$P$ is a prime ideal $\iff$ for $r, s \in R$ such that $rRs \subset P$, then $r \in P$ or $s \in P$.

Question

If $P$ is an ideal in a not necessarily commutative ring $R$, then the following conditions are equivalent: a) $P$ is a prime ideal, b) If $r,s\in R$ such that $rRs\subset P$ then $r\in P$ or $s\in P$.

I tried to prove it like this:

a)$\Rightarrow$ b) let $P$ is prime ideal. If $rRs\subset P$ then $\{r\}\subset P$ or $Rs\subset P$ then $r\in P$ or $Rs\in P$. If $1\in R$ then for $Rs\subset P$,$s\in P$ but if $1\notin R$ How can I prove it?

For b)$\Rightarrow$ a)
I couldn't find anything.

Answer 1

For $a)\Rightarrow b)$, you just need to use the prime condition once more, i.e. $Rs\subset P$ implies $R\subset P$ or $( s) \subset P$, and the first possibility is absurd as $P$ is proper. Thus $( s)\subset P$, i.e. $s\in P$.

For $b)\Rightarrow a)$, take $A,B\subset R$ ideals such that $AB\subset P$. If $A\subset P$ we are done. Otherwise, we have some $g\in A$ such that $g\notin P$. Now consider $gRs\subset AB\subset P$ for every $s\in B$. Then the assumption in $b)$ shows that $s\in P$ for every $s\in B$, i.e. $B\subset P$.

Answer 2

If $rRs\subset P$ then $\{r\}\subset P$ or $Rs\subset P$

This really makes me suspect you are using the definition of prime ideal for commutative rings (that $rs\in P$ implies $r\in P$ or $s\in P$) which is not intended.

If you have any (noncommutative) prime ring that isn't a domain (so that $P=\{0\}$ is prime), then there exists $r,s\neq 0$ such that $rs=0$, but $r\notin \{0\}$ and $Rs\nsubseteq \{0\}$ because $s\neq 0$.

So you can see that the step you're taking there is not valid.

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A correct definition of a prime ideal $P$ in a noncommutative ring is:

If $A$ and $B$ are right ideals such that $AB\subseteq P$, then $A\subseteq P$ or $B\subseteq P$.

Using the right definition:

$a \implies b$

Suppose $rRs\subseteq P$. Then $(rR)(sR)\subseteq P$, and so at least one of $Rr$ and $Rs$ is contained in $P$, and therefore at least one of $r,s$ is in $P$.

$\neg a\implies \neg b$

Since we have supposed that a) does not hold, there exists right ideals $A,B$ such that $AB\subseteq P$ but neither is contained in $P$. Find $a\in A\setminus P$ and $b\in B\setminus P$. Then $aRb=a(Rb)\subseteq AB\subseteq P$ shows that b) does not hold.