Lemma 15.29.9. Let $A$ be a ring. Let $I \subset J \subset A$ be ideals. Assume that $J/I \subset A/I$ is generated by an $H_1$-regular sequence. Then $I \cap J^2 = IJ$.

**Proof.**
To prove this choose $g_1, \ldots , g_ m \in J$ whose images in $A/I$ form a $H_1$-regular sequence which generates $J/I$. In particular $J = I + (g_1, \ldots , g_ m)$. Suppose that $x \in I \cap J^2$. Because $x \in J^2$ can write

with $a_{ij} \in A$, $a_ j \in I$ and $a \in I^2$. Then $\sum a_{ij}g_ ig_ j \in I \cap (g_1, \ldots , g_ m)$ hence by Lemma 15.29.8 we see that $\sum a_{ij}g_ ig_ j \in I(g_1, \ldots , g_ m)$. Thus $x \in IJ$ as desired. $\square$

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