Show that a finitely generated module over a Noetherian local ring is flat iff it is free using Nakayama and Tor.
Solution
1. If \(M\) is free, then \(M\) is flat.
Every free module is a direct sum of copies of \(R\), and tensoring with a free module preserves injections because it is a direct sum of copies of the original map.
2. Conversely, suppose \((R,\mathfrak m)\) is Noetherian local and \(M\) is finitely generated and flat. Let\[ n=\dim_{R/\mathfrak m}(M/\mathfrak m M). \]Choose elements \(m_1,\dots,m_n\in M\) whose residue classes form a basis of \(M/\mathfrak m M\).
Since \(M\) is finitely generated, \(M/\mathfrak m M\) is a finite-dimensional vector space over the residue field \(R/\mathfrak m\).
3. The map\[ \pi:R^n\longrightarrow M, \qquad e_i\longmapsto m_i, \]is surjective.
The classes of the \(m_i\) generate \(M/\mathfrak mM\). Hence \[ M=Rm_1+\cdots+Rm_n+\mathfrak mM. \] Nakayama’s lemma gives \(M=Rm_1+\cdots+Rm_n\), so \(\pi\) is surjective.
4. Let \(K=\ker\pi\). Then \(K\) is finitely generated.
The module \(R^n\) is finitely generated over the Noetherian ring \(R\), so every submodule of \(R^n\), in particular \(K\), is finitely generated.
5. Tensoring\[ 0\longrightarrow K\longrightarrow R^n\longrightarrow M\longrightarrow0 \]with \(k:=R/\mathfrak m\) gives an exact sequence\[ 0\longrightarrow K/\mathfrak mK \longrightarrow k^n \longrightarrow M/\mathfrak mM \longrightarrow0. \]
The long exact Tor sequence begins \[ \operatorname{Tor}_1^R(M,k)\longrightarrow K\otimes_Rk \longrightarrow R^n\otimes_Rk \longrightarrow M\otimes_Rk\longrightarrow0. \] Because \(M\) is flat, \(\operatorname{Tor}_1^R(M,k)=0\). Also \(K\otimes_Rk\cong K/\mathfrak mK\), \(R^n\otimes_Rk\cong k^n\), and \(M\otimes_Rk\cong M/\mathfrak mM\).
6. The map \(k^n\to M/\mathfrak mM\) in 5 is an isomorphism, so\[ K/\mathfrak mK=0. \]
By construction in 2, the images of \(e_1,\dots,e_n\) are exactly the chosen basis of \(M/\mathfrak mM\). Hence the induced map \(k^n\to M/\mathfrak mM\) is an isomorphism. Exactness in 5 then forces its kernel \(K/\mathfrak mK\) to vanish.
7. Nakayama’s lemma gives \(K=0\). Hence \(\pi:R^n\to M\) is an isomorphism, so \(M\) is free.
By 4, \(K\) is finitely generated, and by 6, \(K=\mathfrak mK\). Nakayama’s lemma therefore gives \(K=0\). Since \(\pi\) is already surjective, it is an isomorphism.
8. Therefore a finitely generated module over a Noetherian local ring is flat if and only if it is free.
Combine 1 and 7.