separability is required for integral closures to be finitely generated

The parent theorem assumed that L was a separable extensionMathworldPlanetmath of K. Here is an example that shows that separability is in fact a necessary condition for the result to hold.

Let k=(/p)(b1,b2,) where the bi are indeterminatesMathworldPlanetmath.

Let B be the subring of k[[t]] (power series in t over k) consisting of


such that {c0,c1,c2,} generates a finite extensionMathworldPlanetmath of (/p)(b1p,b2p,). Then every element of B is a power of t times a unit (first factor out the largest power of t. What’s left is a(1+c1t+c2t2+); its inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmath is a-1(1-c1t-)). Hence the ideals of B are powers of t, so B is a PID (in fact, it is a DVR).

Now, let u=b0+b1t+b2t2+. Now, uB because it uses all of the bi and thus the coefficientsMathworldPlanetmath do not define a finite extension of (/p)(b1p,b2p,). However, u is integral over B: bipBupB which implies that the degree of u over the field of fractionsMathworldPlanetmath is p. Hence a basis for K(u)/K is {1,u,u2,,up-1}. There are other elements integral over B:


Clearly the denominators are getting bigger, so the integral closureMathworldPlanetmath of B cannot be finitely generatedMathworldPlanetmathPlanetmathPlanetmath as a B-module.

Title separability is required for integral closures to be finitely generated
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