Prove the only invertible element from a certain ring is $1$

Question

Let $(A, + ,\cdot)$ be a ring that is not a field, such that $x^2=x$ for all non-invertible $ x \in A$. Prove $x^2=x, \forall x \in A$

The goal is to prove the only invertible element is $1$.

If $x$ is non-invertible and not null, then $-x$ is also non-invertible, therefore $(-x)^2=-x$ and $2x=0$. Also $(1+x)^2=1 + 2x + x^2=1 +x$ therefore $1+x$ is also non-invertible (otherwise, from $(1 +x)^2=1 + x$ it follows $1 + x =1$ therefore $x=0$).

I couldn't get further than that. Any help is appreciated.

But you haven't proved that $1$ is the only invertible element. So how can you use it? — Sahiba Arora, Jul 29 '17 at 10:27
I didn't. I just simplified $(1 +x)^2=1 + x$ multiplying by $(1+x)^{-1}$ — , Jul 29 '17 at 10:29
You have shown that $1+x$ is not invertible; given an arbitrary element $u$ you can then show both $xu^2=xu$ and $(1+x)u^2=(1+x)u$, from which you can derive $u^2=u$. — Colin McQuillan, Jul 29 '17 at 15:17
@Colin, still, it was a shot at the right direction. I've tried to see how you got it and got my answer from it. — Ennar, Jul 29 '17 at 16:15

score 3 · Accepted Answer · answered Jul 29 '17 at 15:41

Fix $x\neq 0$ non-invertible. We have that for any $u$ invertible $ux$ is non-invertible and non-zero, which implies $(ux)^2 = ux$ and $xux = x$.

On the other hand, if $u\neq 1$ is invertible, then $u-1$ must be invertible as well: assume the contrary, then $u = 1+y$ for non-zero, non-invertible $y$, but this is contradiction since you proved that $1+y$ is non-invertible.

Combining the results of the two paragraphs, we have that for any invertible $u\neq 1$ it holds that $$x(u-1)x = x \implies xux - x^2 = x\implies x - x = x\implies x = 0,$$ a contradiction with our choice of $x$. Thus, $u = 1$ is the only invertible element of the ring.

Prove the only invertible element from a certain ring is $1$

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