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I have been trying to prove the following theorem:

Every maximal ideal in $\mathbb{Z}[x]$ has the form $(p, f(x))$ where $p$ is prime integer and $f$ is primitive integer polynomial that is irreducible modulo $p$.

Idea: I tried to establish a homomorphism $\phi: \mathbb{Z}[x] \rightarrow \mathbb{F}$. Since $\mathbb{F}$ is a field it has characteristic $p$ and so integer prime p are mapped to 0 in $\mathbb{F}$. Hence $p\in \ker \phi$. Next we consider $\phi': \mathbb{Z}[x] \rightarrow \mathbb{Z_p}[x]$ and pick an arbitrary maximal ideal $M\in \mathbb{Z}[x]$. So, $\phi'(M)$ is maximal as long as $p \in M$ by correspondence. But now I am stuck at this stage and do not know how to proceed. I guess we might have to use primitivity given in problem but dont know how.

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    $\begingroup$ Once you have concluded that every maximal ideal of $\mathbb{Z}[x]$ must contains some prime number $p$, that maximal ideal under taking quotient is a maximal ideal of $\mathbb{F}_p[x]$. But what are the maximal ideals of $\mathbb{F}_[x]$? Every maximal ideal of $\mathbb{F}_p[x]$ is of the form $(f(x))$ where $f$ monic irreducible polynomial. So the preimage of this ideal is the maximal of $\mathbb{Z}[x]$. We can choose a monic $G(x)\in \mathbb{Z}[x]$ whose image in $\mathbb{F}_p[x]$ is $f$, then you can show $(p,G)$ is the preimage of $(f)$. $\endgroup$ – user119882 Apr 5 '14 at 15:58
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    $\begingroup$ "Since F is a field it has characteristic p..." It would probably be good to mention how you eliminate the characteristic zero case... $\endgroup$ – rschwieb Apr 6 '14 at 2:38
  • $\begingroup$ @user119882 Why can you conclude that the maximal ideal under taking quotient is a maximal ideal of $\mathbb F_p[x]$? $\endgroup$ – Bach Jun 28 at 3:10
  • $\begingroup$ See This $\endgroup$ – Bach Jun 28 at 3:14
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Since maximal ideals are prime ideals, and according to this post, it suffices to exclude the situation 2.(which is the only non-trivial case), i.e., we need to show that $(f(x))$ is not a maximal ideal when $f(x)$ is irreducible with $\deg(f(x))>1$.

To see this, note that if we assume that $(f(x))$ is a maximal ideal of $\mathbb Z[x]$, then there should not be any non-unit ideal in $\mathbb Z[x]$ containing $f(x)$.

However, consider $(p,f(x))\supsetneq (f(x))$ where $p$ is a prime such that $p$ does not divide the leading coefficient of $f(x)$. It is trivial to see that $(p,f(x))\ne\mathbb Z[x]$, since $$\frac{\mathbb Z[x]}{(p,f(x))}\cong\frac{\mathbb F_p[x]}{(f(x))}\ne 0.$$ Contradiction!

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  • $\begingroup$ No idea who's p in this answer (an arbitrary prime? then f could be a non-zero constant modulo p), but this has been proved many times. The last time I seen this is only few days ago. $\endgroup$ – user26857 Jun 28 at 7:03
  • $\begingroup$ @user26857 I have edited my post, I mean $f(x)$ is a non-constant irreducible polynomial. I just want to remove this question from the unanswered list, plus I haven't seen anyone else using the same method as mine. $\endgroup$ – Bach Jun 28 at 7:11
  • $\begingroup$ @user26857 Also, $p$ does not divide the leading coefficient of $f(x)$ and I have added this point to my answer. $\endgroup$ – Bach Jun 28 at 7:21
  • $\begingroup$ math.stackexchange.com/questions/1796878/… $\endgroup$ – user26857 Jun 28 at 11:41
  • $\begingroup$ @user26857 Okay, now I see. $\endgroup$ – Bach Jun 28 at 11:57

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