Looking for counterexamples (non-domains)

Question

I stumbled upon this question Why doesn't $xa = x$ for all $x \in R$ imply that $a$ is the unit of $R$? and understood the given answers. But it got me thinking about a counterexample, and I was hoping to find something along these lines:

We have for $a\in R$ and for all $x\in R$

$$xa=x$$ so $$x(a-1)=0$$

$R$ has no zero divisors (for instance, $R$ is a domain - careful, this assumes $1\neq 0$ and commutativity): then the above implies that $x=0$ or $a=1$. But see the link above to understand why $a=1$ does not work here.

$R$ HAS zero divisors: Here I was hoping to find an example where $a-1\neq 0$ is a zero divisor, hence showing that $a$ need not be a unit. I tried thinking about $\mathbb{Z}/n\mathbb{Z}$ and $2 \times 2$ matrices, but then the condition $xa=x$ does not hold for all $x$.

Should there be further thinking about if $1\neq 0$?

We can also say that $R=\{0\}$, taking $R$ without $1$. Then this is a counterexample since $a=0$.

Answer 1

After a week of inactivity on the question it also looks like you are not going to make any adjustments, so I answer it here as-is. It seems like there is no way to rescue the posed question anyhow.

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It seems like you may not have understood the linked question's solutions.

One of the solutions to the question you linked to demonstrates, independently of the ring being a domain or not begin a domain, that if $R$ has an identity $1$, then there is no other element $a$ acting like the identity on the right.

To say it again, more explicitly:

If $R$ has an identity $1$ and there is an $a\in R$ such that $x(a-1)=0$ for all $r\in R$, then $a=1$ (i.e. $a-1=0$.)

In light of the fact that $x$ can be $1$, this is obvious, right?

If you are searching for a counterexample in the form of a nonzero rng that has an element $a$ satisfying $xa=x$ for all $x\in R$, and $R$ doesn't have an identity (and hence no invertible elements at all), then that was also already provided by the solution.

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We can also say that $R=\{0\}$, taking $R$ without $1$. Then this is a counterexample since $a=0$.

And $\{0\}$ is a ring with identity, contrary to your statement. How do you figure it would be a counterexample? A counterexample would require finding a nonzero element with your property, and this ring has no nonzero elements.