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.

  • $\begingroup$ 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 $a \in M$. $\endgroup$ Nov 9, 2019 at 18:50
  • $\begingroup$ @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.... $\endgroup$
    – Math Guy
    Nov 9, 2019 at 21:34
  • $\begingroup$ 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? $\endgroup$ Nov 9, 2019 at 21:35
  • $\begingroup$ @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. $\endgroup$
    – Math Guy
    Nov 9, 2019 at 21:37
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 9, 2019 at 21:54

1 Answer 1


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 $(ba-ab)^2 = 0$.

  • $\begingroup$ Thank you ! Could you tell me how you came up with this solution? $\endgroup$
    – Math Guy
    Nov 10, 2019 at 11:23
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    $\begingroup$ @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). $\endgroup$ Nov 10, 2019 at 13:02

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