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Cosets in rings?

  1. Mar 4, 2007 #1
    Does cosets exist in rings?

    i.e R = Ring, a in R

    set {a*R}

    or

    set {a+R}


    The above two sets looks very similar to cosets in groups but there are two operations in rings so potentially two different cosets both involving the same ring R and element a. If the above two sets are not cosets than what are they called?
     
  2. jcsd
  3. Mar 4, 2007 #2

    matt grime

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    The former is the right ideal generated by R, the latter is the right coset of a in R (and since addition is commutative, the left coset).

    I is a right ideal if Ix is in x for all x in R. Clearly aR is a right ideal. Similarly Ra is a left ideal, and RaR is an ideal (that is a two sided ideal).
     
  4. Mar 4, 2007 #3
    In this case aR is also a left ideal? Because x(aR) with x and a in R means xa is in R. (xa)R is a subset of R.

    Similarly Ra is also a right ideal.
     
  5. Mar 4, 2007 #4

    StatusX

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    No, because (xa)R is not necessarily a subset of aR.
     
  6. Mar 4, 2007 #5
    Good point. In fact x(Ra)=Ra, for x and a in R which is the ring.
     
  7. Mar 4, 2007 #6

    Hurkyl

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    Also, a + R is a rather boring coset, since a + R = R. a + I is more interesting, for ideals I.
     
  8. Mar 4, 2007 #7

    matt grime

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    No. x(Ra) is a subset of Ra. It is most definitely not equal to it in general (x=0, for instance).
     
  9. Mar 4, 2007 #8

    matt grime

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    Doh. Can't believe I missed this. Elements don't have cosets, sugroups/ideals, have cosets
     
  10. Mar 4, 2007 #9
    Ra gives R. xR gives R again. so x(Ra)=xR=R=Ra. In this way x(Ra)=Ra. I agree that it is quite boring but at least it's correct? You think not Matt? R is the ring and a and x are elements in R. All elements in R are closed under multiplication.
     
    Last edited: Mar 4, 2007
  11. Mar 4, 2007 #10

    StatusX

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    Closure under multiplication means xR is a subset of R. To get the other inclusion, you need the existence of inverses. Thus x+R=R, or xR=R when x is a unit, but in general this isn't true. If the example x=0 isn't enough to convince yourself that xR is not always R, you need to go back to the definitions.
     
  12. Mar 6, 2007 #11
    I should have said x must be non zero.

    If x is a unit then x cannot be a zero divisor. This latter point is the important thing because if x was a zero divisor than some non zero entities in R may not be produced from xR. If x is not a zero divisor and is not a unit than xR=R if R does not contain the multiplicative identity, 1?

    x+R=R will work all the time with x in R wouldn't it? Without any limitations on x?
     
    Last edited: Mar 6, 2007
  13. Mar 6, 2007 #12

    matt grime

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    zero or not zero, I don't see what this has to do with anything. Rx is still an ideal

    No, clearly this is false. (Incidentally, if R doesn't have a 1 then it can't have units). If you just remembered that the integers are a ring you'd have a lot more insight into the general situation; what is 2Z? It is ring without 1, call it R. What about the ideal 4Z inside 2Z, it is the ideal 2R.
     
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