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Thread: Ideals

  1. #1
    MHF Contributor chiph588@'s Avatar
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    Ideals

    I'm new to this topic, so any help would be appriciated!

    Note: these are all in a communitive ring $\displaystyle R $.

    1. $\displaystyle a \mid b \Longleftrightarrow (b) \subseteq (a) $.

    2. $\displaystyle u \in R $ is a unit $\displaystyle \Longleftrightarrow (u) = R $.

    3. $\displaystyle a $ and $\displaystyle b $ are associative $\displaystyle \Longleftrightarrow (a) = (b) $.

    4. $\displaystyle p $ is prime $\displaystyle \Longleftrightarrow ab \in (p) $ implies that either $\displaystyle a \in (p) $ or $\displaystyle b \in (p) $.
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  2. #2
    Senior Member vincisonfire's Avatar
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    1. $\displaystyle (a) = \{ ra : r \in R \} $
    If $\displaystyle a \mid b $ then $\displaystyle b =qa $ for some $\displaystyle q \in R $ Do you see why $\displaystyle (b) \subseteq (a)
    $ ?
    2. If $\displaystyle u \in R
    $ is a unit then there exist $\displaystyle u^{-1} \in R $ such that $\displaystyle uu^{-1} = 1 $.
    $\displaystyle (u) = \{ ru : r \in R \} $ If $\displaystyle r = qu^{-1} , q \in R $ then we see that it is all the multiple of 1 and thus the ring itself.
    3. I don't know what associative is sorry.
    4. $\displaystyle (p) = \{ rp : r \in R \} $
    $\displaystyle ab \in (p) \implies ab = rp $ for some $\displaystyle r \in R $. Since p is prime, either a or b has to have p in its prime factorization. It follows that $\displaystyle

    a \in (p)
    $ or $\displaystyle

    b \in (p)
    $
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  3. #3
    MHF Contributor chiph588@'s Avatar
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    for #1.

    $\displaystyle (a) = \{ ra : r \in R \} $

    Since $\displaystyle b = qa , \; (b) = \{ (rq)a : r \in R \} $

    So my question is do we know $\displaystyle rq \in R $?
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  4. #4
    Senior Member vincisonfire's Avatar
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    $\displaystyle (a) = \{ ra : r \in R \} $ This is all the multiples of a.
    $\displaystyle a \mid b $ So b is a multiple of a.
    $\displaystyle (b) = \{ rb : r \in R \} $ Multiples of b or multiples of multiples of a. So yes we know that $\displaystyle rq \in R $. You could also simply use the fact that a ring is closed under multiplication.
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    ah yes, I forgot about closure!
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