# Problem - Rings (Herstien)

• Sep 3rd 2009, 09:18 AM
aman_cc
Problem - Rings (Herstien)
Q: $R$ is a ring with following property
1. It has a unit element, 1.
2. $(ab)^2 = a^2b^2 \forall a,b \in R$

Prove R is commutative.
(Source: Herstein - Problem#22 Pg 168 Ch 3 Ring Theory)

I did it in a very round about way. First proving bab = abb, aba=aab and then the final result ab = ba.

I essentially used
(a+1)(b+1)(a+1)(b+1) = (a+1)(a+1)(b+1)(b+1)
holds true for this ring.

Not too happy with my attempt. I think I just got the final result by a fluke. Is there any better/quick method / structured approach for this problem.
• Sep 3rd 2009, 09:59 AM
Chris L T521
Quote:

Originally Posted by aman_cc
Q: $R$ is a ring with following property
1. It has a unit element, 1.
2. $(ab)^2 = a^2b^2 \forall a,b \in R$

Prove R is commutative.
(Source: Herstein - Problem#22 Pg 168 Ch 3 Ring Theory)

I did it in a very round about way. First proving bab = abb, aba=aab and then the final result ab = ba.

I essentially used
(a+1)(b+1)(a+1)(b+1) = (a+1)(a+1)(b+1)(b+1)
holds true for this ring.

Not too happy with my attempt. I think I just got the final result by a fluke. Is there any better/quick method / structured approach for this problem.

$(ab)^2=a^2b^2\implies a(ba)b=a(ab)b$. Since $1\in R$, units exist. Thus, $a(ba)b=a(ab)b\implies a^{-1}a(ba)bb^{-1}=a^{-1}a(ab)bb^{-1}$ $\implies 1\cdot(ba)\cdot 1= 1\cdot(ab)\cdot1\implies ba=ab$.

Hence, $R$ is commutative.
• Sep 3rd 2009, 11:44 AM
Taluivren
Quote:

Originally Posted by Chris L T521
$(ab)^2=a^2b^2\implies a(ba)b=a(ab)b$. Since $1\in R$, units exist. Thus, $a(ba)b=a(ab)b\implies a^{-1}a(ba)bb^{-1}=a^{-1}a(ab)bb^{-1}$ $\implies 1\cdot(ba)\cdot 1= 1\cdot(ab)\cdot1\implies ba=ab$.

Hence, $R$ is commutative.

problem is that when a ring R has unit element 1 it doesn't mean that all elements of R are units (i.e. they all have multiplicative inverses) - consider ring $\mathbb{Z}$ for example.

i like the way how aman cc proved commutativity, it is correct (supposing he got bab = abb from identity (1+a)b(1+a)b = (1+a)(1+a)bb etc.). i don't know simpler proof.
• Sep 3rd 2009, 09:19 PM
aman_cc
yes - i don't think we can use multiplicative inverse
• Sep 3rd 2009, 09:44 PM
NonCommAlg
Quote:

Originally Posted by aman_cc
Q: $R$ is a ring with following property
1. It has a unit element, 1.
2. $(ab)^2 = a^2b^2 \forall a,b \in R$

Prove R is commutative.
(Source: Herstein - Problem#22 Pg 168 Ch 3 Ring Theory)

I did it in a very round about way. First proving bab = abb, aba=aab and then the final result ab = ba.

I essentially used
(a+1)(b+1)(a+1)(b+1) = (a+1)(a+1)(b+1)(b+1)
holds true for this ring.

Not too happy with my attempt. I think I just got the final result by a fluke. Is there any better/quick method / structured approach for this problem.

it seems that you've done some unnecessary steps! expand $((a+1)b)^2=(a+1)^2b^2$ to get $bab=ab^2.$ so $(b+1)a(b+1)=a(b+1)^2,$ which after expanding will give us $ba=ab.$
• Sep 3rd 2009, 11:59 PM
aman_cc
Absolutely NonCommAlg. Thanks all. So I was not too way off from the track.