Proving that sum of two elements is not invertible in a ring 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?
 A: 
Lemma. In a ring $A$, if $a$ is invertible and $x$ is not invertible, then $ax$ is not invertible.

Proof. Suppose $ax$ is invertible. Then $x=a^{-1}(ax)$ is the product of two invertible elements, hence invertible. QED
In particular $-x=(-1)x$ is not invertible as soon as $x$ is not invertible, because $-1$ is invertible.
Finally, if $x\in D$ and $a$ is invertible, then $ax\ne0$ and $ax$ is not invertible. Therefore $ax\in D$. Since you proved that $1+x$ is not invertible for $x\in D$, the same applies for $1+a^{-1}x$, as $a^{-1}x\in D$. But
$$
a+x=a(1+a^{-1}x)
$$
hence also $a+x\in D$. The text has a typo, because $a^{-1}(1+ax)\ne a+x$ in general.
A: First, if $-x$ is not in D, then $-(-x)^{-1}$ is the inverse of $x$, but this is a contradiction because $x$ is not invertible.
As for the last part, if $a$ invertible and $x$ is not, then $ax$ cannot be invertible, otherwise $((ax)^{-1}a)x=1$ and, so, $x$ would be invertible. In the same way you have that $a^{-1}x$ is not invertible and that $a+x=a(1+a^{-1}x)$ is in $D$, because $a$ is invertible and $1+a^{-1}x$ is not.
