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Let $a$ belong to a ring R. Let $S=\{x \in R | ax=0\}$. Show that S is a subring of R.

My approach is such:

Let $c,d \in S$ so $(c-d)x=cx-dx=0-0=0 and (cd)x=(cx)d=(0)d=0$. Therefore by the subring test, S is a subring of R. Q.E.D

Is this correct or not so much?

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    In f$a$ct, more than be$i$nga subring, it is a right ideal.2012-10-24

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You didn’t look carefully enough at the definition of $S$: $a$ is a fixed element of $R$, and to show that some $r\in R$ belongs to $S$, you must show that $ar=0$.

Suppose that $x,y\in S$. Then $a(x-y)=ax-ay=0-0=0$, so $x-y\in S$. Can you finish up by showing correctly that if $x,y\in S$, then $xy\in S$? (And remember, you can’t assume that $R$ is commutative, as you did in the step $(cd)x=c(xd)$ in your first attempt.)

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    @Whyser: Yes, that’s exactly right.2012-10-23
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frist,we show S is nonempty subset of R. since 0 belongs S therefore S is non empty. now, let a,b belongs to S ,then (a-b)x=ax-bx=0 and (ab)x=ax*bx=0 [ ¥ x € R] hence (a-b) and (ab) belongs to S then by subring test S is a subring of R.