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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.

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If this is homework, what have you tried? Otherwise, Google "x3=x commutative ring" and you'll get several solutions, including – 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
2… – m_t_ Jan 31 '13 at 16:07

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 $xy-yx=0$.

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

$\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

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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 $x \in R.$

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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.

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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. $ab=ba$.

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If you want to feed a computer algebra system, you will be glad that there are dots. – Martin Brandenburg Sep 23 '14 at 19:25

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.

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