Given $(R, +, \cdot)$ a commutative and unitary ring, $a \in U(R)$ and $b \in R$ such that $b^2 = 0$ prove that $(a+b) \in U(R)$
My take on this problem has been the following:
$a \in U(R) \Rightarrow \exists a^{-1} \in R$ such that $a \cdot a^{-1} = a^{-1} \cdot a = 1_R$
Now looking at $a^2$ we see that $\exists a^{-1} \cdot a^{-1} = a^{-2}$ such that $a^2 \cdot a^{-2} = a^{-2} \cdot a^2 = 1_R \Rightarrow a^2 \in U(R)$
Then we take $c=(a+b)\cdot(a-b)= a^2 -ab + ba -b^2$ where $b^2=0$ and $ab =ba$ $\Rightarrow c=a^2 \Rightarrow c\in U(R)$
Now all there is to prove is that given a product of two elements in $U(R)$ then each one of the two elements is a unit.
Thanks for the help in advance
