Local flatness criterion for a module finite over a larger Noetherian local algebra

We first show that $\mathfrak{m}{\otimes }_{R}M\to M$ is injective. Let $z={\sum }_{i=1}^r{a}_i\otimes x_i$ lie in its kernel, with $a_i\in \mathfrak{m}$ and ${\sum }_i{a}_i{x}_i=0$. In the flat $R\mathrm{/}I$-module $M\mathrm{/}IM$, [F1] gives ${\bar{y}}_j\in M\mathrm{/}IM$ and ${\bar{b}}_{ij}\in R\mathrm{/}I$ such that ${\bar{x}}_i={\sum }_j{\bar{b}}_{ij}{\bar{y}}_j,{\sum }_i{\bar{a}}_i{\bar{b}}_{ij}=0\text{for every }j\mathrm{.}$ Choose lifts $y_j\in M$ and $b_{ij}\in R$. Then $x_i-{\sum }_j{b}_{ij}{y}_j\in IM$ and ${\sum }_i{a}_i{b}_{ij}\in I$. Expanding $z$ with these equations expresses it as the image of an element $w\in I{\otimes }_{R}M$: for a term $a_i\otimes (cm)$ with $c\in I$, move $c$ to the first tensor factor to obtain $a_{i}c\otimes m$, and the remaining terms already have first factor ${\sum }_i{a}_i{b}_{ij}\in I$. The image of $w$ in $M$ is the image of $z$, namely zero. The given injectivity of $I{\otimes }_{R}M\to M$ makes $w=0$, hence $z=0$.

Apply [F2] to $0\to \mathfrak{m}\to R\to k:=R\mathrm{/}\mathfrak{m}\to 0$. Step 1.1 gives ${\mathrm{Tor}}_1^R(k,M)=0$. A finite-length $R$-module has a finite filtration with quotients $k$; induction on its length using the long exact Tor sequence of [F2] therefore gives ${\mathrm{Tor}}_1^R(N,M)=0$ for every finite-length $N$. Since $R$ is Noetherian local, $R\mathrm{/}{\mathfrak{m}}^n$ and $R\mathrm{/}(J+{\mathfrak{m}}^n)$ have finite length for any ideal $J$ and $n\ge 1$. Consequently both multiplication maps ${\mathfrak{m}}^n{\otimes }_{R}M\to M,(J+{\mathfrak{m}}^n){\otimes }_{R}M\to M$ are injective.

Fix a finitely generated ideal $J\subseteq R$ and set $K=\ker (J{\otimes }_{R}M\to M)$. For each $n\ge 1$, tensor the exact sequence $J\cap {\mathfrak{m}}^n⟶J\oplus {\mathfrak{m}}^n⟶J+{\mathfrak{m}}^n⟶0$ with $M$. The map $(J\oplus {\mathfrak{m}}^n){\otimes }_{R}M\to (J+{\mathfrak{m}}^n){\otimes }_{R}M$ sends $(z,0)$ to zero for $z\in K$: its product in $M$ is zero, and the second injection in step 2.1 detects this. Right exactness of tensor therefore puts $(z,0)$ in the image of $(J\cap {\mathfrak{m}}^n){\otimes }_{R}M$. In particular, $K\subseteq \mathrm{im}((J\cap {\mathfrak{m}}^n){\otimes }_{R}M⟶J{\otimes }_{R}M)\text{for every }n\mathrm{.}$

By Artin–Rees [F3], for $n\ge c$ the image in step 3.1 lies in ${\mathfrak{m}}^{n-c}(J{\otimes }_{R}M)$. The $S$-module $J{\otimes }_{R}M$ is finite: a finite generating set of $J$ gives a surjection $M^r\to J{\otimes }_{R}M$. As $\mathfrak{m}S\subseteq J(S)$, Krull intersection [F3] yields $K\subseteq {\bigcap }_{n\ge c}{\mathfrak{m}}^{n-c}(J{\otimes }_{R}M)=0.$ Thus $J{\otimes }_{R}M\to M$ is injective for every finitely generated $J$, and [F1] makes $M$ flat over $R$. This argument uses finiteness over $S$ only for Krull intersection; $M$ need not be finite over $R$.

The Axiom of Choice enters through the published Krull-intersection boundary and implies the Dependent Choice used for the cited Tor sequence. The remaining choices above are finite. [F2, F3, step 4.1] $\mathrm{\square }$