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.
On
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.
The answer to your question is no. Take a prime ring $R$ and $R'\subseteq R$ a subring which is not prime. Then the inverse image of $(0)$ is $(0)$ which is not a prime ideal of $R'$. An example is the following: $R=\begin{pmatrix} \mathbb Z & n\mathbb Z\\ \mathbb Z & \mathbb Z\end{pmatrix}$ and $R'=\begin{pmatrix} \mathbb Z & n\mathbb Z\\ 0 & \mathbb Z\end{pmatrix}$ with $n\ge 1$. (This example is taken from Lam, Exercises in Classical Ring Theory, Exercise 10.7.)