TITLE: Inverse image of prime ideal in noncommutative ring
QUESTION [6 upvotes]: We say an ideal $P\neq R$ in a ring $R$ is prime, if for any two ideals $I,J$ of $R$ the following implication holds: if $IJ\subseteq P$ then $I\subseteq P$ or $J\subseteq P$.

If $f\colon R \to S$ is a homomorphism of noncommutative rings and $P$ is a prime ideal in $S$, is $f^{-1}(P)$ a prime ideal in $R$?

Of course, the above statement is true for commutative $R$ and $S$. I managed to prove it for noncommutative $R$ and $S$ when $f$ is an epimorphism.

REPLY [3 votes]: By looking at the induced maps on quotient rings, the question is equivalent to: Is every subring of a prime ring also prime? There are lots of counterexamples. For example, if $R$ is an integral domain, then $M_2(R)$ is a prime ring, but the subring of diagonal matrices $\cong R \times R$ is not.