(A_{\mathfrak {p}}/{\mathrm{Ann}}_{A_{\mathfrak {p}}}(M_{\mathfrak {p}})\) is Artinian. Note that \({\mathrm{Ann}}_{A_{\mathfrak {p}}}(M_{\mathfrak {p}})={\mathrm{Ann}}_A(M)_{\mathfrak {p}}\), hence \(A_{\mathfrak {p}}/{\mathrm{Ann}}_{A_{\mathfrak {p}}}(M_{\mathfrak {p}})=(A/{\mathrm{Ann}}_A(M))_{\mathfrak {p}}\) whose prime ideals are one-to-one correspondence to prime ideals of \(A\) between \({\mathrm{Ann}}_A(M)\) and \({\mathfrak {p}}\), i.e. prime ideals in \({\mathrm{Supp}}(M)\) which is contained in \({\mathfrak {p}}\). Such ideal can only be \({\mathfrak {p}}\) itself, since \(M\) is torsion, every prime ideal in \({\mathrm{Supp}}(M)\) has height \(\geq 1\). Hence \(A_{\mathfrak {p}}/{\mathrm{Ann}}_{A_{\mathfrak {p}}}(M_{\mathfrak {p}})\) is Artinian.)

Since localization is exact, for any prime ideal \({\mathfrak {p}}\) of \(A\), the \(0\to M_{\mathfrak {p}}'\to M_{\mathfrak {p}}\to M_{\mathfrak {p}}''\to 0\) is exact. Let \({\mathfrak {p}}\) runs over all minimal prime ideals of \(A\), we obtain that \(M\) is \(A\)-torsion if and only if \(M'\) and \(M''\) are \(A\)-torsion. Also, we have \(\ell _{A_{\mathfrak {p}}}(M_{\mathfrak {p}})=\ell _{A_{\mathfrak {p}}}(M_{\mathfrak {p}}')+\ell _{A_{\mathfrak {p}}}(M_{\mathfrak {p}}'')\), hence \(\operatorname{char}_A(M)=\operatorname{char}_A(M')\operatorname{char}_A(M'')\) holds.

(i) Clear.

(ii) Let \({\mathfrak {m}}\) be the maximal ideal of \(A\). If \(M\) is finite, then there exists \(r\in {\mathbb {N}}\) such that \({\mathfrak {m}}^rM=0\), hence \({\mathrm{supp}}(M)\subset \{ {\mathfrak {m}}\} \). On the other hand, if \({\mathrm{supp}}(M)\subset \{ {\mathfrak {m}}\} \), then there exists \(r\in {\mathbb {N}}\) such that \({\mathfrak {m}}^rM=0\), hence \({\mathfrak {m}}^r\subset {\mathrm{Ann}}_A(M)\), therefore \(M\) is a finitely generated \(A/{\mathfrak {m}}^r\)-module, which must be finite.

Clear from definition and Proposition 3.3.

Since the height one support of \(\ker (f)\) and \(\operatorname{coker}(f)\) are contained in \(\Sigma \), and since \(S^{-1}\ker (f)=\ker (S^{-1}f)\), \(S^{-1}\operatorname{coker}(f)=\operatorname{coker}(S^{-1}f)\) (localization is exact), we only need to prove that if \(M\) is a finitely generated torsion \(A\)-module whose height one support is contained in \(\Sigma \), then \(S^{-1}M=0\) if and only if \(M\) is pseudo-null (equivalently, \(M_{\mathfrak {q}}=0\) for all \({\mathfrak {q}}\in \Sigma \)): “\(\Rightarrow \)”: Clear. “\(\Leftarrow \)”: For all \({\mathfrak {q}}\in \Sigma \), \(M_{\mathfrak {q}}=0\) means that \({\mathrm{Ann}}(M)\not\subset {\mathfrak {q}}\), since \({\mathfrak {q}}\) are prime ideals, we have \({\mathrm{Ann}}(M)\not\subset \bigcup _{{\mathfrak {q}}\in \Sigma }{\mathfrak {q}}\), so \(S^{-1}M=0\).

Let \(\Sigma =\{ {\mathfrak {q}}_1,\cdots ,{\mathfrak {q}}_r\} =\{ {\mathfrak {q}}\in {\mathrm{Supp}}(M)\mid {\mathrm{ht}}({\mathfrak {q}})=1\} \) (by Proposition 3.1 this is a finite set), and let \(S=A\setminus \bigcup _{i=1}^r{\mathfrak {q}}_i\). Then \(S^{-1}M\) is a finitely generated \(S^{-1}A\)-module, and is torsion (for example, since \(\operatorname{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}A) =S^{-1}\operatorname{Hom}_A(M,A)=0\)).

