Is a Ring with the Property $r^3 = r$ Always Commutative?

  • MHB
  • Thread starter Euge
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    2016
In summary, the property $r^3 = r$ in a ring means that for any element $r$, raising it to the third power results in the same value. However, this does not always guarantee commutativity, only if $r$ is equal to its own inverse. A ring can have this property without being commutative, as the idempotent property only guarantees commutativity under specific conditions. An example of a non-commutative ring with this property is the ring of 2x2 matrices with real entries, where $r$ is the identity matrix. Another property that guarantees commutativity is $r^2 = e$, where $e$ is the multiplicative identity element, known as the inv
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Euge
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The following problem is a nice exercise that both the undergrad and grad members here can enjoy:

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Let $R$ be a ring such that for all $r\in R$, $r^3 = r$. Prove $R$ is commutative.

-----Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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Surprisingly, no one answered this week's problem! You can read my solution below.
First, I'll show that the squares in $R$ commute with every element of $R$. Let $r,s\in R$. Let $t = r^2$. Then $t^2 = r^4 = r^3r = rr = t$, and so $$(st - tst)^2 = stst - st^2st - tstst + tst^2st = stst - stst - tstst + tstst = 0$$ and $$(ts - tst)^2 = tsts - tstst - tst^2s + tst^2st = tsts - tstst - tsts + tstst = 0$$ Therefore, $0 = (st - tst)^3 = st - tst$ and $0 = (ts - tst)^3 = ts - tst$, whence $st = tst = ts$.

Now given $a,b\in R$, we have
$$ab = a^3b = a^2(ab) = aba^2 = ab^3a^2 = (ab)b^2a^2 = b^2aba^2 = b(ba)^2a = (ba)^2ba = (ba)^3 = ba$$
Hence, $R$ is commutative.
 

FAQ: Is a Ring with the Property $r^3 = r$ Always Commutative?

What does the property $r^3 = r$ mean for a ring?

The property $r^3 = r$ means that for any element $r$ in the ring, the result of raising $r$ to the third power is equal to $r$ itself.

Is a ring with the property $r^3 = r$ always commutative?

No, a ring with the property $r^3 = r$ is not always commutative. This property, known as the idempotent property, only guarantees that the ring is commutative if $r$ is equal to its own inverse.

Can a ring have the property $r^3 = r$ without being commutative?

Yes, a ring can have the property $r^3 = r$ without being commutative. This is because the idempotent property only guarantees commutativity under specific conditions, as mentioned in the previous answer.

What is an example of a ring with the property $r^3 = r$ that is not commutative?

An example of a ring with the property $r^3 = r$ that is not commutative is the ring of 2x2 matrices with real entries, where $r$ is the identity matrix.

Are there any other properties of a ring that guarantee commutativity?

Yes, there are other properties of a ring that guarantee commutativity, such as the property $r^2 = e$, where $e$ is the multiplicative identity element. This property is known as the involution property and is stronger than the idempotent property.

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