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Multiplicative closure for subring test?

  1. Dec 7, 2011 #1
    Everything about the subring test is straightforward from the subgroup test, but the multiplicative operation of the subring, S, of ring, R, needs to be closed wrt multiplication, *. How do you prove S is closed wrt * if the only assumption about * is associativity and distributivity over addition in R? Please let me know if I need to clarify.

    EDIT: I found the answer. Apparently * is assumed to be closed in a ring. My algebra book made no mention of this and it seems many sources don't! The ring axioms on proofwiki.org assume * is closed in a ring, thus a subring, making my question easy to answer.
     
    Last edited: Dec 7, 2011
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  3. Dec 7, 2011 #2

    mathwonk

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    ask a complete well formulated question if you want an answer, i.e. yes.
     
  4. Dec 7, 2011 #3
    Assume (R,+,*) is a ring.
    Assume (S,+,*) is a subring of R.
    Prove that if a,b [itex]\in[/itex] S, then a*b [itex]\in[/itex] S.

    This is difficult because the properties of rings, thus subrings, are that (R,+) and (S,+) are an Abelian group and Abelian subgroup, respectively. Also, * distributes over + in R, thus S. However, * on R is associative, thus * on S is associative, but * is not necessarily closed wrt S, though it is wrt R.

    Does that help?

    EDIT: I found the answer. Apparently * is assumed to be closed in a ring. My algebra book made no mention of this and it seems many sources don't! The ring axioms on proofwiki.org assume * is closed in a ring, thus a subring, making my question easy to answer.
     
    Last edited: Dec 7, 2011
  5. Dec 7, 2011 #4

    mathwonk

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    the words " S is a ring" implies your problem as you seem to realize.

    have you noticed that once you stated your problem clearly you also solved it?
     
  6. Dec 9, 2011 #5
    When the phrase "binary operation on a set" is used, closure is always implicit.
     
  7. Dec 9, 2011 #6
    First, the sources I've seen don't explicitly state 'binary' operation in their ring axioms. Second, the implicit closure should come into question when subgroups and subrings are being considered, for it is their closure with respect to binary operations within a subset of a group/ring that needs to be verified.
     
  8. Dec 9, 2011 #7
    I'll extend my previous comment: binary operations are implicitly closed, and it is not uncommon to use the term "operation" in place of "binary operation."

    And you're absolutely correct: operations are closed if you have a ring, but if "I have a (sub)ring (of another ring)" is what you are trying to prove, then you certainly can't assume closure.

    When proving a subring relation, however, you only need to show that the operations are closed, that the set contains all its negatives, and that it is nonempty. Distributivity, associativity, commutativity, etc all come for free if you have a subset, and if it's closed, nonempty, and contains negatives, then it's easy to show that it contains the zero. (a is one of the elements, since it is nonempty. The set then also contains -a, and so it contains a+(-a)=0.)
     
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