Show that a finitely generated module over a Noetherian local ring is flat iff it is free.
Solution
Let \((R,\mathfrak m)\) be a Noetherian local ring and let \(M\) be finitely generated.
Free modules are flat. Conversely, suppose \(M\) is flat. Choose \(m_1,\dots,m_r\in M\) whose images form a basis of the \(R/\mathfrak m\)-vector space \(M/\mathfrak mM\). Nakayama’s lemma says they generate \(M\), so there is an exact sequence \[ 0\to K\to R^r\to M\to0. \] Because \(M\) is flat, tensoring with \(k=R/\mathfrak m\) preserves exactness on the left. Hence \[ 0\to K/\mathfrak mK\to k^r\to M/\mathfrak mM\to0. \] The last map is an isomorphism by construction, so \(K/\mathfrak mK=0\). Since \(R\) is Noetherian, \(K\subset R^r\) is finitely generated; Nakayama therefore gives \(K=0\). Thus \(M\cong R^r\) is free.