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Math Help - Well Defined Subgroups

  1. #1
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    Post Well Defined Subgroups

    Let R = < R, +, *> (* means multiplication) be a ring, and let I < R be an additive subgroup of < R, + >.
    Consider the set of cosets

    R/I = {a + I: a is an element of R}

    equipped with its own operations + and * defined by

    (a + I) + (b + I) = (a + b) + I
    (a + I) (b + I) = ab + I

    How do we prove that the operations + and  are well-defined <---> the
    additive subgroup I satis es the following conditions:

    ab is an element of I for all a in R and b in I

    ba is an element of I for all a in R and b in I

    ?
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  2. #2
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    Quote Originally Posted by DanielThrice View Post
    Let R = < R, +, *> (* means multiplication) be a ring, and let I < R be an additive subgroup of < R, + >.
    Consider the set of cosets

    R/I = {a + I: a is an element of R}

    equipped with its own operations + and * defined by

    (a + I) + (b + I) = (a + b) + I
    (a + I) (b + I) = ab + I

    How do we prove that the operations + and are well-defined


    You have to probe that a+I=a'+I\,,\,b+I=b'+I\Longrightarrow

    \Longrightarrow (a+I)+(b+I)=(a'+I)+(b'+I)\,,\,\,(a+I)(b+I)=(a'+I)(  b'+I).

    I don't understand what you meant with the following part below

    Tonio


    <---> the
    additive subgroup I satises the following conditions:

    ab is an element of I for all a in R and b in I

    ba is an element of I for all a in R and b in I

    ?
    .
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    .
    Tonio, the question is basically asking `prove that R/I makes sense if and only if I is an ideal'.

    To the OP, Tonio has told you how to prove something is well defined. So do this. To prove the other part of the `if and only if', you need to first assume that a+I=a^{\prime}+I and b+I=b^{\prime}+I but (a+I)(b+I) \neq (a^{\prime}+I)(b^{\prime} + I).

    That is, ab+I \neq a^{\prime}b^{\prime} + I. This implies that ab-a^{\prime}b^{\prime} \not\in I.

    You want to prove that there exists some r \in R and x \in I such that xr \not\in I or rx \not\in R. To do this, use your assumption that ab-a^{\prime}b^{\prime} \not\in I, as well as that a-a^{\prime} \in I and b-b^{\prime} \in I. There is a "trick" involved here, which I will let you figure out for yourself...
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  4. #4
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    Progress

    So, proving that something is "well defined" essentially means proving that the definition makes sense?

    Here, "a+ I" represents a set of elements, the coset a is in, and a is a "representative" member of that set.

    "(a+ I)+ (b+ I)= (a+ b)+ I" really says "select a member of each coset and add them. The sum of the two cosets is the coset a+ b is in".
    What if we used different representatives?
    That is, suppose c was in coset a+ I (which we could also call c+ I) and d was in coset b+ I (so that we could also have called that set d+I). We would not expect c+ d to be the same as a+ b but we must have that c+ d and a+ b are in the same coset. If this is, then this addition is well defined right? Is it?
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  5. #5
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    Yest, it is!

    And, here, saying that a and c are in the same coset means that a= c+ x for some x in I. Saying that b and d are in the same coset means that b= d+ y for some y in I. Finally, then ab= (c+ x)(d+ y)= cd+ cx+ dy+ xy. Then ab and cd are in the same coset (and so addition is "well defined" if and onlyif cx+ dy+ xy is in I.
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