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Ring Theory: Equivalence Relations

  1. Jan 25, 2004 #1

    I have a question regarding equivalence relations from my ring theory course.


    Which of the following are equivalence relations?

    e) "is a subset of" (note that this is not a proper subset) for the set of sets S = {A,B,C...}.

    Example: A "is a subset of" B.

    Now I know that for a binary relation to be an equivalence relation the relation must be symmetric, reflexive, and transitive.

    I would initially say that e) would be an equivalence relation since the following:

    e) This is not a proper subset so assume that each of the sets in S is a set equal to S. This should mean that the relation would be symmetric, reflexive, and transitive.

    But since the possibility of the sets in S being proper subsets of S exist then the relation must have the following restriction: A R B and B R A (where R is the relation and A,B are elements of S) iff A = B. Would this mean R is anti-symmetric?

    I would then guess that unless the relation is unconditionally reflexive, transitive, and symmetric then the relation could not be an equivalence relation. Would this be a correct assessment?

    Any input is appreciated. Thankyou.
  2. jcsd
  3. Jan 25, 2004 #2


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    Right; this is not an equivalence relation.
  4. Jan 25, 2004 #3
    Yes. Thankyou.

    Actually I know that it is not an equivalence relation, but I was wondering if my reasoning was correct. I should have been more specific. Sorry.
  5. Jan 25, 2004 #4


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    Yes; because it fails to be symmetric, it cannot be an equivalence relation.
  6. Jan 25, 2004 #5
    Thanks again Hurkyl.
  7. Jan 26, 2004 #6
    does every relation having two properties of equivalence relations but not all three have a minimal relation that contains it as well as being an equivalence relation?

    in particular, i know that if one adds the condition "or is equal to" to the relation "is a proper subset of", it becomes an equivalence relation equal to "is a subset of".
  8. Jan 26, 2004 #7

    matt grime

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    'Is a subset of' is not an equivalence relation for the same reason as 'is a proper subset of'

    Here's an example of making an equivalence by fiat. (genuine maths)

    Let S be the set of (finite dimensional) simple modules for a group algebra. define a relation on modules by M~N if there is some indecomposable module that has both M and N in its composition series. This is symmetric and reflexive, but not transitive. So just make it so by fiat. The equivalence classes are the blocks of the algebra.
  9. Jan 26, 2004 #8
    oops. no, it's not symmetric, now is it.

  10. Jan 26, 2004 #9


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    Unless I misunderstand what you're suggesting, the description you're giving is somewhat preposterous, since what it's really doing is defining ~ so that M~N if there is a path from M~N in a graph generated by assigning each module to a node, and placing edges betwen pairs of modules that share a composition series.

    "So just make it by fiat" is actually a change to the definition of '~' which results from globbing the modules together.
  11. Jan 26, 2004 #10


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    I'm not sure what you mean. There is always the trivial relation: a~b is true for all a and b, and the other trival relation, a~b iff a = b. Obviously both are equivalance relations.

    Consider that there are three properties:

    Transitive, reflexive and symetric:

    reflexive and symetric:
    a~b iff a and b have a common factor > 1.

    So transitive and symetric:
    a~b iff a and b are both not 0.
    Any transitive and symetric relation can be extended to an equivalance by adding that a~a if a=a.

    transitive and reflexive:
    a~b iff a ≤ b
  12. Jan 27, 2004 #11

    matt grime

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    It certainly isn't preposterous, and is one of the many equivalent definitions of blocks of algebras.

    It's a standard, simple way to make a symmetric reflexive relation transitive.

    Oh, perhaps you are not noticing the strong constraint that there is an *indecomposable* module with them as composition factors. You might perhaps need to know more about group algebras and projective modules, but this defintitely isn't 'globbing the modules together' merely some of them: those that lie in the same block; correspond to the same indecomposable idempotents of the identity functor; do something old fashioned with Brauer Characters.
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