Let $(A,+,.)$ be a ring such that A is not a field and $x^2=x, \forall $ non-invertible $ x\in A $. Prove that:
a) $a+x$ is not invertible for all $a,x\in A$ with $a$ invertible and $x\ne0$,$x$ noninvertible
b) $x^2=x,\forall x\in A$
I've seen the proof but I don't understand everything.
Let D be the set of all non-zero and non-invetible elements of A. If $x$ is and element in D, then $-x$ is in D so $2x=0$. (I don't know why $-x$ should be in D).
$(1+x)^2=1+x$ so $1+x$ is non-invertible. Let $a$ be an invertible element in A.
And I don't understand what happens next:
$ax$ and $1+ax$ are in D so $a+x=a^{-1}(1+ax)$ is in D.
Can somebody explain this to me, please?