Local flatness criterion by regular parameters

AC gives DC, so the Dependent-Choice suppliers [F1], [F11] and [F12] are available. Given a nonempty set $X$, an entire relation $R$ on $X$ and $a\in X$, apply [F15] to the family of nonempty subsets of $X$ to obtain $g$ with $g(T)\in T$ for every nonempty $T\subseteq X$, and put $f(x):=g(\{y\in X:xRy\})$, a function $X\to X$ because $R$ is entire. By [F13] there is $x\text{\hspace{0.17em}⁣}:\mathbb{N}\to X$ with $x_0=a$ and $x_{n+1}=f(x_n)$; then $x_{n}R{x}_{n+1}$ for all $n$ because $f(x_n)\in \{y:x_{n}Ry\}$. This is exactly the statement of [F14]. Supply a free resolution of $M$ by mapping the free module on its underlying set onto $M$, then repeating this construction on each successive kernel; recursion gives the required resolution for [F1].

Ideals of finite colength. Let $J\subseteq R$ be an ideal containing ${\mathfrak{m}}^n$ for some $n\ge 0$. For $n=0$, $J=R$ and $R\mathrm{/}J=0$. For $n\ge 1$, $R\mathrm{/}J$ has finite length over $R$: the ring $R\mathrm{/}{\mathfrak{m}}^n$ carries the finite chain $0\subseteq {\mathfrak{m}}^{n-1}\mathrm{/}{\mathfrak{m}}^n\subseteq \cdots \subseteq R\mathrm{/}{\mathfrak{m}}^n$ whose successive quotients are ${\mathfrak{m}}^i\mathrm{/}{\mathfrak{m}}^{i+1}={\mathfrak{m}}^i\mathrm{/}(\mathfrak{m}\cdot {\mathfrak{m}}^i)$; each ${\mathfrak{m}}^i$ is finitely generated over the Noetherian ring $R$ by [F9], so each ${\mathfrak{m}}^i\mathrm{/}{\mathfrak{m}}^{i+1}$ is a finitely generated module over the field $\kappa =R\mathrm{/}\mathfrak{m}$, hence a finite-dimensional $\kappa$-vector space, which has a finite composition series with simple factors $\kappa$ and therefore finite length by [F5]; a finite extension of modules of finite length has finite length with additive length by [F6], so ${\mathrm{\ell }}_R(R\mathrm{/}{\mathfrak{m}}^n)<\mathrm{\infty }$, and $R\mathrm{/}J$, being a quotient of $R\mathrm{/}{\mathfrak{m}}^n$, has finite length as well.

Vanishing on finite length. Suppose ${\mathrm{Tor}}_1^R(\kappa ,M)=0$ with $\kappa =R\mathrm{/}\mathfrak{m}$. Then ${\mathrm{Tor}}_1^R(N,M)=0$ for every $R$-module $N$ of finite length ${\mathrm{\ell }}_R(N)$: prove this by induction on ${\mathrm{\ell }}_R(N)$. For ${\mathrm{\ell }}_R(N)=0$ we have $N=0$. If ${\mathrm{\ell }}_R(N)\ge 1$, choose a proper submodule $N^{\mathrm{\prime }}\subset N$ that is maximal for inclusion, which exists because $N$ has finite length; then $N\mathrm{/}{N}^{\mathrm{\prime }}$ is simple by [F7], so choosing $0\mathrm{\ne }s\in N\mathrm{/}{N}^{\mathrm{\prime }}$ presents it as $(R\mathrm{/}\mathrm{Ann}(s))\cdot s$, and $\mathrm{Ann}(s)$ is a maximal ideal of $R$, because for a proper ideal $J⊋\mathrm{Ann}(s)$ the submodule $Js$ is nonzero and hence all of $N\mathrm{/}{N}^{\mathrm{\prime }}$, forcing $Js=N\mathrm{/}{N}^{\mathrm{\prime }}$ and $1\in J$. By [F8] the only maximal ideal is $\mathfrak{m}$, so $N\mathrm{/}{N}^{\mathrm{\prime }}\cong \kappa$. By [F6] ${\mathrm{\ell }}_R(N^{\mathrm{\prime }})={\mathrm{\ell }}_R(N)-1$, so the induction hypothesis applies to $N^{\mathrm{\prime }}$, and the exact sequence ${\mathrm{Tor}}_1^R(N^{\mathrm{\prime }},M)\to {\mathrm{Tor}}_1^R(N,M)\to {\mathrm{Tor}}_1^R(\kappa ,M)=0$ from [F1] has vanishing outer terms, whence ${\mathrm{Tor}}_1^R(N,M)=0$.

