Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $R$ be a ring, where $a^{3} = a$ for all $a\in R$. Prove that $R$ must be a commutative ring.

Please guide me with a proof. Thank you for your kindness.

share|improve this question
If this is homework, what have you tried? Otherwise, Google "x3=x commutative ring" and you'll get several solutions, including mathematik.uni-bielefeld.de/~sillke/PUZZLES/herstein. –  lhf Sep 24 '11 at 11:05
Thank you @lhf. –  Aj I Sep 24 '11 at 11:22
You can also take a look at this MathOverflow question. –  Pierre-Yves Gaillard Sep 24 '11 at 11:29
An exercise in Herstein's textbook Topics in Algebra. Herstein said that, of all the mail he got concerning that textbook, the vast majority was about this single exercise. –  GEdgar Jan 31 '13 at 16:05
mathoverflow.net/questions/29590/… –  mt_ Jan 31 '13 at 16:07

6 Answers 6

To begin with

$$ 2x=(2x)^3 =8 x^3=8x \ . $$

Therefore $6x=0 \ \ \forall x$.


$$ (x+y)=(x+y)^3=x^3+x^2 y + xyx +y x^2 + x y^2 +yxy+ y^2 x + y^3 $$ and

$$ (x-y)=(x-y)^3=x^3-x^2 y - xyx -y x^2 + x y^2 +yxy+ y^2 x -y^3 $$

Subtracting we get

$$ 2(x^2 y +xyx+yx^2)=0 $$

Multiply the last relation by $x$ on the left and right to get

$$ 2(xy+x^2yx+xyx^2)=0 \qquad 2(x^2yx+xyx^2+yx)=0 \ . $$

Subtracting the last two relations we have

$$ 2(xy-yx)=0 \ . $$

We then show that $3( x+x^2)=0 \ \ \forall x$. You get this from

$$ x+x^2=(x+x^2)^3=x^3+3 x^4+3 x^5+x^6=4(x+x^2) \ . $$

In particular

$$ 3 (x+y +(x+y)^2) =3( x+x^2+ y+ y^2+ xy+yx)=0 \, $$

we end-up with $3(xy+yx)=0$. But since $6xy=0$, we have $3(xy-yx)=0$. Then subtract $2(xy-yx)=0$ to get $xy-yx=0$.

share|improve this answer

$\rm(1)\quad ab=0\: \Rightarrow\: ba = 0\ \ via\ \ ba = (ba)^3 = b\ ab\ ab\ a = 0$

$\rm(2)\quad c^2 = c\: \Rightarrow\: c\: $ central $ $ [which means that we have $\rm\ \color{#C00}{xc = cx}\ $ for all $\rm\:x$]

$\rm\begin{eqnarray}Proof:\quad c(x-cx) &=&\rm 0\:\Rightarrow\: (x-cx)c = 0\ \ by\ (1),\ \ so\ \ \color{#C00}{xc} = cxc\\ \rm (x-xc)c &=&\rm 0\:\Rightarrow\: c(x-xc) = 0\ \ by\ (1),\ \ so\ \ \color{#C00}{cx} = cxc\end{eqnarray}$

$\rm(3)\quad x^2\:$ central via $\rm\:c = x^2\:$ in $(2)$

$\rm(4)\quad c^2 = 2c\:\Rightarrow\: c\:$ central. $\ $ Proof: $\rm\:c = c^3 = 2c^2\:$ central by $(3)$.

$\rm(5)\quad x+x^2\:$ central via $\rm\:c = x + x^2\:$ in $(4)$

$\rm(6)\quad x = (x+x^2)-x^2\:$ central via $(3),(5).\quad$ QED

share|improve this answer

I happen to have come across a recent set of exercises on many of the small-$n$ cases of Jacobson's Theorem. It also happens that my solution is different than those contained in the link of @lhf above.

So we have that $a^3 = a \quad\forall a \in R$, and so $2a = (a+a)^3 =8a$, thus $6a = 0$.

Now consider the ideals $2R$ and $3R$. The intersection of $2R$ and $3R$ is trivial, as if $a \in 2R \cap 3R$, then $a = 2r = 3s$ for some $r,s$. Thus $3a = 6r = 0 = 6s = 2a$, and so $(3-2)a = a= 0$. So $2R \oplus 3R = R$. Further, if $a \in 2R$, $b \in 3R$, then $ab, ba \in 2R \cap 3R$ and so $ab = ba = 0$. So we only worry about commutativity in each ideal separately.

