# Proving this basic fact about the annihilator in abstract algebra

1. Apr 17, 2013

### jdinatale

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

2. Apr 17, 2013

### micromass

Staff Emeritus
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.

3. Apr 17, 2013

### jdinatale

Because my book defines the annihilator of X in Y as $$A=\{y\in Y~\vert~yx =0~\text{for all}~x\in X\}$$

4. Apr 17, 2013

### micromass

Staff Emeritus
And what are X and Y?

5. Apr 17, 2013

### jdinatale

"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.

6. Apr 17, 2013

### micromass

Staff Emeritus
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.