Rings

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  1. JBKProductions's Avatar
    1 13-04-2012  17:01
    • Overlord in Training
    • Posts: 2,106
    Rings
    Let R be an integral domain and let x, y \in R Show that (x) = (y) if and only if \exists \lambda \in R^* such that x = \lambda y
    Where (x) = \{rx: r \in R \}, R^* = \{x: x \ is \ a \ unit \} .
    Proving from left to right.
    In part of the answer it says x \in (x) = (y) \Rightarrow x = \lambda y for some \lambda \in R and y \in (y) = (x) \Rightarrow y = \mu x for some \mu \in R . Thus x = \lambda y = \lambda \mu x \Rightarrow x-\lambda \mu x = 0_R \Rightarrow x(1_R - \lambda \mu) = 0_R \Rightarrow x=0_R or (1_R - \lambda \mu) = 0_R .
    If x = 0_R \Rightarrow y = \mu 0_R = 0_R \Rightarrow x = 1_R y . I don't understand the last implication, how does it imply \lambda = 1_R from the case when x= 0_R ?
  2. nuodai's Avatar
    2 13-04-2012  17:05
    • PS Helper
    • TSR Legend
    Re: Rings
    If x=0 then it implies that y=0, and so x=1.y. In fact, for any \lambda it'd be true that x=\lambda y in this case; they just gave a concrete example of a value of \lambda that works.
  3. JBKProductions's Avatar
    3 13-04-2012  17:08
    • Overlord in Training
    • Posts: 2,106
    Re: Rings
    Oh I see, thanks.
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