Let $R$ be an associative ring. Let $C(R)=\{a\in R: ab=ba\; \forall b\in R\}$. Let $a^2=0 \Rightarrow a=0\; \forall a\in R$. How to prove that for element $c\in R,\ c^2=c \Rightarrow c\in C(R)$?
How to prove that for element $c\in R,\ c^2=c \Rightarrow c\in C(R)$?
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linear-algebra
ring-theory
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0Let $x, y\in R$ and try your conditions on $x+y\in R$. – 2017-02-16
1 Answers
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Let $c \in R$, such that $c^2=c$. Let $b \in R$. Note that $$(c(b-bc))^2 = cbcb-cbcbc-cbc^2b+cbc^2bc = cbcb-cbcbc-cbcb+cbcbc = 0$$, so $c(b-bc) = 0$, i.e. $cb = cbc$.
Analogously get $((b-cb)c)^2 = 0$ and $bc = cbc$. Thus $bc = cb$.