## 15.16 Flattening stratification

Let $R \to S$ be a ring map and let $M$ be an $S$-module. For any $R$-algebra $R'$ we can consider the base changes $S' = S \otimes _ R R'$ and $M' = M \otimes _ R R'$. We say $R \to R'$ flattens $M$ if the module $M'$ is flat over $R'$. We would like to understand the structure of the collection of ring maps $R \to R'$ which flatten $M$. In particular we would like to know if there exists a universal flattening $R \to R_{univ}$ of $M$, i.e., a ring map $R \to R_{univ}$ which flattens $M$ and has the property that any ring map $R \to R'$ which flattens $M$ factors through $R \to R_{univ}$. It turns out that such a universal solution usually does not exist.

We will discuss universal flattenings and flattening stratifications in a scheme theoretic setting $\mathcal{F}/X/S$ in More on Flatness, Section 38.21. If the universal flattening $R \to R_{univ}$ exists then the morphism of schemes $\mathop{\mathrm{Spec}}(R_{univ}) \to \mathop{\mathrm{Spec}}(R)$ is the universal flattening of the quasi-coherent module $\widetilde{M}$ on $\mathop{\mathrm{Spec}}(S)$.

In this and the next few sections we prove some basic algebra facts related to this. The most basic result is perhaps the following.

Lemma 15.16.1. Let $R$ be a ring. Let $M$ be an $R$-module. Let $I_1$, $I_2$ be ideals of $R$. If $M/I_1M$ is flat over $R/I_1$ and $M/I_2M$ is flat over $R/I_2$, then $M/(I_1 \cap I_2)M$ is flat over $R/(I_1 \cap I_2)$.

Proof. By replacing $R$ with $R/(I_1 \cap I_2)$ and $M$ by $M/(I_1 \cap I_2)M$ we may assume that $I_1 \cap I_2 = 0$. Let $J \subset R$ be an ideal. To prove that $M$ is flat over $R$ we have to show that $J \otimes _ R M \to M$ is injective, see Algebra, Lemma 10.39.5. By flatness of $M/I_1M$ over $R/I_1$ the map

$J/(J \cap I_1) \otimes _ R M = (J + I_1)/I_1 \otimes _{R/I_1} M/I_1M \longrightarrow M/I_1M$

is injective. As $0 \to (J \cap I_1) \to J \to J/(J \cap I_1) \to 0$ is exact we obtain a diagram

$\xymatrix{ (J \cap I_1) \otimes _ R M \ar[r] \ar[d] & J \otimes _ R M \ar[r] \ar[d] & J/(J \cap I_1) \otimes _ R M \ar[r] \ar[d] & 0 \\ M \ar@{=}[r] & M \ar[r] & M/I_1M }$

hence it suffices to show that $(J \cap I_1) \otimes _ R M \to M$ is injective. Since $I_1 \cap I_2 = 0$ the ideal $J \cap I_1$ maps isomorphically to an ideal $J' \subset R/I_2$ and we see that $(J \cap I_1) \otimes _ R M = J' \otimes _{R/I_2} M/I_2M$. By flatness of $M/I_2M$ over $R/I_2$ the map $J' \otimes _{R/I_2} M/I_2M \to M/I_2M$ is injective, which clearly implies that $(J \cap I_1) \otimes _ R M \to M$ is injective. $\square$

Comment #3420 by Giulia on

Typo: first line, I guess that the module is M rather than N.

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