Proving this basic fact about the annihilator in abstract algebra

jdinatale
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Maybe I'm misinterpreting the question, I'm not sure how to prove that n_0 i = 0.

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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

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

and so on.
 
micromass said:
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

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

and so on.

Because my book defines the annihilator of X in Y as A=\{y\in Y~\vert~yx =0~\text{for all}~x\in X\}
 
And what are X and Y?
 
micromass said:
And what are X and Y?

"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.
 
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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