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Let $R$ be a commutative ring and let $a\in R$.

Show that $I=\{x\in R\mid ax=0\}$ is an ideal.

For all $b \in R$, $$bI=b\{x\in R\mid ax=0\}=\{bx\in R\mid a(bx)=0\} =\{xb\in R\mid a(x)=0\}=Ib\;,$$ thus I can say $I$ is an ideal of $R$, right? Thanks

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    To be an ideal, $I$ doesn't need to commute with elements of $R$. It has to satisfy the property $RI\subseteq I$.2012-02-01
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    Alternatively, what is the kernel of the homomorphism $\varphi : R\to R : x \mapsto ax$?2012-02-01
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    Just a note: this is the [_annihilator_](http://en.wikipedia.org/wiki/Annihilator_(ring_theory)) of $a$.2012-02-01

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