Note that the prime ideals \({\mathfrak {P}}\) of \(S^{-1}A\) are one-to-one correspondence to prime ideals \({\mathfrak {p}}\) of \(A\) satisfying \({\mathfrak {p}}\cap S=\varnothing \) (i.e. \({\mathfrak {p}}\subset \bigcup _{i=1}^r{\mathfrak {q}}_i\), i.e \({\mathfrak {p}}\subset {\mathfrak {q}}_i\) for some \(i\)), by \({\mathfrak {P}}=S^{-1}{\mathfrak {p}}\) and \({\mathfrak {p}}={\mathfrak {P}}\cap A\). In particular, \(S^{-1}A\) is of dimension \(\leq 1\).

By structure theorem of finitely generated torsion modules over a PID, there exist primes \({\mathfrak {p}}_1,\cdots ,{\mathfrak {p}}_s\) of \(A\) satisfying \({\mathfrak {p}}_i\cap S=\varnothing \), and positive integers \(k_1,\cdots ,k_s\), such that there exists an isomorphism \(g:S^{-1}M\xrightarrow \sim \bigoplus _{i=1}^sS^{-1}(A/{\mathfrak {p}}_i^{k_i})\) of \(S^{-1}A\)-modules. Since \(S^{-1}M\) is torsion, the \({\mathfrak {p}}_i\)’s must be of height one. Since \(\operatorname{Hom}_{S^{-1}A}(S^{-1}M,\bigoplus _{i=1}^sS^{-1}(A/{\mathfrak {p}}_i^{k_i})) =S^{-1}\operatorname{Hom}_A(M,\bigoplus _{i=1}^sA/{\mathfrak {p}}_i^{k_i})\), by multiplying an element of \(S\) to \(g\) if necessary (this doesn’t change the fact that \(g\) is an isomorphism), we may find an \(A\)-linear map \(f:M\to \bigoplus _{i=1}^sA/{\mathfrak {p}}_i^{k_i}\) such that \(g=S^{-1}f\). Now by (i) we know that \(f\) is a pseudo-isomorphism.

Conversely, if \(({\mathfrak {p}}_i^{k_i})_{i=1}^s\) is such that there exists a pseudo-isomorphism \(M\to \bigoplus _{i=1}^s A/{\mathfrak {p}}_i^{k_i}\), then enlarging \(S\) if necessary, by Lemma 3.9, its localization \(S^{-1}M\to \bigoplus _{i=1}^sS^{-1}(A/{\mathfrak {p}}_i^{k_i})\) is an isomorphism of \(S^{-1}A\)-module, hence by structure theorem of finitely generated torsion modules over a PID, the \(({\mathfrak {p}}_i^{k_i})_{i=1}^s\) is unique up to ordering.

(a)\(\Rightarrow \)(b): Clear.

(b)\(\Rightarrow \)(a): Let \(\Sigma =\{ {\mathfrak {q}}_1,\cdots ,{\mathfrak {q}}_r\} =\{ {\mathfrak {q}}\in {\mathrm{Supp}}(M)\cup {\mathrm{Supp}}(N)\mid {\mathrm{ht}}({\mathfrak {q}})=1\} \) (by Proposition 3.1 this is a finite set), and let \(S=A\setminus \bigcup _{i=1}^r{\mathfrak {q}}_i\). Since \(M_{\mathfrak {p}}\cong N_{\mathfrak {p}}\) for all height one primes \({\mathfrak {p}}\) of \(A\), the \(S^{-1}M\) and \(S^{-1}N\), being finitely generated torsion modules over a PID \(S^{-1}A\), are isomorphic. Say \(g:S^{-1}M\xrightarrow \sim S^{-1}N\) is an isomorphism of \(S^{-1}A\)-modules. Since \(\operatorname{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}N)=S^{-1}\operatorname{Hom}_A(M,N)\), by multiplying an element of \(S\) to \(g\) if necessary (this doesn’t change the fact that \(g\) is an isomorphism), we may find an \(A\)-linear map \(f:M\to N\) such that \(g=S^{-1}f\). Now by Lemma 3.9 we know that \(f\) is a pseudo-isomorphism.

If \(A\) is a Noetherian regular domain (more generally, a Noetherian integrally closed domain), then for each height one prime \({\mathfrak {p}}\) of \(A\), \(A_{\mathfrak {p}}\) is a PID. For any finitely many height one primes \({\mathfrak {p}}_1,\cdots ,{\mathfrak {p}}_r\) of \(A\), let \(S:=A\setminus \bigcup _{i=1}^r{\mathfrak {p}}_i\), then \(S^{-1}A\) is a semilocal integral domain, whose maximal ideals are \(S^{-1}{\mathfrak {p}}_1,\cdots ,S^{-1}{\mathfrak {p}}_r\), and we have \((S^{-1}A)_{S^{-1}{\mathfrak {p}}_i}=A_{{\mathfrak {p}}_i}\), therefore by (i) we know that \(S^{-1}A\) is a PID.