Injectivity for finite colength ideals. Let $J\subseteq R$ be any ideal. Since the free module $R$ with the resolution concentrated in degree $0$ satisfies ${\mathrm{Tor}}_1^R(R,M)=0$ by [F10], the long exact sequence of [F1] for $0\to J\to R\to R\mathrm{/}J\to 0$ exhibits $\ker (J{\otimes }_{R}M\to M)$ as the image of ${\mathrm{Tor}}_1^R(R\mathrm{/}J,M)\to J{\otimes }_{R}M$. Hence if $J\supseteq {\mathfrak{m}}^n$ and ${\mathrm{Tor}}_1^R(\kappa ,M)=0$, then $R\mathrm{/}J$ has finite length by step 1.2 and ${\mathrm{Tor}}_1^R(R\mathrm{/}J,M)=0$ by step 2.1, so $J{\otimes }_{R}M\to M$ is injective.

The diagram chase. Let $I\subseteq R$ be a finitely generated ideal and let $K:=\ker (I{\otimes }_{R}M\to M)$. For every $n$ the sequence $0\to I\cap {\mathfrak{m}}^n\to I\oplus {\mathfrak{m}}^n\to I+{\mathfrak{m}}^n\to 0$, with maps $x\mapsto (x,-x)$ and $(a,b)\mapsto a+b$, is exact, so after tensoring with $M$ and using [F20] the sequence $(I\cap {\mathfrak{m}}^n){\otimes }_{R}M\to I{\otimes }_{R}M\oplus {\mathfrak{m}}^n{\otimes }_{R}M\to (I+{\mathfrak{m}}^n){\otimes }_{R}M\to 0$ is exact; the vertical maps to $M$ are the multiplication maps, which are injective on ${\mathfrak{m}}^n{\otimes }_{R}M$ and on $(I+{\mathfrak{m}}^n){\otimes }_{R}M$ because both ideals contain ${\mathfrak{m}}^n$ and step 3.1 applies. Given $k\in K$, its image $(k,0)$ in the middle maps to zero in $(I+{\mathfrak{m}}^n){\otimes }_{R}M$ and hence, by exactness at the middle, equals $(x,-x)$ for some $x\in (I\cap {\mathfrak{m}}^n){\otimes }_{R}M$; then $x$ maps to $k$ in $I{\otimes }_{R}M$ and to $0$ in ${\mathfrak{m}}^n{\otimes }_{R}M$, so $x\in \ker ((I\cap {\mathfrak{m}}^n){\otimes }_{R}M\to M)$ and $K$ lies in the image of $(I\cap {\mathfrak{m}}^n){\otimes }_{R}M\to I{\otimes }_{R}M$.

Concluding $K=0$. Apply Artin–Rees [F3] over the Noetherian ring $R$ to the finite module $R$, its submodule $I$ and the ideal $\mathfrak{m}$. It gives $c\ge 0$ such that $I\cap {\mathfrak{m}}^n={\mathfrak{m}}^{n-c}(I\cap {\mathfrak{m}}^c)\subseteq {\mathfrak{m}}^{n-c}I$ for every $n\ge c$. Put $N:=I{\otimes }_{R}M$, a finite $S$-module because $I$ is finite over $R$ and $M$ is finite over $S$. By step 4.1 and this inclusion, $K\subseteq (\mathfrak{m}S{)}^{n-c}N$ for every $n\ge c$: an elementary tensor $ra\otimes m$ with $r\in {\mathfrak{m}}^{n-c}$ equals $r(a\otimes m)$. The local homomorphism gives $\mathfrak{m}S\subseteq \mathfrak{n}=J(S)$, so Krull intersection [F4] on the finite $S$-module $N$ gives ${\bigcap }_{j\ge 0}(\mathfrak{m}S{)}^{j}N=0$. Hence $K=0$.

Since $I\subseteq R$ was an arbitrary finitely generated ideal and $I{\otimes }_{R}M\to M$ is injective, [F2] shows that $M$ is flat over $R$. This proves the first assertion.

The regular-parameter case. Let $x_1,\dots ,x_d$ be a regular system of parameters of $R$ and let ${\bar{x}}_1,\dots ,{\bar{x}}_d\in S$ be their images, extending to a regular system of parameters $y_1,\dots ,y_e$ of $S$. By [F19] the tuple $(y_1,\dots ,y_e)$ is $S$-regular, hence so is its initial segment ${\bar{x}}_1,\dots ,{\bar{x}}_d$. By [F16] the Koszul complex $K_R(x_1,\dots ,x_d;R)$ is a free resolution of $\kappa =R\mathrm{/}\mathfrak{m}$, and tensoring its defining formulas, [F17], with $-{\otimes }_{R}S$ replaces each $x_i$ by ${\bar{x}}_i$ and reproduces the Koszul complex $K_S({\bar{x}}_1,\dots ,{\bar{x}}_d;S)$ of [F17] term by term, so $K_R(x){\otimes }_{R}S\cong K_S(\bar{x};S)$ as complexes. Therefore, using [F10] for the right-resolution construction and [F12] (available by step 1.1) to identify it with the balanced Tor of [F11], ${\mathrm{Tor}}_1^R(\kappa ,S)=H_1(K_R(x){\otimes }_{R}S)=H_1(K_S(\bar{x};S))=0$ by [F18], as $\bar{x}$ is an $S$-regular sequence. The module $S$ is a finite $S$-module, so the first assertion of this lemma, applied to $M=S$, gives that $S$ is flat over $R$.