# 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?

• I have copied it correctly. If 1+x was invertible, that would have implied x=0 which is not in D. – Gaboru Feb 17 at 22:36

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.

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.

• Bu how $a+x=a^{-1}(1+ax)$ since $a$ is not equal to $a^{-1}$ – Gaboru Feb 17 at 22:52
• I edited the answer – Francesco Feb 17 at 22:57