In $3R$, we have both $a^3 = a$ and $2a = 2 \cdot 3r = 0$ (some $r$). Then $1 + a = (1 + a)^3 = 1 + 3a + 3a^2 + a^3 = 1 + a + a^2 +a = 1 + a^2 \implies a^2 = a$. So what? In that case, we also have $(1 + a) = (1 + a)^2 = 1 + 2a + a^2 = 1 + 2a + a$, and so $2a = 0$ (yes, we have this in our ideal, but this is true in general in Boolean rings). Continuing, $(a + b) = (a + b)^2 = a^2 + ab + ba + b^2 = a + ab + ba + b$, so $ab = -ba = -ba + 2ba = ba$.

For $2R$, we have both $a^3 = a$ and $3a = 0$. Then we have that $a + b = (a + b)^3 = a^3 + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b^3 $$= a + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b$ on the one hand, and $a - b = (a - b)^3 = a^3 - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b^3$ $= a - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b$.

Taking the difference between these, we see $2(a^2b + aba + ba^2) = 0$, and so $a^2 b + aba + ba^2 = 0$. Multiply by $a$, and we get $a^3b + a^2ba + aba^2 = ab + (a^2b + aba)a = ab + (-ba^2)a = ab - ba = 0$. Thus $ab = ba$.

As both ideals commute separately and in products, $R$ commutes in general.

share|improve this answer

Re-posting it from here. Note that $R$ is not necessarily unital.

Some general facts:

We call a ring $R$, J-ring (Jacobson ring), if for any $x \in R$ there is a natural number $n(x) >1$ s.t. $x^{n(x)}=x$. (In fact, Jacobson has proven that any J-ring is commutative, for the proof you may take a look at Non-commutative Rings written by Herestein)

Lemma 1: If $R$ be a J-ring, then $N(R)= \{0 \}$ where $N(R)$ denotes the nilradical of $R$.

Proof: Let $0\not= x\in N(R)$. Then there is a smallest natural number greater $1$ s.t. $x^m=0$. Since $R$ is a J-ring, there is an $n>1$ s.t. $x^n=x$. Let $m=nq+r$ where $0 \leq r <n$. Therefore,


However, $q+r<m$, which is a contradiction, since $m$ was chosen to be the smallest number satisfying $x^m=0$.

Lemma 2: Suppose that in a ring $R$, $N(R)= \{0 \}$, then any idempotent element $a$ i.e. $a^2=a$, lies in the center $Z(R)$.

Proof: Suppose that $x \in R$. Then


Since $N(R)= \{0 \}$, then we have $axa-ax=0 \rightarrow axa=ax$. With the same approach and by considering $(axa-xa)^2$ we will obtain $axa=xa$. Hence, $ax=xa$ and since $x$ was an arbitrary element of $R$ then $a \in Z(R)$.

Lemma 3: In a J-ring $R$, we have $x^{n(x)-1} \in Z(R).$

Proof: $(x^{n(x)-1})^2=x^{2n(x)-2}=x^{n(x)}x^{n(x)-2}=xx^{n(x)-2}=x^{n(x)-1}$. Thus $x^{n(x)-1}$ is an idempotent element of $R$ and by Lemma 1 & 2. we get the result.

In particular, in your question, $n=n(x)=3$ and $x^2 \in Z(R),$ for any $x \in R.$ Moreover


Exercise: The same question with $x^4=x$ for any $x \in R.$

share|improve this answer

Here are some hints, i'll develop the answers if necessary :

Let $R$ be a ring such that $x^3=x$ for any $x$ in $R$.

1) Determine the nilpotent elements in $R$.

2) Show that any idempotent in $R$ is central (i.e. for any $e\in R$ such that $e^2=e$ and any $x\in R$, we have $ex=xe$). Deduce from this that if $x\in R$, then $x^2$ belongs to the center of $R$.

3) Conclude that $R$ must be commutative.

share|improve this answer

This is not the best proof, but let me add this one because it hasn't been mentioned yet. The proof rests on a single non-commutative polynomial identity (whose verification can be done quickly with a computer algebra system).

Define $f(x,y):=(x+y)^3-x^3-y^3 \in \mathbb{Z} \langle x,y \rangle$. Then we have

$$f\bigl(x,y+(x \cdot y-y \cdot x)\bigr) - f(x,y) - f\bigl(x,(x \cdot y-y \cdot x)\bigr) - (x \cdot f(x,y) - f(x,y) \cdot x)$$ $$ = -x^3 \cdot y + y \cdot x^3$$ Thus, if $R$ is a ring with $a^3=a$ for all $a \in R$, we have $f(a,b)=0$ for $a,b \in R$, and hence also $-a^3 \cdot b + b \cdot a^3=0$ resp. $ab=ba$.

share|improve this answer
If you want to feed a computer algebra system, you will be glad that there are dots. –  Martin Brandenburg Sep 23 at 19:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.