Let \(R\) be a PID and \(M\) be an \(R\dash\)module. Let \(p\) be a prime element of \(R\). The module \(M\) is called \(\generators{p}\dash\)primary if for every \(m \in M\) there exists \(k > 0\) such that \(p^k m = 0\).
Suppose M is \(\generators{p}\dash\)primary. Show that if \(m \in M\) and \(t \in R, ~t \not\in \generators{p}\), then there exists \(a \in R\) such that \(atm = m\).
A submodule \(S\) of \(M\) is said to be pure if \(S \cap r M = rS\) for all \(r \in R\). Show that if \(M\) is \(\generators{p}\dash\)primary, then \(S\) is pure if and only if \(S \cap p^k M = p^k S\) for all \(k \geq 0\).
Solution
(a) Existence of \(a \in R\) such that \(atm = m\): Since \(M\) is \(\langle p \rangle\)-primary, there exists an integer \(k > 0\) such that \(p^k m = 0\). Since \(R\) is a PID, every prime element is irreducible. Because \(p \nmid t\), we have \(\gcd(t, p^k) = 1\). By Bézout’s identity in the PID \(R\), there exist elements \(a, b \in R\) such that: \[ at + b p^k = 1. \] Multiplying both sides by \(m\) (using the module action): \[ (at + b p^k) m = 1 \cdot m \implies a t m + b(p^k m) = m. \] Since \(p^k m = 0\), we obtain \(atm + 0 = m\), so \(atm = m\).
(b) Characterization of purity for \(\langle p \rangle\)-primary modules:
The forward direction (\(\implies\)): if \(S\) is pure, then \(S \cap rM = rS\) holds for every \(r \in R\); taking \(r = p^k\) for each \(k \ge 0\) gives exactly \(S \cap p^k M = p^k S\).
For the reverse direction (\(\Longleftarrow\)): Assume \(S \cap p^k M = p^k S\) for all \(k \geq 0\). Let \(r \in R\). If \(r = 0\), \(S \cap 0M = 0 = 0S\), which holds. So assume \(r \neq 0\). Since \(R\) is a PID (hence a UFD), we can factor \(r\) as: \[ r = u p^k t, \] where \(u \in R^\times\) is a unit, \(k \geq 0\), and \(t \in R\) satisfies \(\gcd(t, p) = 1\), i.e. \(t \notin \langle p \rangle\). Note that multiplying by a unit preserves submodules, so \(r M = p^k t M\) and \(r S = p^k t S\).
We always have \(rS \subseteq S \cap rM\). To show \(S \cap rM \subseteq rS\): Let \(s \in S \cap rM\). Then \(s = rm = u p^k t m\) for some \(m \in M\). In particular, \(s = p^k(u t m) \in p^k M\). Since \(s \in S\), we have \(s \in S \cap p^k M\). By our hypothesis for \(p^k\), \(S \cap p^k M = p^k S\). Thus, there exists \(s' \in S\) such that: \[ s = p^k s'. \] Now, consider \(s' \in S\). Since \(M\) is \(\langle p \rangle\)-primary, \(S\) is also \(\langle p \rangle\)-primary. Since \(t \notin \langle p \rangle\), by part (a) there exists \(a \in R\) such that: \[ a t s' = s'. \] Therefore: \[ s = p^k s' = p^k (a t s') = p^k t (a s') = (u^{-1} r) (a s') = r (u^{-1} a s'). \] Since \(s' \in S\) and \(S\) is an \(R\)-submodule, \(u^{-1} a s' \in S\). Hence \(s \in rS\).
This proves \(S \cap rM = rS\) for all \(r \in R\), so \(S\) is pure.