Lemma 10.128.9. Let $R$, $S$, $S'$ be local rings and let $R \to S \to S'$ be local ring homomorphisms. Let $M$ be an $S'$-module. Let $\mathfrak m \subset R$ be the maximal ideal. Assume

1. $R \to S'$ is essentially of finite presentation,

2. $R \to S$ is essentially of finite type,

3. $M$ is of finite presentation over $S'$,

4. $M$ is not zero,

5. $M/\mathfrak mM$ is a flat $S/\mathfrak mS$-module, and

6. $M$ is a flat $R$-module.

Then $S$ is essentially of finite presentation and flat over $R$ and $M$ is a flat $S$-module.

Proof. As $S$ is essentially of finite presentation over $R$ we can write $S = C_{\overline{\mathfrak q}}$ for some finite type $R$-algebra $C$. Write $C = R[x_1, \ldots , x_ n]/I$. Denote $\mathfrak q \subset R[x_1, \ldots , x_ n]$ be the prime ideal corresponding to $\overline{\mathfrak q}$. Then we see that $S = B/J$ where $B = R[x_1, \ldots , x_ n]_{\mathfrak q}$ is essentially of finite presentation over $R$ and $J = IB$. We can find $f_1, \ldots , f_ k \in J$ such that the images $\overline{f}_ i \in B/\mathfrak mB$ generate the image $\overline{J}$ of $J$ in the Noetherian ring $B/\mathfrak mB$. Hence there exist finitely generated ideals $J' \subset J$ such that $B/J' \to B/J$ induces an isomorphism

$(B/J') \otimes _ R R/\mathfrak m \longrightarrow B/J \otimes _ R R/\mathfrak m = S/\mathfrak mS.$

For any $J'$ as above we see that Lemma 10.128.8 applies to the ring maps

$R \longrightarrow B/J' \longrightarrow S'$

and the module $M$. Hence we conclude that $B/J'$ is flat over $R$ for any choice $J'$ as above. Now, if $J' \subset J' \subset J$ are two finitely generated ideals as above, then we conclude that $B/J' \to B/J''$ is a surjective map between flat $R$-algebras which are essentially of finite presentation which is an isomorphism modulo $\mathfrak m$. Hence Lemma 10.128.4 implies that $B/J' = B/J''$, i.e., $J' = J''$. Clearly this means that $J$ is finitely generated, i.e., $S$ is essentially of finite presentation over $R$. Thus we may apply Lemma 10.128.8 to $R \to S \to S'$ and we win. $\square$

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