separability is required for integral closures to be finitely generated
The parent theorem assumed that was a separable extension of . Here is an example that shows that separability is in fact a necessary condition for the result to hold.
Let where the are indeterminates.
Let be the subring of (power series in over ) consisting of
such that generates a finite extension of . Then every element of is a power of times a unit (first factor out the largest power of . What’s left is ; its inverse is ). Hence the ideals of are powers of , so is a PID (in fact, it is a DVR).
Now, let . Now, because it uses all of the and thus the coefficients do not define a finite extension of . However, is integral over : which implies that the degree of over the field of fractions is . Hence a basis for is . There are other elements integral over :
Clearly the denominators are getting bigger, so the integral closure of cannot be finitely generated as a -module.
Title | separability is required for integral closures to be finitely generated |
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Canonical name | SeparabilityIsRequiredForIntegralClosuresToBeFinitelyGenerated |
Date of creation | 2013-03-22 17:02:15 |
Last modified on | 2013-03-22 17:02:15 |
Owner | rm50 (10146) |
Last modified by | rm50 (10146) |
Numerical id | 6 |
Author | rm50 (10146) |
Entry type | Example |
Classification | msc 13B21 |
Classification | msc 12F05 |