The exercise asks me two things:
first, I need to prove that
$$a) \ \ P \mbox{ is a prime ideal} \iff A/P \mbox{ is integral domain}$$ and then
b) The only prime ideals of $\mathbb Z$ are $\{0\}$ and the principal ideals $p\mathbb Z$ for $p$ prime.
The first one I managed to prove like this:
$$\rightarrow A/P \mbox{ integral domain} \iff (I+a)(I+b) = I \implies I+a = I \mbox{ or } I+b = I \implies \\ a\in I \mbox{ or } b\in I \\ \leftarrow a,b\in I \implies (I+a)(I+b) = I+ab = I. \mbox {I prime} \implies a\in I \mbox{ or } b\in I \implies I+a = I \mbox { or } I+b \in I \implies A/I \mbox{ is integral domain}$$
Now, for the question $b$ it asks me to prove that whenever $ab\in \mathbb Z$ we have $ab$ prime, therefore, $a=1, b=p$ for $p$ prime? Isn't it 'proven' already? What can I do to argument it? I can't understand
Hint
All ideals of $\mathbb{Z}$ are of the form $n\mathbb{Z}$. So to say $P=k\mathbb{Z}$ is a prime ideal of $\mathbb{Z}$ is to show that for every pair of integers $a,b$ if $ab \in P$, then either $a \in P$ or $b \in P$. In other words, you have to prove that if $ab=kl$ for some $l \in \mathbb{Z}$, then either $a=ks$ or $b=kt$ for some $s,t\mathbb{Z}$.
Ask yourself? if $k$ was not a prime then can you find $a,b \in \mathbb{Z}$ such that it violates the last condition written above?
OR you can ask yourself when will the quotient ring $\mathbb{Z}/k\mathbb{Z}$ be an integral domain?