Suppose that $a$ and $b$ belong to a commutative ring $R$. If $a$ is a unit of $R$ and $b^{2}=0$ . Show that $a+b$ is a unit of $R$

Question

Suppose that $a$ and $b$ belong to a commutative ring $R$. If $a$ is a unit of $R$ and $b^{2}=0$ . Show that $a+b$ is a unit of $R$

I try it..

consider $(a+b)(b)=ab+b^{2}=ab=1$ Since a is unit

Answer 1

No, the product of two units is not $1$, but you are on the right track.

In fact, a more general statement is true: if $a$ is a unit, and $b$ is nilpotent, then $a+b$ is a unit.

The hint by Mark gives $(a+b)(a-b)=a^2$. How can you modify the second factor such that product is equal to $1$? (or said more simply: if the product of two elements is a unit, then both elements are actually units)

Answer 2

consider $(a+b)(a-b)=a^{2}-b^{2}$

therefore we get $(a+b)(a-b)(a^{-2})=1$ which is $(a+b)(a^{-1}-ba^{-2})=1$