Let \(A\) be a commutative ring and \(M\) a finitely generated \(A\)-module. Define \[\begin{align*} \Ann(M) = \{a \in A: am = 0 \text{ for all } m \in M\} .\end{align*}\] Show that for a prime ideal \(\mathfrak p \subset A\), the following are equivalent:
\(\Ann(M) \not\subset \mathfrak p\)
The localization of \(M\) at the prime ideal \(\mathfrak p\) is \(0\).
\(M \otimes_A k(\mathfrak p) = 0\), where \(k(\mathfrak p) = A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}\) is the residue field of \(A\) at \(\mathfrak p\).
Solution
Let \(S=A\setminus\mathfrak p\), so \(A_{\mathfrak p}=S^{-1}A\) and \(M_{\mathfrak p}=S^{-1}M\).
1. If\[ \operatorname{Ann}(M)\not\subset\mathfrak p, \]then\[ M_{\mathfrak p}=0. \]
Choose \[ s\in\operatorname{Ann}(M)\setminus\mathfrak p. \] Then \(s/1\) is a unit in \(A_{\mathfrak p}\), while it annihilates every element of \(M_{\mathfrak p}\). If \(m/1\in M_{\mathfrak p}\), then \[ \frac{s}{1}\frac{m}{1}=0. \] Multiplying by the inverse of \(s/1\) gives \(m/1=0\). Hence \(M_{\mathfrak p}=0\).
2. Conversely, if\[ M_{\mathfrak p}=0, \]then\[ \operatorname{Ann}(M)\not\subset\mathfrak p. \]
Let \(m_1,\dots,m_r\) generate \(M\). Since \(m_i/1=0\) in \(M_{\mathfrak p}\), for each \(i\) there exists \(s_i\in S\) such that \[ s_i m_i=0. \] Put \[ s=s_1\cdots s_r. \] Because \(\mathfrak p\) is prime and no \(s_i\) belongs to \(\mathfrak p\), we have \(s\notin\mathfrak p\). Moreover \(s\) annihilates every generator \(m_i\), hence all of \(M\). Thus \[ s\in\operatorname{Ann}(M)\setminus\mathfrak p. \]
3. There is a natural isomorphism\[ M\otimes_A k(\mathfrak p) \cong M_{\mathfrak p}\otimes_{A_{\mathfrak p}} k(\mathfrak p) \cong M_{\mathfrak p}/\mathfrak pA_{\mathfrak p}M_{\mathfrak p}. \]
Since every element of \(S\) maps to a unit in the residue field \(k(\mathfrak p)\), tensoring with \(k(\mathfrak p)\) factors through localization, giving the first isomorphism. The second is the standard tensor-quotient identity \[ N\otimes_R R/I\cong N/IN \] with \(R=A_{\mathfrak p}\), \(I=\mathfrak pA_{\mathfrak p}\), and \(N=M_{\mathfrak p}\).
4. If \(M_{\mathfrak p}=0\), then\[ M\otimes_A k(\mathfrak p)=0. \]
This is immediate from 3.
5. If\[ M\otimes_A k(\mathfrak p)=0, \]then\[ M_{\mathfrak p}=0. \]
By 3, \[ M_{\mathfrak p}/\mathfrak pA_{\mathfrak p}M_{\mathfrak p}=0. \] The \(A_{\mathfrak p}\)-module \(M_{\mathfrak p}\) is finitely generated because \(M\) is finitely generated over \(A\). The ring \(A_{\mathfrak p}\) is local with maximal ideal \(\mathfrak pA_{\mathfrak p}\). Nakayama’s lemma therefore yields \[ M_{\mathfrak p}=0. \]
6. Hence the three conditions are equivalent:\[ \boxed{ \operatorname{Ann}(M)\not\subset\mathfrak p \iff M_{\mathfrak p}=0 \iff M\otimes_A k(\mathfrak p)=0. } \]