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. $\square$
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