# Thread: Algebra, Problems For Fun (6)

1. ## Algebra, Problems For Fun (6)

Let $\displaystyle R$ be a unitary ring. Suppose that $\displaystyle ab=0 \Longrightarrow ba=0,$ for any $\displaystyle a,b \in R.$ Prove that $\displaystyle ab=1 \Longrightarrow ba=1,$ for any $\displaystyle a,b \in R.$

2. So... in an integral domain, "one-sided invertible $\displaystyle \Leftrightarrow$ invertible" ?

3. Originally Posted by clic-clac
So... in an integral domain, "one-sided invertible $\displaystyle \Leftrightarrow$ invertible" ?
that's right!

4. Originally Posted by NonCommAlg
Let $\displaystyle R$ be a unitary ring. Suppose that $\displaystyle ab=0 \Longrightarrow ba=0,$ for any $\displaystyle a,b \in R.$ Prove that $\displaystyle ab=1 \Longrightarrow ba=1,$ for any $\displaystyle a,b \in R.$
Hi NonCommAlg.

_____$\displaystyle ab=1$

$\displaystyle \implies\ ab-1=0$

$\displaystyle \implies\ b(ab-1)=b0=0=0b=(ab-1)b$

$\displaystyle \implies\ bab-b=ab^2-b$

$\displaystyle \implies\ bab=ab^2$

$\displaystyle \implies\ bab-ab^2=0$

$\displaystyle \implies\ (ba-ab)b=0$

$\displaystyle \implies\ b(ba-ab)=0$

$\displaystyle \implies\ ab(ba-ab)=a0=0$

$\displaystyle \implies\ 1(ba-ab)=0$

$\displaystyle \implies\ ba-ab=0$

$\displaystyle \implies\ ba=ab=1$

5. Originally Posted by TheAbstractionist
Hi NonCommAlg.

_____$\displaystyle ab=1$

$\displaystyle \implies\ ab-1=0$

$\displaystyle \implies\ b(ab-1)=b0=0=0b=(ab-1)b$

$\displaystyle \implies\ bab-b=ab^2-b$

$\displaystyle \implies\ bab=ab^2$

$\displaystyle \implies\ bab-ab^2=0$

$\displaystyle \implies\ (ba-ab)b=0$

$\displaystyle \implies\ b(ba-ab)=0$

$\displaystyle \implies\ ab(ba-ab)=a0=0$

$\displaystyle \implies\ 1(ba-ab)=0$

$\displaystyle \implies\ ba-ab=0$

$\displaystyle \implies\ ba=ab=1$
that's correct. good work! an almost similar way: $\displaystyle 0=b(1-ab)=b-bab=(1-ba)b=b(1-ba).$ therefore $\displaystyle 1-ba=ab(1-ba)=0.$