Let $R$ be a rng, so that $1\not\in R$. I am trying to show that following are equivalence of definition of prime ideal $P$;
(i) $AB\subseteq P$ with $A,B\subseteq R$ implies $A\subseteq P$ or $B\subseteq P$
(ii) $aRb\subseteq P$ with $a,b\in R$ implies that $a\in P$ or $b\in P$
I can show (ii) implies (i) by saying suppose that $AB\subseteq P$ with $A\not\subseteq P$ and $B\not \subseteq P$ then $ARB\subseteq P$ and then fixing the elements of $A$ and $B$ in turn and applying (ii) gives the result.
I can't seem to show (i) implies (ii) though. If $R$ has a $1$ this is simple but without it I am unsure how to proceed as the ideal generated by an element is not equal to $aR$.
From (i) we know that if $(a)$ and $(b)$ are principal ideals in $R$ such that $(a)(b)\subseteq P$ then we have $(a)\subseteq P$ or $(b)\subseteq P$ which implies that $a\in P$ or $b\in P$. This is also true for the product of any finite number of principal ideals. Indeed if $(a_1)(a_2)...(a_n)\subseteq P$, then at least one of them is the subset of $P$, of course if we have (i).
Now assume that (i) is true and suppose that $aRb\subseteq P$. Then $$RaRRbR\subseteq RaRbR\subseteq RPR\subseteq P.$$ Since $RaR$ and $RbR$ are ideals of $R$, (i) implies that $RaR\subseteq P$ or $RbR\subseteq P$. Now suppose that $RaR\subseteq P$, then $$(a)^3\subseteq R(\mathbb Za+Ra+aR+RaR)R\subseteq RaR\subseteq P.$$ By (i), $(a)\subseteq P$ which implies that $a\in P$. Similarly, if $RbR\subseteq P$ it implies $b\in P$.