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Proving this basic fact about the annihilator in abstract algebra

  1. Apr 17, 2013 #1
    Maybe I'm misinterpreting the question, I'm not sure how to prove that n_0 i = 0.

    aaa_zps530385fc.png
     
  2. jcsd
  3. Apr 17, 2013 #2

    micromass

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    I don't get why you multiply both on the left and on the right. I would think that all modules here are left R-modules. So you should always multiply with R on the left. In particular, we have

    [tex]A=\{m\in M~\vert~im=0~\text{for all}~i\in R\}[/tex]

    and so on.
     
  4. Apr 17, 2013 #3
    Because my book defines the annihilator of X in Y as [tex]A=\{y\in Y~\vert~yx =0~\text{for all}~x\in X\}[/tex]
     
  5. Apr 17, 2013 #4

    micromass

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    And what are X and Y?
     
  6. Apr 17, 2013 #5
    "If X is a submodule of M, the annihilator of X in Y is defined to be..."

    Here X is a submodule and Y is the ring, the "R" in the R-module.
     
  7. Apr 17, 2013 #6

    micromass

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    OK, so if you say "the annihalator of I in M", then how does this fit this definition??

    In your definition, you have "the annihalator of [some submodule] in [ring]". But if you write "the annihalator of I in M", then I see "the annihaltor of [some ideal] in [module]". Of course an ideal is a module too, but the problem remains that this doesn't fit the definition. So I think there should have been another definition.
     
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