Prove that prototype of prime ideal in homomorphism $f : A \to B$ is prime ideal. Prove that preimage of prime ideal in homomorphism $f : A \to B$ is prime ideal.
So, it's easy to prove that it wrong in other way (image of prime ideal could't be prime in $f : \mathbb{Z} \to \mathbb{Q}$).
But can we prove in other way (that preimage of prime ideal is alsoprime ideal) more formally?
 A: I guess you mean preimage of a prime ideal is again prime?
This can be proved as follows (for commutative rings):

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*If $J$ is an ideal of $B$ and $I = f^{-1}(J)$ is its preimage, then $f$ induces an injective homomorphism of rings $\overline{f}: A/I \rightarrow B/J$.


*An ideal $I$ of a ring $A$ is prime if and only if the quotient $A/I$ is integral (note that the zero ring is not integral by convention).


*If $f:A \rightarrow B$ is an injective homomorphism of rings and $B$ is integral, then $A$ is also integral.
A: Let $p$ be a prime Ideal of $B$ and $f:A \to B$ be a homomorphism of (commutative) rings. We want to show that $f^{-1}(p)$ is also a prime ideal.
First we show that it is an ideal.

*

*Notice that since $f(0)=0 \in p$ it follows that $0 \in f^{-1}(p).$


*Let $a,b \in f^{-1}(p).$ Then there are $p_1=f(a), p_2=f(b)$ and hence $f(a+b)=f(a)+f(b)=p_1+p_2 \in P,$ thus $a+b \in f^{-1}(p)$.


*Let $r \in A,a \in f^{-1}(p)$ then there exists a $p_1 \in P$ with $f(a)=p_1$ and thus $f(ra)=f(r)p_1 \in P$, hence $ra \in f^{-1}(p)$. This means that $f^{-1}(p)$ is an ideal of $A$, since it is a subset and the necessary properties hold. We now show that it is prime as well.
Let $a,b \in A$ with $ab \in f^{-1}(p)$. We have to show that either $a$ or $b$ is in $f^{-1}(p)$. We have that $f(a)f(b)=f(ab) \in p$ and since $p$ is prime we have that $f(a)$ or $f(b)$ is in $p$. Notice that $f^{-1}(p)$ also has to be a proper ideal, since otherwise $1 \in f^{-1}(p)$ implies that $1 \in p$ since $f$ is a ring homomorphism, contradicting that $p$ is prime.
