Let $S$ be a multiplicatively closed subset of a ring $R$ (not assuming commutativity or identity). Suppose that $R$ has right Ore condition with respect to $S$ show that
$T = \{a \in R : as = 0 \ for\ some \ s \in S (\text{s depending on a)}\} $
is an ideal of R.
I'm having difficulty seeing how to answer this. Closure under left multiplication is obvious but I am not seeing how to use the right Ore condition. If anyone could give me a starting point or an answer that would be great! Thanks in advance.
The right Ore condition says that for all $s_1 \in S, r_1 \in R$ there exist $s_2, r_2$ such that $r_1s_2=s_1r_2$. Let $t \in T, r \in R$. Then there exists an $s \in S$ such that $ts=0$. Now there exists an $\bar s \in S$ such that $ts=\bar sr$, so $\bar s r=0$. Now there exists again an $\hat s \in S$ such that $0=\bar s r= r \hat s $. We have therefore found an $\hat s$ with $tr\hat s =0$ , so $tr \in T$ for any $r \in R$, so the closure under right multiplication is proved.