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Let $R$ be a ring (without assuming identity or commutativity), and $P$ a proper ideal of $R$. Show that the following are equivalent:

(a) For ideals $A,B$: $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$.

(b) For right ideals $T,S$: $TS\subseteq P$ implies $T\subseteq P$ or $S\subseteq P$.

(c) For elements $a,b\in R$: $aRb\in P$ implies $a\in P$ or $b\in P$.

(b) $\implies $ (a) is trivial

(c) $\implies$ (b) is easy: Assume (c) and let $T,S$ be as in (b), with $T\not\subseteq P$. Fix $r\in T\backslash P$ and let $s\in S$. Then $rRs\in P$ since $T$ is a right ideal and by assumption $s\in P$.

I haven't been able to prove (a) $\implies $ (c) since my first idea was to realise $a,b$ as two sided ideals, however, the ideal generated by $a$ in a ring without unity is $\langle a\rangle =RaR + aR+Ra+\mathbb Z a$, which is a bit unwieldly. I am not convinced that multiplying the ideals $\langle a\rangle\cdot \langle b\rangle$ is assured to fall inside $P$ despite having $aRb\in P$. I am mostly concerned about the $a\cdot b$ terms, since there is no identity, it's not in $aRb$. Also the fact that each ideal $\langle a\rangle$ could well be the whole ring.

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  • $\begingroup$ There's an ambiguity in the use of $R$ in (b). $\endgroup$ – Hagen von Eitzen Apr 16 at 13:23
  • $\begingroup$ The question I am about to ask is probably on other people's minds: What's the source? Are you sure the question cares about not having identity? I have read a lot of ring theory, and I don't recall a single resource that bothered to ask specifically about prime ideals in noncommutative rings without identity. $\endgroup$ – rschwieb Apr 16 at 13:24
  • $\begingroup$ Fixed the ambiguity, thank you @HagenvonEitzen $\endgroup$ – George Apr 16 at 13:25
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    $\begingroup$ @George Well, your first thought was certainly the most natural. Even though it's "unwieldy" it seems instructive to write it down. We would like for $(a)(b)\subseteq P$ to apply the assumption from $a)$ but the problem is after multiplying, there are parts of the product that one can't be sure are in $P$, namely $RabR+Rab+abR+ab\mathbb Z$. It feels like everything would be well if $aRb$ were replaced with $aRb+ab\mathbb Z$ in c), but as it stands, it is hard to see how just $aRb$ will suffice. $\endgroup$ – rschwieb Apr 16 at 13:41
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    $\begingroup$ @rschwieb That's what I was thinking. The question doesn't say more other than it is also true, but perhaps missed the fact that it needed slightly stronger requirements. If you feel this is unlikely to be true without the further assumption, then I could email my professor. $\endgroup$ – George Apr 16 at 13:46

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