# Rings, and zero divisors

• Apr 2nd 2009, 09:24 PM
lttlbbygurl
Rings, and zero divisors
Let R be a ring that contains at least two elements. Suppose for each nonzero a in R there exists a unique b in R such that aba = a.

a. show that R has no divisors of 0
b. show that bab = b
c. show that R has unity
d. show that R is a division ring

B I worked out but I'm not sure if it works..
we're given aba = a and we need bab=b
abab= ab
multiply by inverse of a to get bab=b

thanks!
• Apr 2nd 2009, 10:27 PM
NonCommAlg
Quote:

Originally Posted by lttlbbygurl

Let R be a ring that contains at least two elements. Suppose for each nonzero a in R there exists a unique b in R such that aba = a.

a. show that R has no divisors of 0.

suppose $\displaystyle xy=0$ but $\displaystyle x \neq 0$ and $\displaystyle y \neq 0.$ so there exists a unique $\displaystyle z \in R$ such that $\displaystyle xzx=x.$ but then: $\displaystyle x(z+y)x=xzx + xyx=x.$ therefore $\displaystyle z+y=z,$ and hence $\displaystyle y=0.$ contradiction.

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b. show that bab = b.
we have $\displaystyle baba=ba$ and thus $\displaystyle (bab - b)a=0.$ since $\displaystyle a \neq 0,$ the first part of the problem gives us: $\displaystyle bab-b=0.$

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c. show that R has unity.
let $\displaystyle 0 \neq a \in R.$ so there exists $\displaystyle b \in R$ such that $\displaystyle aba=a.$ the claim is that $\displaystyle ab=1_R.$ so let $\displaystyle c \in R.$ then $\displaystyle caba=ca$ and hence $\displaystyle (cab - c)a=0.$ therefore $\displaystyle cab=c,$ by the first part of the problem.

also by the second part, $\displaystyle bab=b$ and thus $\displaystyle babc=bc,$ which gives us $\displaystyle b(abc - c)=0.$ thus $\displaystyle abc = c.$

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d. show that R is a division ring.
let $\displaystyle 0 \neq x.$ so there exists $\displaystyle y \in R$ such that $\displaystyle xyx=x$ and thus $\displaystyle (xy - 1)x=0,$ which gives us: $\displaystyle xy=1.$