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I Thinking about the Definition of a Unit of a ring R ... ...

  1. Feb 12, 2017 #1
    I have been thinking around the definition of a unit in a ring and trying to fully understand why the definition is the way it is ... ...

    Marlow Anderson and Todd Feil, in their book "A First Course in Abstract Algebra: Rings, Groups and Fields (Second Edition), introduce units in a ring with 1 in the following way ... ...


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    So ... an element ##a## of a ring ##R## with ##1## is a unit if there is an element ##b## of ##R## such that
    ##ab = ba = 1## ... ...


    So ... if, in the case where ##R## was noncommutative, ##ab = 1## and ##ba \neq 1## then ##a## would not be a unit ... is that right?

    Presumably it is not 'useful' to describe ##a## as a 'left unit' in such a case ... that is, presumably one-sided units are not worth defining ... is that right?



    Could someone please comment on and perhaps clarify/correct the above ...


    Hope someone can help ...

    Peter
     

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  3. Feb 12, 2017 #2

    fresh_42

    Staff: Mentor

    I'm not aware of an example and still try to find an argument, why this should be impossible. At least I've found some strange implications.

    Let us assume ##ab=1## and ##ba \neq 1##.

    Then ##a(ba-1)=0## by associativity and you don't want to have a unit being a zero divisor.
    In addition, ##a(ba+b-1)=1=ab## and ##ba+b-1 \neq b## and you don't want to have two different right inverses either.

    So although I haven't found a direct contradiction (yet!?), the resulting ring has some really weird properties, at least ##a \in R## has.
     
  4. Feb 12, 2017 #3
    Thanks for the post fresh_42 ...

    But I need some help to see why a(ba-1) = 0 ...

    Can you help ...

    Peter
     
  5. Feb 12, 2017 #4

    Stephen Tashi

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    Science Advisor

    Yes.

    The term "left inverse" is in use. Perhaps that's why we don't hear about "left units".
     
  6. Feb 12, 2017 #5

    lavinia

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    I think there are many rings with one sided units.
     
  7. Feb 12, 2017 #6

    pwsnafu

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    Consider the ring of all operators ##C^\infty \to C^\infty##. What is the inverse of the derivative?
     
  8. Feb 13, 2017 #7

    fresh_42

    Staff: Mentor

    ##a\cdot (ba-1)=a\cdot (b\cdot a)-a\cdot 1=(a\cdot b) \cdot a - a= 1 \cdot a - a = 0## so ##a## is a unit (##ab=1##) and a zero divisor ##a\cdot (ba-1)=0##.

    ##a(ba+b-1)=a\cdot (ba-1)+a\cdot b=0+1=1=ab## by the above and the definition of ##b##. So ##(ba+b-1)## and ##b## are both right inverse to ##a##. But if they were equal, that is ##ba+b-1=b## then ##ba-1=0## or ##ba=1## which we excluded. Thus they are actually two different right inverses.
     
  9. Feb 14, 2017 #8

    lavinia

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    A classic example occurs in the ring of linear maps on the vector space of infinite tuples of numbers in some field.

    The linear map ##R## that shifts the sequence once to the right and puts zero at the beginning of the sequence is inverted by the linear map ##L## that shifts the sequence once to the left and drops off the first number in the sequence. But ##L## is not invertible since it is not injective.

    - I will look for some examples that do not involve infinite dimensional space such as this one or the example given above of the derivative operator. I think there are examples for some group rings over certain algebraic extensions of the rationals.
     
    Last edited: Feb 16, 2017
  10. Feb 15, 2017 #9
    Thanks to all who have posted ... it has been most helpful ... I am really grateful for the help ...

    Peter
     
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