Faithful flatness of finite-type algebraic-to-analytic local rings

AC implies DC: for a nonempty set $Y$ and a serial relation $R\subseteq Y\times Y$, the family of successor sets $\{\{z\in Y:yRz\}:y\in Y\}$ consists of nonempty sets. [A1] gives a choice function on this family; composing it with the successor-set map gives a self-map $s:Y\to Y$ with $yRs(y)$. For any prescribed $y_0$, [F10] gives the sequence $y_{k+1}=s(y_k)$. For any $A$-module $N$, let $F_0$ be free on its underlying set and let ${ϵ}_0:F_0\twoheadrightarrow N$ send each standard basis vector to the element it names, using [F11]. A free module on any set is projective: given a surjection $E\twoheadrightarrow Q$ and a map from the free module to $Q$, [A1] chooses a preimage in $E$ for the image of each basis vector, and [F11] extends that assignment to a lift. Put $K_0=\ker {ϵ}_0$; recursively, for each $j\ge 0$, take $F_{j+1}$ free on the set $K_j$, let ${ϵ}_{j+1}:F_{j+1}\twoheadrightarrow K_j$ send each basis vector to the element it names, and set $K_{j+1}=\ker {ϵ}_{j+1}$. The differential $F_{j+1}\to F_j$ is ${ϵ}_{j+1}$ followed by the inclusion $K_j\hookrightarrow F_j$. The sequence is exact by the definitions of the kernels, each $F_j$ is projective by the lifting argument, and [F10] supplies the recursion. Thus every $A$-module has the projective-resolution data required by [F8] and [F9], and AC supplies their DC hypothesis.

By [F1], around $x$ the scheme has a finite-type affine chart. [F2] makes $\mathbb{C}$ a field, so its only ideals are $0$ and $\mathbb{C}$ and each is finitely generated; DC from step 1.1 and the ideal characterization in [F3] therefore make $\mathbb{C}$ Noetherian. The finite-variable corollary in [F2] now makes its chart polynomial ring Noetherian, and [F3] makes the chart's defining ideal $I$ finitely generated. Translate the coordinates to put $x$ at the origin, adding a dummy coordinate with relation $w_1=0$ if the chart has no polynomial coordinates. Apply [F4] to obtain the local map $A\to B$ and its common completion $T=\mathbb{C}\mathrm{⟦}{w}_1,\dots ,w_n\mathrm{⟧}\mathrm{/}I\mathbb{C}\mathrm{⟦}{w}_1,\dots ,w_n\mathrm{⟧}$.

By [F5] under [A1], the completion maps $A\to T$ and $B\to T$ are faithfully flat. For a nonzero $a\in A$, the ideal $(a)$ is a nonzero $A$-module. Faithfulness means $(a){\otimes }_{A}T\mathrm{\ne }0$: otherwise the nonexact sequence $0\to (a)\to 0$ would become exact after tensoring. Flatness makes the multiplication map $(a){\otimes }_{A}T\to A{\otimes }_{A}T=T$ injective. If $a$ mapped to zero in $T$, this map would be zero, a contradiction. Hence $A\to T$ is injective. If an element of $A$ maps to zero in $B$, its image in $T$ is zero as well, so injectivity of $A\to T$ makes it zero. Therefore $A\to B$ is injective.

Let $U\hookrightarrow V$ be any injection of $A$-modules and let $K$ be the kernel of $B{\otimes }_{A}U\to B{\otimes }_{A}V$. Tensoring $0\to K\to B{\otimes }_{A}U\to B{\otimes }_{A}V$ over $B$ with $T$ preserves its kernel because $T$ is flat over $B$ by [F5] under [A1]. By associativity, the resulting map is $T{\otimes }_{A}U\to T{\otimes }_{A}V$, which is injective because $T$ is flat over $A$. Thus $K{\otimes }_{B}T=0$; faithful flatness of $B\to T$ gives $K=0$. The flatness criterion [F6] now makes $B$ flat over $A$.

The map $A\to B$ is local by [F4]. If $J⊊A$ is a proper ideal, then $J$ lies in the maximal ideal of the local ring $A$, whose image lies in the maximal ideal of $B$; hence $JB⊊B$. Combining this with the flatness in step 3.2 and [F7] shows that $A\to B$ is faithfully flat.

Let $M$ be any $A$-module and $m\in M$ nonzero. Its annihilator $J=\{a\in A:am=0\}$ is a proper ideal, and the map $A\mathrm{/}J\to M$ sending $1+J$ to $m$ is injective. Flatness from step 3.2 preserves this injection. The balanced tensor relations give inverse maps $(A\mathrm{/}J){\otimes }_{A}B\to B\mathrm{/}JB$, $(a+J)\otimes b\mapsto ab+JB$, and $B\mathrm{/}JB\to (A\mathrm{/}J){\otimes }_{A}B$, $b+JB\mapsto (1+J)\otimes b$. By [F7], $B\mathrm{/}JB\mathrm{\ne }0$, so $m\otimes 1\mathrm{\ne }0$. Therefore $M\to M{\otimes }_{A}B$ is injective for every $M$.

Put $C=B\mathrm{/}A$, which is defined because step 3.1 proves $A\hookrightarrow B$. For any left $A$-module $M$, apply [F8] under the DC and resolution construction in step 1.1 to the short exact sequence of right $A$-modules $0\to A\to B\to C\to 0$. The long exact sequence gives ${\mathrm{Tor}}_1^A(A,M)\to {\mathrm{Tor}}_1^A(B,M)\to {\mathrm{Tor}}_1^A(C,M)\to A{\otimes }_{A}M\to B{\otimes }_{A}M$. The first two Tor groups vanish: $A$ is free over itself and $B$ is flat by step 3.2. The final map is injective by step 5.1, so exactness forces ${\mathrm{Tor}}_1^A(C,M)=0$ for every $M$. By [F9], $C=B\mathrm{/}A$ is a flat $A$-module.