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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.

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    $\begingroup$ 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. $\endgroup$
    – lhf
    Commented Sep 24, 2011 at 11:05
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    $\begingroup$ You can also take a look at this MathOverflow question. $\endgroup$ Commented Sep 24, 2011 at 11:29
  • $\begingroup$ I remember we solved this problem in class as a fun application of Jacobson's density theorem. Somehow this seems better (if a bit overkill) than the ad hoc calculations that give the other solutions. $\endgroup$ Commented Sep 24, 2011 at 13:57
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    $\begingroup$ 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. $\endgroup$
    – GEdgar
    Commented Jan 31, 2013 at 16:05
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    $\begingroup$ mathoverflow.net/questions/29590/… $\endgroup$ Commented Jan 31, 2013 at 16:07

7 Answers 7

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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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    $\begingroup$ If you subtract the right sides of the expanded $(x+y)^3$ and $(x-y)^3$, you'll get $2y$ on the left side and not $0$. $\endgroup$
    – Berci
    Commented Sep 2, 2020 at 12:15
  • $\begingroup$ I think I missed something, but can we just stop at $2(xy - yx) = 0$ and conclude right away that it implies $xy = yx$, thus commutativity? Also, is there any reason that the coefficient 2 can't be eliminated? Thank you. $\endgroup$
    – A.Magnus
    Commented Apr 7, 2023 at 16:06
  • $\begingroup$ @A.Magnus Yes, we can't eliminate 2 because it can happen that 2x is 0 but x is some non zero element of the ring. In fact there are non trivial rings in which x+x is 0 for all elements x $\endgroup$ Commented Jan 11 at 18:19
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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\: $ is central $\ $ [which means that $\rm\ \color{#C00}{xc = cx}\ $ for all $\rm\:x$]

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

$\rm(3)\quad x^2\:$ 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 $(3),(5)\,$ by centrals closed under subtraction.$\ $ 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,

$$x^m=x^{nq+r}=(x^n)^qx^r=x^qx^r=x^{q+r}=0$$

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 $x \in R.$

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  • $\begingroup$ Why until today nobody has realized that Lemma 2 is wrong? Counterexample $M_2(F)$, where $F$ is a field. $\endgroup$
    – freshman
    Commented Feb 23, 2020 at 22:28
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    $\begingroup$ @freshman I don’t think that’s a counterexample because $M_2(F)$ has nilpotent elements while in lemma 2 the assumption is that the nilradical is just 0. $\endgroup$
    – WLOG
    Commented May 6, 2020 at 15:13
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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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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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    $\begingroup$ "we see $2(a^2b + aba + ba^2) = 0$, and so $a^2 b + aba + ba^2 = 0$" How do we know that the ring $R$ is an integral domain? $\endgroup$
    – Cookie
    Commented Sep 5, 2016 at 0:13
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    $\begingroup$ @Cookie We don't, the reason we can this is because we are inside ideal 2R, here 3 times any element is 0. So 2x = -x $\endgroup$ Commented Jan 11 at 18:23
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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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  • $\begingroup$ I know this is very old, but by any chance you might see this, I wanted to ask how to go to the last step, I showed x^2 is in the center. From here, I tried a bunch of things and the only relevant looking thing I proved is x(xy-yx)x=yx-xy $\endgroup$ Commented Jan 13 at 20:02
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Here goes a proof that provides motivation to find a method for the more general theorem of Jacobson:

If $R$ is a ring such that for every $a\in R$ there exists $n(a)>1$ such that $a^{n(a)}=a$, then $R$ is commutative.

This method is clearly a big overkill for the case $n(a)=n=3$, since it relies heavily on the structure of rings, but as I said it has pedagogical value:

First observe that for a division ring $D$ the conclusion is rather easy, since the commutativity of $D$ arises from the reversion of the order of the product when we apply inverses:

$$a^3=a\Rightarrow a^2=1\Rightarrow a^{-1}=a \ \ \ (a\neq 0)$$ $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba \ \ \ (a,b\neq 0)$$

Of course, trivially $0\cdot a=a\cdot 0$ for every $a\in R$.

So a natural question is: can we reduce our problem to that of division rings? In other words: can our ring be written from a family of division rings in such a way that

1) the division rings satisfy $a^3=a$, and

2) our ring inherits commutativy from them?

A reasonable answer is that 1) would be satisfied if $R$ is epimorphic on every division ring $D_i$, while 2) would be satisfied if $R$ is injected in the sum or product of the $D_i$.

Hence we see that we if $R$ is a subdirect product of division rings then we can get our desired scheme of proof.

Now the following reasoning requires some experience: as we know, primitive rings are strongly related to division rings (via Schur's lemma, the Jacobson density theorem, etc.). Can we at least prove that our ring is a subdirect product of primitive ones?

In that case, $R$ would be semiprimitive, and we know that conversely every semiprimitive ring is a subdirect product of primitive ones. So, is $R$ semiprimitive?

We need to show that the Jacobson radical $J(R)$ is $0$. We resort to the characterization of $J(R)$ by elements. If $a\in J(R)$ then $a^2\in J(R)$ and $1-a^2$ is invertible, but we know $a^3=a\Rightarrow a(1-a^2)=0$, so $a=0$.

Therefore $R$ is a subdirect product of primitive rings $R_i$. Does the inherited identity $a^3=a$ force $R_i$ to be a division ring?

What can be said in special cases? If $R_i$ is artinian, then $R_i=$Mat$_n(D)$ for some $n\in\mathbb{N}$ and some division ring $D$. If $n>1$ then there are nonzero nilpotent elements in $R_i$, that cannot satisfy $a^3=a$. Therefore $n=1$ and $R_i=D$.

What if $R_i$ is not artinian? Then by a consequence of the Density theorem (Litoff's theorem for primitive rings) we have there exists a division ring $D$ such that for every $n\in\mathbb{N}$ there is an epimorphism from $R_i$ to Mat$_n(D)$. So the same argument as before applies: $a^3=a$ in $R_i$ would imply $a^3=a$ in Mat$_n(D)$ for every $n$, but that is impossible, so $R_i$ is artinian and therefore a division ring.

We have arrived at the end of our proof. Now, if we want to prove Jacobson's theorem in full generality, it is natural to try the same scheme of proof (due to its degree of abstraction) and indeed it works, the only difficulty being in showing that $a^{n(a)}=a$ forces a division ring to be a field. This is how Jacobson's commutativity theorem is proved by Herstein in chapter 3 of Noncommutative rings (second edition, 1971).

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