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=(ℤ/p⁢ℤ)⁢(b1,b2,…) where the bi are indeterminatesMathworldPlanetmath.

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

{∑i=0∞ci⁢ti}

such that {c0,c1,c2,…} generates a finite extensionMathworldPlanetmath of (ℤ/p⁢ℤ)⁢(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+c1⁢t+c2⁢t2+⋯); its inverseMathworldPlanetmathPlanetmathPlanetmathPlanetmath is a-1⁢(1-c1⁢t-⋯)). Hence the ideals of B are powers of t, so B is a PID (in fact, it is a DVR).

Now, let u=b0+b1⁢t+b2⁢t2+⋯. Now, u∉B because it uses all of the bi and thus the coefficientsMathworldPlanetmath do not define a finite extension of (ℤ/p⁢ℤ)⁢(b1p,b2p,…). However, u is integral over B: bip∈B⇒up∈B 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:

b1+b2⁢t+b3⁢t2+⋯=u-b0t=-b0t+1t⁢u
b2+b3⁢t+b4⁢t2+⋯=u-b0-b1⁢tt2=-b0-b1⁢tt2+1t2⁢u
⋮

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