For an $R$-finite module over a local map, flatness modulo $I$ and injectivity of $I\otimes M\to M$ imply flatness

Let $\mathfrak{m}$ be the maximal ideal of $R$. We first prove that the multiplication map $\mathfrak{m}{\otimes }_{R}M\to M$ is injective. Take an element $z={\sum }_{i=1}^n{f}_i\otimes x_i$ in its kernel, so $f_i\in \mathfrak{m}$ and ${\sum }_i{f}_i{x}_i=0$. Reducing modulo $I$, the module $M\mathrm{/}IM$ is flat over $R\mathrm{/}I$, so [L1] applied over $R\mathrm{/}I$ yields elements ${\bar{y}}_1,\dots ,{\bar{y}}_t\in M\mathrm{/}IM$ and coefficients ${\bar{a}}_{ij}\in R\mathrm{/}I$ with ${\bar{x}}_i={\sum }_{j=1}^t{\bar{a}}_{ij}{\bar{y}}_j\text{and}{\sum }_{i=1}^n({\bar{f}}_i){\bar{a}}_{ij}=0\text{ for every }j\mathrm{.}$ Choose lifts $y_j\in M$ and $a_{ij}\in R$. Then $x_i-{\sum }_j{a}_{ij}{y}_j\in IM$ and ${\sum }_i{f}_i{a}_{ij}\in I$ for every $j$. Writing each $x_i-{\sum }_j{a}_{ij}{y}_j$ as a finite sum of terms $bm$ with $b\in I$, one sees that $z$ is the image in $\mathfrak{m}{\otimes }_{R}M$ of an element of $I{\otimes }_{R}M$ whose product in $M$ is also $0$. Hypothesis 2 makes that element zero, hence $z=0$.

Let $k=R\mathrm{/}\mathfrak{m}$. Choose elements $x_1,\dots ,x_r\in M$ whose images form a $k$-basis of $M\mathrm{/}\mathfrak{m}M$. By [L2], these elements generate $M$, so they define a surjection $\pi :F:=R^r\twoheadrightarrow M\mathrm{.}$ Let $K:=\ker (\pi )$. Because $R$ is Noetherian, [L3] makes $K$ finitely generated.

Tensoring the exact sequence $K\to F\to M\to 0$ with the ideal $\mathfrak{m}$ gives an exact sequence $\mathfrak{m}{\otimes }_{R}K⟶\mathfrak{m}{\otimes }_{R}F⟶\mathfrak{m}{\otimes }_{R}M⟶0$ by [L4]. Step 1.1 identifies $\mathfrak{m}{\otimes }_{R}M$ with its image $\mathfrak{m}M\subseteq M$, and [L4] identifies $\mathfrak{m}{\otimes }_{R}F$ with $\mathfrak{m}F$. Under these identifications, the kernel of $\mathfrak{m}F\to \mathfrak{m}M$ is exactly $K\cap \mathfrak{m}F$, while the image of $\mathfrak{m}{\otimes }_{R}K$ is $\mathfrak{m}K$. Therefore $K\cap \mathfrak{m}F=\mathfrak{m}K\mathrm{.}$

The induced map $F\mathrm{/}\mathfrak{m}F\to M\mathrm{/}\mathfrak{m}M$ sends the standard basis of $k^r$ to the chosen basis from step 1.2, so it is an isomorphism. Its kernel is $(K+\mathfrak{m}F)\mathrm{/}\mathfrak{m}F\cong K\mathrm{/}(K\cap \mathfrak{m}F)=K\mathrm{/}\mathfrak{m}K$ by step 2.1. Hence $K\mathrm{/}\mathfrak{m}K=0$, so $\mathfrak{m}K=K$. Now [L5] gives $K=0$. Therefore $\pi$ is an isomorphism, $M\cong R^r$ is free, and in particular $M$ is flat over $R$.

Thus $M$ is flat over $R$.