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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

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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
  • @Ionică It makes this problem trivial, reducing it to computing two gcds by the Euclidean algorithm (for integers and polynomials). Similar ideas work more generally, see the links in this answer. – Bill Dubuque Jun 15 '15 at 16:59
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See Theorem $2.6$ in the lecture notes 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 $x^2=x$ and that $R$ is commutative.

Dietrich Burde
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Note $\,\ \overbrace{2^6\!-2}^{\large\color{#0a0}j} = 0 = \overbrace{3^6\!-3}^{\large\color{#0a0} k},\,$ so by Bezout their gcd $\,\color{#c00}2 = (\color{#0a0}{j,k}) = n\color{#0a0}j\!+\!m\color{#0a0}k = 0$.

Similarly $\ f(x) = x^6\!-x = 0 = f(x\!+\!1),\,$ so over $\,\color{#c00}{\Bbb F_2}\,$ their gcd $= x^2-x = 0.\ \ \bf\small QED$

Bill Dubuque
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