# Proof for maximal ideals in $\mathbb{Z}[x]$ [duplicate]

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.

• 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 $(f)$. Apr 5, 2014 at 15:58
• "Since F is a field it has characteristic p..." It would probably be good to mention how you eliminate the characteristic zero case... Apr 6, 2014 at 2:38
• @user119882 Why can you conclude that the maximal ideal under taking quotient is a maximal ideal of $\mathbb F_p[x]$?
– Bach
Jun 28, 2019 at 3:10
• See This
– Bach
Jun 28, 2019 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!

• 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. Jun 28, 2019 at 7:03
• @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.
– Bach
Jun 28, 2019 at 7:11
• @user26857 Also, $p$ does not divide the leading coefficient of $f(x)$ and I have added this point to my answer.
– Bach
Jun 28, 2019 at 7:21
• math.stackexchange.com/questions/1796878/… Jun 28, 2019 at 11:41