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Math Help - Problem - Rings (Herstien)

  1. #1
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    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.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by aman_cc View Post
    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.
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    (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.
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  4. #4
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    yes - i don't think we can use multiplicative inverse
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  5. #5
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    Quote Originally Posted by aman_cc View Post
    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.
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  6. #6
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    Absolutely NonCommAlg. Thanks all. So I was not too way off from the track.
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