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In a ring: $\, x^6 = x\,$ for all $x\ \Rightarrow\, x^2 = x\,$ for all $x$

I found a short and interesting problem:

Given a ring $(R, +, \cdot)$ and knowing that $x ^ 6 = x\ (\forall x\in R)$, prove that $x ^ 2 = x\ (\forall x \in R)$.

While it is short, I cannot figure out how to solve it. If it would be the reverse, then the solution were simple: $(x ^ 2) ^ 3 = x$.

Given this information, can be the problem be solved? If so, which is the simplest way to solve it?

Bill Dubuque
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Ionică Bizău
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  • You mean that this should hold for all $x$ in the ring, or just for some $x$? – Tobias Kildetoft Jun 15 '15 at 10:53
  • @TobiasKildetoft The problem says that for all. – Ionică Bizău Jun 15 '15 at 10:55
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    @Mathmo123 The condition should be "$x^6 = x$ for all $x\in R$", but the "for all" was forgotten, I think. – Daniel Fischer Jun 15 '15 at 10:59
  • @Mathmo123 The information about ring yes, but what I'm thinking at is could it be that it's referring to zero divisors only? I think that it would make it easier, however is there a possible solution without supposing that $x$ is a zero divisor? – Ionică Bizău Jun 15 '15 at 11:02
  • Well, if $x$ is not a zero-divisor, then this says that $x^5 = 1$, so these are units, and the set of units has exponent $5$. It may be relevant that it is not possible for a ring to have precisely $5$ units (though I am not sure). – Tobias Kildetoft Jun 15 '15 at 11:03
  • Is the assertion you're looking to prove known to be true ? – Gabriel Romon Jun 15 '15 at 11:04

3 Answers3

11

Consider the case where the ring has a unit (if not, then one could consider $R$ as a $\mathbb{Z}$-algebra, but the details would change in that case).

Observe $2^6=2$ and $3^6=3$. In other words, $64=2$ and $729=3$. So $62=0$ and $726=0$. Since $\gcd(62,726)=2$, it follows that $2=0$ (by repeated subtraction).

Therefore, we have a ring where $2=0$. Now, consider $(x+1)^6=(x+1)$. The LHS expands as $$ x^6+6x^5+15x^4+20x^3+15x^2+6x+1=x+1. $$ Simplifying the even coefficients, it follows that $$ x^6+x^4+x^2+1=x+1. $$ Since $x^6=x$, we know that $x^4+x^2=0$ or that $x^4=x^2$. Since $x^4=x^2$, by multiplying by $x^2$, we have $x^6=x^4$ so $x^6=x^2$, but since $x^6=x$, $x=x^2$.

(There are a few places in this calculation, where one should be careful to make sure that I'm not cheating, but the ideas should work, at least in the case where $R$ has a unit.)

Michael Burr
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  • @MichaelBurr Thanks! I understood the way to solve it! However, can you please explain what Simplifying the even coefficients means? Why did the coefficients disappear? Also, why $x ^ 4 = x ^ 2$ instead of $x ^ 4 = - x ^ 2$? – Ionică Bizău Jun 15 '15 at 12:52
  • @IonicăBizău Since $2=0$, we know that all even coefficients are $0$, for example $3x^2=x^2+2x^2=x^2$. Since $2=0$, $1=-1$, so $x^2=-x^2$. – Michael Burr Jun 15 '15 at 12:59
  • @Ionică It's clearer using the Bezout identity for gcds - see my answer. – Bill Dubuque Jun 15 '15 at 15:57
  • @BillDubuque That's also interesting. Thanks! (Y) – Ionică Bizău Jun 15 '15 at 16:57
  • @IonicăBizău I added the gcd computation to my answer. This makes it it is easy to see that equalities derived in the above answer are equvalent to applying the Euclidean algorithm. $\ \ $ – Bill Dubuque Jun 01 '25 at 06:39
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Note $\ \overbrace{2^6\!-2}^{\large\color{#0a0}j} = 0 = \overbrace{3^6\!-3}^{\large\color{#0a0} k},\,$ so by $\rm\color{darkorange}{Bezout}$ their gcd $\,\color{#d0f}2 = (\color{#0a0}{j,k}) =\:\! n\color{#0a0}j+m\color{#0a0}k = 0$.

