TITLE: A proof that shows surjective homomorphic image of prime ideal is prime
QUESTION [16 upvotes]: Let $A, B$ be commutative rings with $1_{A}, 1_{B}$. Suppose that $\mathfrak{p} \neq (1)$ is a prime ideal in $A$ with $\mathfrak{p} \supseteq \ker{\varphi}$ where $\varphi: A \rightarrow B$ is a surjective homomorphism. I want to show $\varphi(\mathfrak{p})$ is a prime ideal.
I tried to solve it directly, but I got stuck, and I realized that it is easy to see from non-unital ring isomorphism $\mathfrak{p}/\ker{\varphi} \simeq \varphi(\mathfrak{p})$, but it is strange that I could not write more direct basic set-theoretic proof.

I would like to see if someone can write it down (though the word "direct" may be ambiguous).
If prime ideal does not contain kernel of surjective homomorphism, it seems like primeness is not preserved, but there is no reason for me to believe this yet, so I want to know if this is true and why / why not.


Add: For #2 above, now I think $\varphi(\mathfrak{p})$ is always prime as long as $\varphi$ is surjective since $\mathfrak{p}/(\ker{\varphi} \cap \mathfrak{p}) \simeq \varphi(\mathfrak{p})$. I know this may be a technical question, but it is important for me to make this accurate since many arguments that I come up with depend on this.

Add 2: With the answer that I selected, what I added above is not true.

Add 3: What I considered was $\varphi|_{\mathfrak{p}}:\mathfrak{p} \rightarrow \varphi(\mathfrak{p})$. We can think of any ideal as ring without $1$, so this gives $\mathfrak{p}/(\ker\varphi \cap \mathfrak{p}) = \mathfrak{p}/\ker\varphi|_{\mathfrak{p}} \simeq \varphi(\mathfrak{p})$. The problem here is that $\mathfrak{p}/(\ker\varphi \cap \mathfrak{p})$ is not an ideal of $A/\ker\varphi$ unless $\ker\varphi \subseteq \mathfrak{p}$.

REPLY [13 votes]: Yet another, more abstract proof.
If $\phi : A \to B$ is a surjective homomorphism, then for every ideal $\mathfrak{b} \subseteq B$ we get an isomorphism $\overline{\phi} : A/\phi^{-1}(\mathfrak{b}) \to B/\mathfrak{b}$. In particular, if $\phi^{-1}(\mathfrak{b}) \subseteq A$ is a prime (primary, maximal, reduced, etc.) ideal, then the same holds for $\mathfrak{b}$. Now, if $\mathfrak{a} \subseteq A$ is an ideal, then $\phi^{-1}(\phi(\mathfrak{a}))=\mathfrak{a} + \ker(\phi)$. Hence, if $\mathfrak{a}+\ker(\phi)$ is prime, then also $\phi(\mathfrak{a})$ is prime. The claim is for the special case $\ker(\phi)\subseteq \mathfrak{a}$.