Let $A$ be an integral domain and $I$ be an ideal of $A$ generated by a regular sequence. Is $I$ flat (as an $A$-module)?
2026-03-27 10:30:12.1774607412
Flat ideals in a ring
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No, the ideal I needn't be flat. Consider the polynomial ring $k[X,Y]$ in two indeterminates over the field $k$. The ideal $I=(X,Y)$ is generated by the regular sequence $X,Y$ but it is not flat. Indeed, since I is finitely presented, it would be projective, of rank 1 hence free : the ideal $I$ would then be principal, which it is not.