Lifting idempotents modulo nil ideals

Posted: November 1, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: , , ,

Definition 1. Let I be a two-sided ideal of a ring R. We say that an idempotent element a \in R/I can be lifted if a=e + I for some idempotent element e \in R.

Definition 2. A subset S of a ring R is called nil if every element of S is nilpotent.

Problem. If I is a nil ideal of R, then every idempotent of R/I can be lifted.

Solution. Suppose that a=r+ I is an idempotent element of R/I. Then r-r^2 \in I and thus (r-r^2)^n=0, for some integer n \geq 1. Therefore

r^n - r^{n+1} \sum_{i=1}^n (-1)^{i-1} \binom{n}{i}r^{i-1} = 0.

So if we let

s =\sum_{i=1}^n (-1)^{i-1} \binom{n}{i}r^{i-1},

then r^n=r^{n+1}s and rs=sr. Now let e=(rs)^n. Using the fact that r^n=r^{n+1}s we will get e^2=e. Also, since r+ I = r^2+I, we have r+I=r^k + I for all k \geq 1. Therefore

r+I=r^n+I=r^{n+1}s+I=(r^{n+1}+I)(s+I)=rs+I.

Thus

a=r+I=r^n + I=(r+I)^n=(rs+I)^n=(rs)^n+I=e+I. \Box

Example. Let R be a (left) Artinian ring and let J be the Jacobson radical of R. Since J is nilpotent, and hence nil, every idempotent of R/J can be lifted.

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Comments
  1. fu don't ask says:

    The power of $r$ in the first summation should be $i-1$.

    But if $R$ has no multiplicative identity,
    what is $r^0$ means?
    Here is another solution,

    http://goo.gl/zEK01L

    http://goo.gl/FpncJD

    • Yaghoub Sharifi says:

      Yes, you’re right. It should be i-1. I just fixed it, Thank you!
      In this blog all rings have 1 unless otherwise specified. Thanks for the link though! 🙂

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