# If $a^3 =a$ for all $a$ in a ring $R$, then $R$ is commutative.

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

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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 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

To begin with

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

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

Also

$$(x+y)=(x+y)^3=x^3+x^2 y + xyx +y x^2 + x y^2 +yxy+ y^2 x + y^3$$ 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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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 $$= a + a^2b + aba + ab^2 + ba^2 + bab + b^2a + b on the one hand, and a - b = (a - b)^3 = a^3 - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b^3 = a - a^2b - aba + ab^2 - ba^2 + bab + b^2a - b. Taking the difference between these, we see 2(a^2b + aba + ba^2) = 0, and so a^2 b + aba + ba^2 = 0. Multiply by a, and we get a^3b + a^2ba + aba^2 = ab + (a^2b + aba)a = ab + (-ba^2)a = ab - ba = 0. Thus ab = ba. As both ideals commute separately and in products, R commutes in general. - Re-posting it from here. Note that R is not necessarily unital. Some general facts: We call a ring R, J-ring (Jacobson ring), if for any x \in R there is a natural number n(x) >1 s.t. x^{n(x)}=x. (In fact, Jacobson has proven that any J-ring is commutative, for the proof you may take a look at Non-commutative Rings written by Herestein) Lemma 1: If R be a J-ring, then N(R)= \{0 \} where N(R) denotes the nilradical of R. Proof: Let 0\not= x\in N(R). Then there is a smallest natural number greater 1 s.t. x^m=0. Since R is a J-ring, there is an n>1 s.t. x^n=x. Let m=nq+r where 0 \leq r <n. Therefore,$$x^m=x^{nq+r}=(x^n)^qx^r=x^qx^r=x^{q+r}=0$$However, q+r<m, which is a contradiction, since m was chosen to be the smallest number satisfying x^m=0. Lemma 2: Suppose that in a ring R, N(R)= \{0 \}, then any idempotent element a i.e. a^2=a, lies in the center Z(R). Proof: Suppose that x \in R. Then$$(axa-ax)^2=(axa-ax)(axa-ax)=axaaxa-axaxa-axaax+axax=axaxa-axaxa-axax+axax=0.$$Since N(R)= \{0 \}, then we have axa-ax=0 \rightarrow axa=ax. With the same approach and by considering (axa-xa)^2 we will obtain axa=xa. Hence, ax=xa and since x was an arbitrary element of R then a \in Z(R). Lemma 3: In a J-ring R, we have x^{n(x)-1} \in Z(R). Proof: (x^{n(x)-1})^2=x^{2n(x)-2}=x^{n(x)}x^{n(x)-2}=xx^{n(x)-2}=x^{n(x)-1}. Thus x^{n(x)-1} is an idempotent element of R and by Lemma 1 & 2. we get the result. In particular, in your question, n=n(x)=3 and x^2 \in Z(R), for any x \in R. Moreover$$xy=(xy)^3=xyxyxy=x(yx)^2y=(yx)^2xy=yxyx^2y=yx^3y^2=yxy^2=y^3x=yx.$$Exercise: The same question with x^4=x for any x \in R. - Here are some hints, i'll develop the answers if necessary : Let R be a ring such that x^3=x for any x in R. 1) Determine the nilpotent elements in R. 2) Show that any idempotent in R is central (i.e. for any e\in R such that e^2=e and any x\in R, we have ex=xe). Deduce from this that if x\in R, then x^2 belongs to the center of R. 3) Conclude that R must be commutative. - This is not the best proof, but let me add this one because it hasn't been mentioned yet. The proof rests on a single non-commutative polynomial identity (whose verification can be done quickly with a computer algebra system). Define f(x,y):=(x+y)^3-x^3-y^3 \in \mathbb{Z} \langle x,y \rangle. Then we have$$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 at 19:25