A finite flat module over a local ring is free

Choose elements $x_1,\dots ,x_r\in M$ whose residue classes form a basis of the vector space $M\mathrm{/}\mathfrak{m}M$. By [L1], they generate $M$. Thus there is a surjection $\phi :R^r\twoheadrightarrow M$ sending the $i$th standard basis vector to $x_i$.

Let $K=\ker \phi$. Because $R$ is Noetherian and $R^r$ is finite, [L3] makes $K$ finitely generated. Tensoring $0\to K\to R^r\to M\to 0$ with the residue field $k=R\mathrm{/}\mathfrak{m}$ remains exact because $M$ is flat, so $0\to K\mathrm{/}\mathfrak{m}K\to k^r\to M\mathrm{/}\mathfrak{m}M\to 0.$ The last map is an isomorphism by the choice of the $x_i$, hence $K\mathrm{/}\mathfrak{m}K=0$.

Nakayama now gives $K=0$. Thus $\phi$ is an isomorphism and $M\cong R^r$ is free. The equational criterion [L2] explains why no hidden relation survives once the residue-field relations vanish.

Therefore every finite flat module over a Noetherian local ring is free.