Preimages of prime ideals when only one of the rings is commutative Given a ring homomorphism between noncommutative rings, the preimage of a prime ideal is not necessarily prime. On the other hand, the preimage of a prime ideal is always prime if both rings are commutative, which is why the usual prime spectrum $\operatorname{Spec}$ is functorial on commutative rings. 
I am just curious what happens if only one of the rings involved is commutative. Suppose we have a pair of rings $R$ and $K$, where $K$ is commutative, and $R$ is not. 
If we have a ring homomorphism $f: R \to K$, is the preimage of every prime ideal in $K$ prime?
Likewise, if we have a ring homomorphism $g: K \to R$, Is the preimage of every prime ideal in $R$ prime?
 A: An ideal $P\subset R$ is prime iff $R/P$ is a prime ring.  So modding out the prime ideal and its preimage, you can assume your homomorphism is injective and the question is whether a subring of a prime ring is prime, when you know one of the rings is commutative.
In the case of $f:R\to K$, a commutative prime ring is just an integral domain, and a subring of an integral domain is an integral domain.  So the answer is yes, the preimage of a prime ideal is prime.  Another way to think about this case is that the inverse image of a completely prime ideal is always completely prime, where $P$ is completely prime if $ab\in P$ implies $a\in P$ or $b\in P$ (the proof is easy).  Since prime ideals in commutative rings are completely prime, the inverse image of a prime ideal in a commutative ring is always prime (in fact, completely prime).
In the case of $g:K\to R$, the answer is no in general.  For instance, if $A$ is an integral domain, then $R=M_2(A)$ is a prime ring, but the subring $A\times A$ of diagonal matrices is commutative and not prime.
