In other words, show that if every element, other than 1, has a left quasi-inverse, then every element, other than 0, has an inverse.
I take it we are allowed to assume that the ring has a multiplicative identity, 1?
Hello !
can you help me proving this exercise please :
An element a in a ring R has a left quasi inverse if there exist an element b in R with a+b-ab=0.
Prove that if every element in a ring R except 1 has a left quasi-inverse, then R is
a division ring..
THANK YOU !
In other words, show that if every element, other than 1, has a left quasi-inverse, then every element, other than 0, has an inverse.
I take it we are allowed to assume that the ring has a multiplicative identity, 1?
here is something to get you started:
a + b - ab = 0
a + (1 - a)b = 0
1 + (1 - a)b = 1 - a
1 = (1 - a) - (1 - a)b
1 = (1 - a)(1 - b).
so 1 - a has a right-inverse, unless a = 1. why does this show U(R) = R\{0}?