proof of characterizations of the Jacobson radical

First, note that by definition a left primitive ideal is the annihilatorMathworldPlanetmath of an irreducible left R-module, so clearly characterizationMathworldPlanetmath 1) is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to the definition of the Jacobson radicalMathworldPlanetmath.

Next, we will prove cyclical containment. Observe that 5) follows after the equivalence of 1) - 4) is established, since 4) is independent of the choice of left or right idealsMathworldPlanetmathPlanetmath.

  1. 1) 2)

    We know that every left primitive ideal is the largest ideal contained in a maximal left ideal. So the intersectionMathworldPlanetmath of all left primitive ideals will be contained in the intersection of all maximal left ideals.

  2. 2) 3)

    Let S={M:M a maximal left ideal of R} and take rR. Let tMSM. Then rtMSM.

    Assume 1-rt is not left invertible; therefore there exists a maximal left ideal M0 of R such that R(1-rt)M0.

    Note then that 1-rtM0. Also, by definition of t, we have rtM0. Therefore 1M0; this contradictionMathworldPlanetmathPlanetmath implies 1-rt is left invertible.

  3. 3) 4)

    We claim that 3) satisfies the condition of 4).

    Let K={tR:1-rt is left invertible for all rR}.

    We shall first show that K is an ideal.

    Clearly if tK, then rtK. If t1,t2K, then


    Now there exists u1 such that u1(1-rt1)=1, hence


    Similarly, there exists u2 such that u2(1-u1rt2)=1, therefore


    Hence t1+t2K.

    Now if tK,rR, to show that trK it suffices to show that 1-tr is left invertible. Suppose u(1-rt)=1, hence u-urt=1, then tur-turtr=tr.

    So (1+tur)(1-tr)=1+tur-tr-turtr=1.

    Therefore K is an ideal.

    Now let vK. Then there exists u such that u(1-v)=1, hence 1-u=-uvK, so u=1-(1-u) is left invertible.

    So there exists w such that wu=1, hence wu(1-v)=w, then 1-v=w. Thus (1-v)u=1 and therefore 1-v is a unit.

    Let J be the largest ideal such that, for all vJ, 1-v is a unit. We claim that KJ.

    Suppose this were not true; in this case K+J strictly contains J. Consider rx+syK+J with xK,yJ and r,sR. Now 1-(rx+sy)=(1-rx)-sy, and since rxK, then 1-rx=u for some unit uR.

    So 1-(rx+sy)=u-sy=u(1-u-1sy), and clearly u-1syJ since yJ. Hence 1-u-1sy is also a unit, and thus 1-(rx+sy) is a unit.

    Thus 1-v is a unit for all vK+J. But this contradicts the assumptionPlanetmathPlanetmath that J is the largest such ideal. So we must have KJ.

  4. 4) 1)

    We must show that if I is an ideal such that for all uI, 1-u is a unit, then Iann(MR) for every irreducible left R-module MR.

    Suppose this is not the case, so there exists MR such that Iann(MR). Now we know that ann(MR) is the largest ideal inside some maximal left ideal J of R. Thus we must also have IJ, or else this would contradict the maximality of ann(MR) inside J.

    But since IJ, then by maximality I+J=R, hence there exist uI and vJ such that u+v=1. Then v=1-u, so v is a unit and J=R. But since J is a proper left ideal, this is a contradiction.

Title proof of characterizations of the Jacobson radical
Canonical name ProofOfCharacterizationsOfTheJacobsonRadical
Date of creation 2013-03-22 12:48:56
Last modified on 2013-03-22 12:48:56
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 31
Author rspuzio (6075)
Entry type Proof
Classification msc 16N20