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Math Help - The unity ..

  1. #1
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    The unity ..

    Hi all ,
    i want to ask you about the unity .. what do we mean about " this ring has unity and that ring without unity.."

    and how i can find the all units in " ring whit unity" for example Z10


    im wait for your answer

    sorry for my bad English language..
    with my respect to you..
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Miral View Post
    Hi all ,
    i want to ask you about the unity .. what do we mean about " this ring has unity and that ring without unity.."

    and how i can find the all units in " ring whit unity" for example Z10


    im wait for your answer

    sorry for my bad English language..
    with my respect to you..
    You of course know that every ring R has two operations, one multiplicative one additive. Since R under addition must be an abelian group there must be some additive identity 0\in R such that 0+a=a+0=a for all a\in R. There need not be a multiplicative identity 1\in R such that a\cdot 1=1\cdot a=a.

    If it has this multiplicative identity it is said to have unity, if it doesn't it is said to be a ring without unity.

    The units of \mathbb{Z}_{10} from my limited experience actually refers to those a\in\mathbb{Z}_{10} such that (a,10)=1
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  3. #3
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    As Drexel 28 said, the "unity" in a ring is the multiplicative identity. A "unit", on the other hand, is a member of the ring that has a multiplicative inverse (which, of course, is only possible in a ring with unity).
    In Z_{10}, the "unity" is 1 and numbers, n, for which (n, 10) (the least common multiple) is not 1 do not have inverses. Since 10= 2(5), the only possible "least common multiples" are 1, 2, 5, and 10. If (n, 10)= 2, then n= 2a so 5(n)= 5(2a)= 10a= 0 (mod 10). Similarly, if (n, 10)= 5, then n= 5a so 2(n)= 2(5a)= 10a= 0 (mod 10). Of course if (n, 10)= 10, n is itself equivalent to 0 (mod 10).

    If mn= 0 with m and n both non-zero, in any ring, then n and m cannot have inverses- if n had an inverse, n^{-1} then we would have (mn)n^{-1}= m= 0(n^{-1}= 0, a contradiction.
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  4. #4
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    Quote Originally Posted by Drexel28 View Post
    You of course know that every ring R has two operations, one multiplicative one additive. Since R under addition must be an abelian group there must be some additive identity 0\in R such that 0+a=a+0=a for all a\in R. There need not be a multiplicative identity 1\in R such that a\cdot  1=1\cdot a=a.

    If it has this multiplicative identity it is said to have unity, if it doesn't it is said to be a ring without unity.

    The units of \mathbb{Z}_{10} from my limited experience actually refers to those a\in\mathbb{Z}_{10} such that (a,10)=1



    Quote Originally Posted by HallsofIvy View Post
    As Drexel 28 said, the "unity" in a ring is the multiplicative identity. A "unit", on the other hand, is a member of the ring that has a multiplicative inverse (which, of course, is only possible in a ring with unity).
    In Z_{10}, the "unity" is 1 and numbers, n, for which (n, 10) (the least common multiple) is not 1 do not have inverses. Since 10= 2(5), the only possible "least common multiples" are 1, 2, 5, and 10. If (n, 10)= 2, then n= 2a so 5(n)= 5(2a)= 10a= 0 (mod 10). Similarly, if (n, 10)= 5, then n= 5a so 2(n)= 2(5a)= 10a= 0 (mod 10). Of course if (n, 10)= 10, n is itself equivalent to 0 (mod 10).

    If mn= 0 with m and n both non-zero, in any ring, then n and m cannot have inverses- if n had an inverse, n^{-1} then we would have (mn)n^{-1}= m= 0(n^{-1}= 0, a contradiction.





    thank you Drexel28 and HallsofIvy

    from your answers i have:
    the multiplicative identity in the ring = The unity

    the units in {Z}_{10} is 1, 3 , 7 , 9
    since {Z}_{10} = {1,2,3,4,5,6,7,8,9}
    and gcd(1,10)=1 & gcd(3,10)=1 & gcd(7,10)=1 & gcd(9,10)=1

    and i have u* u^{-1}= u^{-1}* u = unity

    and since the unity in {Z}_{10} =1

    Thus

    u=1 ----> u^{-1}= 1
    u=3 ----> u^{-1}= 7
    u=7 ----> u^{-1}= 3
    u=9 ----> u^{-1}= 9

    right?

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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Miral View Post

    u=1 ----> u^{-1}= 1
    u=3 ----> u^{-1}= 7
    u=7 ----> u^{-1}= 3
    u=9 ----> u^{-1}= 9

    right?
    I would agree with that. Notice that in terms of groups the units of \mathbb{Z}_{n} form a group under modular multiplication. This can be seen from above, which is quite a nice Cayley table .

    I'm fairly sure that this extends to the general case, but not being a ring theorist, and not (at the moment) being too interested I haven't checked. In particular, if \left(R,+,\cdot\right) is a ring, then \left(U(R),\cdot\right) where U(R) are the units of R form a group that is not necessarily abelian.
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    Quote Originally Posted by Drexel28 View Post
    I would agree with that. Notice that in terms of groups the units of \mathbb{Z}_{n} form a group under modular multiplication. This can be seen from above, which is quite a nice Cayley table .


    Thanks


    I'm fairly sure that this extends to the general case, but not being a ring theorist, and not (at the moment) being too interested I haven't checked. In particular, if \left(R,+,\cdot\right) is a ring, then \left(U(R),\cdot\right) where U(R) are the units of R form a group that is not necessarily abelian.


    ok i got it ^^"

    thank you again ^_*
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