I'm currently trying to solve this problem.
Let $f: R \rightarrow S$ be a surjective ring homomorphism. Let $K = \ker(f)$. Assume $P$ is a prime ideal s.t. $K \subset P$. Show $f(P)$ is a prime ideal in $S$.
I solved the ideal part.
Let $y \in f(P)$, by definition then $y = f(x)$ for some $x \in P$. Now let $s \in S$. Then since $f$ is surjective $\exists r \in R$ s.t. $f(r) = s$. So then since $f$ is a homomorphism we have $f(x)\cdot s = f(x) \cdot f(r) = f(x\cdot r)$. Then since $P$ is ideal and $x \in P$, $x\cdot r \in P$, so $f(x \cdot r) = f(x)\cdot s \in f(P)$ so $f(P)$ is closed under multiplication by an element in $S$.
Now let $y, y' \in f(P)$, again by definition we have $x, x' \in R$ s.t. $f(x) = y, f(x') = y'$. By $f$ homomorphism we have $y - y' = f(x) - f(x') = f(x-x')$. Then by $P$ ideal we have $x-x' \in P$, thus $y-y' \in f(P)$.
Therefore $f(P)$ is an ideal.
The part I'm having trouble with is showing that $P$ prime implies $f(P)$ prime.
I started off let $AB \subset f(P)$, by definition we know we have $C \subset P$ s.t. $f(C) = AB$. From here I need to arrive at showing that either $A \subset f(P)$ or $B \subset f(P)$. I'm not really sure how to move forward from here but I know I somehow need to involve the fact that $K \subset P$ since I haven't used that yet.
Any help would be appreciated.
Note: I'm not assuming that $R$ or $S$ is commutative, I already know how to solve it if they are using the commutative definition of prime.
$f:R\to S$ induces an isomorphism $\overline{f}:R/K \to S$. So it is enough to show that $P/K$ is a (completely) prime ideal of $R/K$.
Every ideal of $R/K$ can be written uniquely in the form $I/K$ for some ideal $I\supset K$ of $R$ (take the preimage under the projection $R\to R/K$). But if $A/K\cdot B/K \subset P/K$, then $AB \subset P+K = P$, and the conclusion follows shortly thereafter.