# Flat ideals in a ring

If $A$ is an integral domain and $I$ is an ideal of $A$ generated by a regular sequence $f_1,\ldots,f_r$. Is $I$ flat (as an $A$-module)?

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 Is $A$ Noetherian? – shamovic Apr 18 '11 at 12:22

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.