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?

  • $\begingroup$ I have copied it correctly. If 1+x was invertible, that would have implied x=0 which is not in D. $\endgroup$ – tyuiop Feb 17 '19 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.

  • $\begingroup$ Bu how $a+x=a^{-1}(1+ax)$ since $a$ is not equal to $a^{-1}$ $\endgroup$ – tyuiop Feb 17 '19 at 22:52
  • $\begingroup$ I edited the answer $\endgroup$ – Francesco Feb 17 '19 at 22:57

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