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Let I be a two sided ideal of R. Prove that I=eR for some central idempotent e ϵ R if and only if R=I+J for some two sided ideal J. When this occurs, show that e and J are uniquely determined by I.

Attempt: e is a central idempotent of R then e^2=e and er = re for all r∈R Suppose I∈R RTP:R ⊕ J for some two sided ideal J. We seek ideal J such that 1.I∩J=0 2.I+J=R

By Lemma R = eR ⨁ (1-e)R if e is central then so is 1-e

Second part
Suppose I is (two-sided)ideal and R=I⊕J with J ideal

How do I show that 1. There exist a unique idempotent e such that I = eR i.e. if I=eR=fR then f=e 2. If R = I⊕J and R = I⊕K with J,K ideals then J=K

Uthando
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1 Answers1

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Assuming $R= I \oplus J$, it follows that there exist unique $e \in I$, $f \in J$ such that $1=e+f$. Recall that for all $x \in I, y \in J$ you have $$xy \in IJ \subseteq I \cap J = 0$$ so that $xy =0$.

Now, $$e=e(e+f)=e^2+ef=e^2$$ so that $e$ is an idempotent. Moreover, for all $x \in I$ $$x= x(e+f) = xe + xf = xe \in eR$$ so that $x \in eR$. By arbitrarity of $x$, it follows that $I=eR$.

Crostul
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  • Nice answer. This argument assumes that $R$ is unital. If it is not, you can adjoin a unit, run this argument, and then intersect the resulting complementary ideal $J$ of $R^+$ with $R$. – Aidan Sims Apr 12 '16 at 11:00
  • @AidanSims Oh, you are right, I always forget that rings may not have a unit. – Crostul Apr 12 '16 at 11:03
  • Actually, on second thoughts, I don't think you can adjoin a unit to run the argument if $R$ is not unital. Consider $X := \mathbb{K} \setminus{0,1}$, where $\mathbb{K}$ is the Cantor middle-third set. Let $R := C_0(X)$, let $I = {f \in R : \operatorname{supp}(f) \subseteq (0,1/3]}$ and let $J = {f \in R : \operatorname{supp}(f) \subseteq [2/3, 1)}$. Then $R = I \oplus J$, but neither $I$ nor $J$ has the form $eR$ for any idempotent $e \in R$. – Aidan Sims Apr 12 '16 at 11:09