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Jacobson ideal

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Let R be a unital ring. Define J(R)={a [itex]\in[/itex] R| 1-ra is a unit for any r [itex]\in[/itex] R}

    Show that J(R) is an ideal in R. (It is called the Jacobson ideal of R)

    2. Relevant equations
    I is ideal of ring R
    , then I satifies
    a+b [itex]\in[/itex] I [itex]\forall[/itex] a,b [itex]\in[/itex] I

    ra [itex]\in[/itex] I [itex]\forall[/itex] r [itex]\in[/itex] R

    3. The attempt at a solution

    I've been trying to use direct definition by having two elements 1-ra, 1-rb [itex]\in[/itex] J(R), then I tried to do (1-ra)+(1-rb) and hope to end up another element which has format 1-rc, but I couldn't get it.

    Similarly, I let some x [itex]\in[itex] R, then try to compute x(1-ra), hope can end up format 1-ry, so it can satisfy second condition of being an ideal of ring R, but I still cannot get that format.

    Unless I haven't use information that 1-ra is unit to help me solve the problem. But not quite sure how to use this bit information.

    Can anyone please help me with this question? Thanks a lot.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 20, 2012 #2


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    Gold Member

    1-ra is not in J(R), a is. If a and b are in J(R), 1-ra and 1-rb are units - you need to prove that 1-r(a+b) is a unit as well to show that a+b is contained in J(R)
  4. Mar 24, 2012 #3
    Hey I have also been working on this problem- got as far as showing that if ua and ub [itex]\in R[/itex] such that ua(1-ra) =1=(1-ra)ua
    and ub(1-rb) =1=(1-rb)u then ubua(1-r(a+b))=1.

    But (1-r(a+b))ubua=(1-raub)a[itex]\neq?1[/itex]

    Does anyone have suggestions on how to go from here?
  5. Mar 25, 2012 #4
    It might be easier to show that J(R) is the intersection of all maximal ideals. This is not hard to show.
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