TITLE: Prime ideals of $\mathbb{Z}[x]$ considered as modules
QUESTION [6 upvotes]: There is a neat description of the prime ideals of the ring $R=\mathbb{Z}[x]$, see here (MSE 174595) for example.
Now I am just interested in the structure of these ideals as modules over $R$.
Of course, all non-zero principal ideals are isomorphic as modules, since they are isomorphic to the ring itself considered as module. This is true for any domain $R$.
But what about the non-principal (prime) ideals, which according to the link above are of the form
$(p, f(x))$
with $p \in \mathbb{Z}$ a prime and $f(x)$ a polynomial whose reduction mod $p$ is irreducible.
Question: Are any of these isomorphic as $R$-modules?
I think I could prove with some computations on relations between the generators that for different primes $p \neq q$, the ideals $(p, x)$ and $(q,x)$ are not isomorphic as $R$-modules. But before I throw in more complicated polynomials, I wanted to ask if there is a smarter way to deal with this, and/or if the question is answered somewhere.
(A related vague question is: Is there a good way how to think about ideals as modules? I find it psychologically difficult, as one is more used to think of their quotients as modules. For example, although I boldly stated the above "of course" fact for principal ideals, part of my brain found it counter-intuitive at first.)

REPLY [2 votes]: To elaborate on Mohan's comment, we first show that if $I=(p,f)$ is such an ideal, then every module homomorphism $\varphi:I\to R$ is given by $\varphi(a)=ra$ for some $r\in R$.  Indeed, $\varphi(pf)=p\varphi(f)=f\varphi(p)$.  Since $R$ is a UFD, $p$ is irreducible, and $p$ does not divide $f$, this implies $p$ divides $\varphi(p)$; let $r\in R$ be such that $\varphi(p)=rp$.  We then have $p\varphi(f)=f\varphi(p)=rpf$, which implies $\varphi(f)=rf$.  Since $p$ and $f$ generate $I$, it follows that $\varphi(a)=ra$ for all $a\in I$.
Now suppose $J$ is another such ideal and $\varphi:I\to J$ is an isomorphism.  Since $J$ is a submodule of $R$, there exists $r\in R$ such that $\varphi(a)=ra$ for all $a\in I$.  But then $r$ divides every element of $J$, which means $r$ is a unit.  It follows that $J$ is equal to $I$.