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Math Help - Ideals and Containment.

  1. #1
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    Ideals and Containment.

    Let R be a ring with a proper left ideal M such that, for any r \in R, either r \in M or r is a unit.

    (a) Prove that if L is a left ideal which contains M then either L = R or L = M.

    (b) Prove that every proper left ideal of R is contained in M.


    For (a) we have M \subseteq L so we need to show that we can take an arbitrary x \in L and rearrange it so it looks like an element of M. So if we take x \in L then x = rl for some r \in R and l \in L (because L is a left ideal). I'm not too sure how to proceed from here... how can we relate L with M?

    Thanks,
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  2. #2
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    Re: Ideals and Containment.

    If L = M, then we're done. So, assume L properly contains M. That means there's an element x that's in L but not in M.
    From the given, what can you say about x?
    What can you say about about an ideal that contains a unit?
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  3. #3
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    Re: Ideals and Containment.

    Quote Originally Posted by Bingk View Post
    If L = M, then we're done. So, assume L properly contains M. That means there's an element x that's in L but not in M.
    From the given, what can you say about x?
    What can you say about about an ideal that contains a unit?

    Suppose  x \in L then that means x = rl for some r \in R and l \in L.

    If r \in M then rl \in L but doesn't that just show that M \subseteq L?

    I'm not sure what I can infer beyond that...


    Now suppose r is a unit in R. Then r^{-1} \in R also since R is a ring.

    Now rL \in L since L is a left ideal. Therefore r.r^{-1} = 1 \in L.

    But for any r \in R, since we have 1 \in L, then r = r.1 \in L.

    And, since r was an arbitrary element in R, then R = L. Was that sound reasoning?
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  4. #4
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    Re: Ideals and Containment.

    From the given, r is in M or r is a unit, but not both, right?

    So, from where I left off, that means x is a unit, since it's not in M.

    After, that yes, your reasoning is sound ... except, I think rL \in L should be rR \subset L (with some re-wording).

    Basic idea:
    x is a unit so it has a multiplicative inverse (denoted 1/x, which is in R).
    Since x is in L and L is an ideal, then xr is in L for all r in R.
    That means x*(1/x) = 1 is in L (since 1/x is in R).
    So, 1r = r is in L for all r in R.
    Which gives L = R.
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