Similarly $\ f(x) = x^6\!-x = 0 = \underbrace{f(x\!+\!1)}_{\large g(x)},\,$ so over $\,\color{#d0f}{\Bbb F_2}\!:\,$ $\,\underbrace{\color{#0af}{x^2\!-\!x} = \gcd(f,g)=0.\ \ \bf\small QED}_{\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\large \text{by $\rm\color{darkorange}{Bezout:}$}\ \ =\,\ h\:\! f\, +\, k\:\!g\, =\, 0+0}$

Note $ $ Mark's answer effectively computes the above gcd of $f(x)$ and $\, g(x)=f(x\!+\!1)$ over $\,\Bbb F_2$ (i.e. $\!\bmod 2)\,$ in an ad-hoc way. Indeed, examining it closely shows that each step corresponds to a step in the extended gcd computation below. The advantage of using the Euclidean gcd algorithm is that no ingenuity is required, and it it works far more generally, and it is very efficient.

$\!\begin{align} [\![1]\!]\:\!\ \ g = (x\!+\!1)^6\!-\!(x\!+\!1)\, &=\, \left<\,\color{#c00}1,\ \ \ \ \color{#0a0}0\,\right>\qquad \ {\rm i.e.}\qquad\ \ \ g\, =\:\, \color{#c00}1\cdot g\, +\, \color{#0a0}0\cdot f\\[.1em] [\![2]\!] \qquad\qquad\ \, f =x^6\!-x\ \, &=\, \left<\,\color{#c00}0,\ \ \ \ \color{#0a0}1\,\right>\qquad\ {\rm i.e.}\qquad\ \ \ f\ =\,\ \color{#c00}0\cdot g\, +\, \color{#0a0}1\cdot f\\[.1em] [\![3]\!]:=[\![1]\!]+[\![2]\!]\qquad\qquad x^4\!+\!x^2 &=\, \left<\,\color{#c00}1,\quad\color{#0a0}{1}\,\right>\qquad\ \:\!{\rm i.e.}\,\ \ x^4\!+\!x^2 =\, \color{#c00}1\cdot g\: +\,\color{#0c0}{1}\cdot f\\[.1em] [\![4]\!]:=[\![2]\!]+(x^2\!+\!1)[\![3]\!]\ \ \ \color{#0af}{x^2\!+\!x}\ \,&=\, \left<\,\color{#c00}{x^2\!+\!1},\, \color{#0a0}{x^2}\,\right>\ \: {\rm i. e.}\,\ \ \underbrace{\color{#0af}{x^2\!+\!x} = (\color{#c00}{x^2\!+\!1})g + \color{#0a0}{x^2} f}_{\textstyle\rm \color{darkorange}{Bezout}} \end{align}$

Note: I omitted computation of the final remainder $(= 0)$ since we don't need it here, since $\,f,g=0 \Rightarrow \color{0af}{x^2\!+\!x} = af+bg = 0$, i.e. we only need $\,(\color{#0af}{x^2\!+\!x})\subseteq (f,g),\,$ not also its reverse.

This method of deriving consequences polynomial identities by taking gcds with substituted arguments often proves useful, e.g. see the literature I cite in this answer.

Bill Dubuque
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5

See Jacobson's Theorem in the page here. If $x^6=x$ for all $x$, then it is easy to show that $x^3+x^5=0$ for all $x$. Also $-x=(-x)^6=x^6=x$, so that $x=x^6=x\cdot x^5=x\cdot x^3=x^4$. A similar step shows that $x^2=x$. Then $R$ is commutative, see here: Let R be a ring. Prove that if $x^2=x$ for each $x \in R$, then R is a commutative ring.

Dietrich Burde
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