# A set of elements in a reduced unity ring

Let $$(A,+,\cdot)$$ be a unity ring with the property that if $$x \in A$$ and $$x^2=0$$ then $$x=0$$. Consider the set $$M=\{a\in A | a^3=a\}$$. Prove that:

a) $$2a\in Z(A)$$, $$\forall a\in M$$, where $$Z(A)$$ denotes the centre of the ring $$A$$;
b) $$ab=ba$$, $$\forall a,b\in M$$.

My attempts revolved around the fact that an idempotent element in a reduced ring is central.
So, since for $$a\in M$$ we have that $$(a^2)^2=a^2$$, it follows that $$a^2\in Z(A)$$, $$\forall a\in M$$.
The next thing I wanted to use in order to solve a) was that $$Z(A)$$ is a subring of $$A$$, so if I had proved that $$(a+1)^2 \in Z(A)$$, $$\forall a\in M$$, then we would have reached the desired conclusion. However, I couldn't prove this and I honestly doubt that it is true.
Another idea that I had was to prove that $$M$$ is a subring of $$A$$. Of course, this didn't work out because I cannot even prove that $$M$$ is closed under addition. Again, I don't know if this is true and it most likely isn't.
As for b), I think that a) should be of use, but I don't know how. It is a well-known problem that a ring with $$x^3=x$$ for any $$x$$ in that ring is commutative, but since $$(M,+,\cdot)$$ is almost definitely not a ring, this doesn't help.
EDIT: Is there any chance that this question is simply wrong? I tended to believe this before asking it here too, but since nobody has made any progress on it until now I am even more inclined to think so.

• If you honestly doubt that $(a+1)^2 \in Z(a)$ for all $a \in M$, then you honestly doubt that part (a) is true, since if part (a) is true (and since you've already noted $a^2 \in Z(a)$ for all $a \in M$), you get $(a+1)^2 = a^2+2a+1 \in Z(a)$ for all $a \in M$. Nov 9, 2019 at 18:50
• @mathworker21 I agree with you. But since I have spent hours and hours trying to prove that $(a+1)^2 \in Z(A)$ for all $a\in M$ and nothing worked out, then I am inclined to believe that this statement may actually be wrong. Yet, since I cannot provide a counterexample, here I am, still hoping that someone better than me at rings will solve it.... Nov 9, 2019 at 21:34
• I don't know. I don't think the proof that any ring with $x^3 = x$ for all $x$ implies the ring is commutative is that easy... I could imagine spending hours failing to find that proof. By the way, where did you find this problem? Nov 9, 2019 at 21:35
• @mathworker21 I know the "classical" one where $x^3=x$ for all $x$ holds in a ring, but since here we do not have a ring it doesn't really help I think. The problem is from a magazine from my country. Nov 9, 2019 at 21:37
• you missed my point. my point was that working on these kinds of problems for hours and failing doesn't mean they are false. (Of course that is always true, but I find the statement more meaningful here). The evidence/example I gave was the $x^3=x$ problem. For that problem, I could imagine working on it for several hours without finding the solution. And of course that problem is true. Nov 9, 2019 at 21:54

We show $$M \subseteq Z(A)$$. This immediately gives (a) and (b).

Lemma 1: $$yx = 0 \implies xzy = 0$$ for any $$z$$.

Proof: $$(xzy)(xzy) = xz(yx)zy = 0$$.

Lemma 2: $$x^2 = x$$ implies $$x \in Z(A)$$.

Proof: For any $$y \in A$$, a short computation shows $$(xy-xyx)(xy-xyx) = 0 = (yx-xyx)(yx-xyx)$$.

Lemma 3: $$a \in M \implies a^2 \in Z(A)$$.

Proof: $$(a^2)^2 = a^4 = a^2$$, so use Lemma 2.

Claim: For any $$a \in M$$, $$a \in Z(A)$$.

Proof: Since $$(a-1)[a(a+1)]=0$$, Lemma 1 implies that for any $$b \in A$$, $$0 = a(a+1)b(a-1) = (a^2b+ab)(a-1).$$ Also, $$a(a+1)(a-1)=0$$ implies $$0 = ba(a+1)(a-1) = (ba^2+ba)(a-1) = (a^2b+ba)(a-1),$$ where the last equality used Lemma 3. Subtracting gives: (1) $$0 = (ba-ab)(a-1)$$. The exact same argument shows: (2) $$0 = (a-1)(ba-ab)$$. (1) immediately implies $$0 = (ba-ab)(a-1)b = (ba-ab)(ab-b)$$, and (2) with Lemma 1 implies $$0 = (ba-ab)b(a-1) = (ba-ab)(ba-b)$$. Subtracting the two results gives $$(ba-ab)^2 = 0$$.

• Thank you ! Could you tell me how you came up with this solution? Nov 10, 2019 at 11:23
• @MathGuy well, I played around with it for a while, so I exhausted many different approaches. Then I eventually stumbled upon something like Lemma $1$, first in the form of $0 = a(a+1)b(a-1)$ and realized it started giving equations I hadn't seen/derived before. So I knew I had something good. Then it was just a matter of finishing up, which was pretty easy (the proof of the claim is rather short). Nov 10, 2019 at 13:02