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A nice little lemma in commutative algebra says the following (see for instance proposition 5.17 in [Atiyah-MacDonald]):

If $A$ is a Noetherian integrally closed domain, $K$ its field of fractions and $L$ a finite separable extension of $K$, then the integral closure $B$ of $A$ in $L$ is a finite $A$-module.

(One way to prove this is via the trace pairing.)

My question is: Do you know an example where $L|K$ is not separable (but still finite) and this fails, i.e. $B$ is not finitely generated as an $A$-module?

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    $\begingroup$ See exercises 11 and 12 of section 5, chapter 3 (pp. 205-206) of Borevich and Shafarevich's Number Theory. In this example $B$ does not have a finite basis as an $A$-module (what Borevich and Shafarevich call a "fundamental basis" of $L/K$), and since $A$ is a PID and $B$ is torsion-free that means $B$ can't be a finitely generated $A$-module. $\endgroup$
    – KConrad
    Commented Mar 22, 2015 at 7:14

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