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.

  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ "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

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!

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.