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

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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I suggest you to write $a(cd)=(ac)d=0d=0$. Is $R$ commutative? – Sigur Oct 23 '12 at 21:34
You should also prove that $S$ is non-empty. – Julian Kuelshammer Oct 23 '12 at 21:35
In fact, more than beinga subring, it is a right ideal. – rschwieb Oct 24 '12 at 14:37

## 2 Answers

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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So would the last step be $a(xy)=(ax)y=(0)y=0$? – Whyser Oct 23 '12 at 21:50
@Whyser: Yes, that’s exactly right. – Brian M. Scott Oct 23 '12 at 21:54

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.

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