Is J(R) the Intersection of All Maximal Ideals in R?

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Homework Statement



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

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


Homework Equations


I is ideal of ring R
, then I satifies
a+b \in I \forall a,b \in I

ra \in I \forall r \in R

The Attempt at a Solution



I've been trying to use direct definition by having two elements 1-ra, 1-rb \in 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 \inR, 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.<br /> <br /> 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.<br /> <br /> Can anyone please help me with this question? Thanks a lot.<br />
 
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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)
 
Hey I have also been working on this problem- got as far as showing that if ua and ub \in R 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\neq?1

Does anyone have suggestions on how to go from here?
 
It might be easier to show that J(R) is the intersection of all maximal ideals. This is not hard to show.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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