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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.)

Torsten Schoeneberg
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    For $I=(p,f(x))$, show that $\mathrm{Hom}_R(I,R)=R$. Then, it will follow that another prime ideal $J$ is isomorphic to $I$ if and only if $I=J$. – Mohan Oct 05 '17 at 10:37
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    @Mohan: Looking at duals, and Hom's in general, sounds like a good idea, and I see $Hom_R(I, R) \simeq R$, but I do not see how to conclude from there. Especially since we have $Hom_R(J, R) \simeq R$ for the nonzero principal ideals $J$ as well. – Torsten Schoeneberg Oct 05 '17 at 21:17
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    @TorstenSchoeneberg It's an easy exercise to show that two ideals $I,J$ in an integral domain $R$ are isomorphic iff there is $a,b\in R$ such that $aI=bJ$. If $R$ is a UFD (or a GCD domain) we may assume $\gcd(a,b)=1$. Since $aI\subseteq(b)$ we get $I\subseteq(b)$. If $I$ contains a prime element $p$, then $b=1$ or $b=p$. If $b\ne 1$, then $I=(p)$. Analogously for $J$. In your case necessarily $a=b=1$. – user26857 Oct 08 '17 at 22:09
  • @user26857: Thank you for this nice method. I had not been aware of that "easy exercise" but I think I have proved it for myself now (it is quite easy -- once the idea is there), and I will not forget it from now on. -- When you write "1" in the end, you mean "unit" (in our case, $\pm 1$). – Torsten Schoeneberg Oct 09 '17 at 20:06
  • @TorstenSchoeneberg Yes, I mean unit. – user26857 Oct 09 '17 at 20:24

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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 $I$.

Eric Wofsey
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    Thank you! Just to make things clearer for myself, I now see two reasons why the situation differs from the principal ideals, which are mutually isomorphic: a) The first statement (every hom. $I \rightarrow R$ is given by multiplication with some $r\in R$) is stronger than just $Hom_R(I,R) \simeq R$, and does not hold for the principal ideals. b) At the end, for "which means $r$ is a unit", we use that $J$ is not contained in a proper principal ideal. – Torsten Schoeneberg Oct 05 '17 at 22:04