Answer to Question #226101 in Abstract Algebra for Gour

Question

Suppose that a and b belong to a commutative ring R with unity. If a is a unit of R and b^2 = 0, show that a + b is a unit of R.

Answer 1

We look at this problem from two perspective.

Case 1 (b=0)

If b=0 then a+b = a. And since a is a unit of R, we can conclude that a+b=a is a unit of R.

Case 2 ($b \neq 0)$

Since a is a unit, let c be the inverse of a.

In order to show that a+b is a unit, it suffices to show that 1+bc is a unit.

Given that $b^2=0$ , we have that $b^3 = 0$ because $b^3=b^2b=0(b)=0$

Also, $(bc)^3 = 0$

But

$1= 1+ (bc)^3 = (1+bc)(1-bc+(bc)^2)$

That is 1+bc is a unit. Hence, a+b is a unit.