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Thread: is every ideal of a ring an annihilator of an element of some module?

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    is every ideal of a ring an annihilator of an element of some module?

    Let $\displaystyle I$ be a right ideal of a ring $\displaystyle R.$ Is there always a right $\displaystyle R-$module $\displaystyle M$ and $\displaystyle m\in M$ such that $\displaystyle I=\mathrm{ann}(m)?$
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    Re: is every ideal of a ring an annihilator of an element of some module?

    Quote Originally Posted by ymar View Post
    Let $\displaystyle I$ be a right ideal of a ring $\displaystyle R.$ Is there always a right $\displaystyle R-$module $\displaystyle M$ and $\displaystyle m\in M$ such that $\displaystyle I=\mathrm{ann}(m)?$
    yes, $\displaystyle M=R/I$ and $\displaystyle m = 1+I.$
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    MHF Contributor Drexel28's Avatar
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    Re: is every ideal of a ring an annihilator of an element of some module?

    Quote Originally Posted by ymar View Post
    Let $\displaystyle I$ be a right ideal of a ring $\displaystyle R.$ Is there always a right $\displaystyle R-$module $\displaystyle M$ and $\displaystyle m\in M$ such that $\displaystyle I=\mathrm{ann}(m)?$
    I assume that our rings are unital, no? Note then that $\displaystyle R/I$ has a natural $\displaystyle R$-module structure. If $\displaystyle x\in I$ then we see that $\displaystyle (r+I)x=rx+I=0$ since $\displaystyle rx\in I$ for all $\displaystyle r\in R$ and so $\displaystyle x\in\text{ann}(R/I)$. Conversely, if $\displaystyle x\in\text{ann}(R/I)$ then $\displaystyle 0=(1+I)x=x+I$, so that $\displaystyle x\in I$, etc.


    EDIT: Misread to try and prove that $\displaystyle \text{ann}(M)=I$, but NonCommAlg read it correctly. Same idea though.
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