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In this post, we will try to find all the maximal ideals of $\Bbb Z[X]$, that is $\mathrm{Maxspec}(\Bbb Z[X])$. Of course, there are some posts in MSE or out, but nowhere I found a complete proof. So, I tried to prove it.

Theorem. The maximal ideals of $\Bbb Z[X]$ are of the form $\langle p, f(X)\rangle \trianglelefteq \Bbb Z[X]$ where $p$ is a prime number and $f(x)\in \Bbb Z[X]$ is a polynomial in $\Bbb Z[X]$ which is irreducible $\Bbb Z_p[X]$.

Proof. Let $M\trianglelefteq\Bbb Z[X]$ a prime ideal of $\Bbb Z[X]$. We assume that $M\cap\Bbb Z \neq \{0\}$.

Now, we have the ring $\Bbb Z[X]$, the subring $\Bbb Z \subset \Bbb Z[X]$ of $\Bbb Z[X]$ and the ideal $M\trianglelefteq \Bbb Z[X]$. So, from the 2nd Isomorphism Theorem for Rings, we have (1) $M+\Bbb Z[X]\subseteq \Bbb Z[X]$ is a subring of $\Bbb Z[X]$, (2) $M\trianglelefteq \Bbb Z$, (3) $M\cap\Bbb Z\trianglelefteq \Bbb Z$ and (4): $$\frac{\Bbb Z}{M\cap \Bbb Z} \cong \frac{M+\Bbb Z}{M}\subseteq \frac{\Bbb Z[X]}{M}.$$ Now, from hypothesis $M$ is maximal, thus $M$ is prime. So, $\Bbb Z[X]/M$ is an integral domain. But, in the integral $\Bbb Z[X]/M$ lives the subring $\Bbb Z/ M\cap \Bbb Z$. So, the ring $\Bbb Z/ M\cap \Bbb Z$ is also an integral domain and hence the ideal $M\cap \Bbb Z$ is prime. We know that $\mathrm{Spec}(\Bbb Z)=\{0\}\cup\{\langle p \rangle \triangleleft \Bbb Z: p\text{ prime}\}$. So, $$M\cap \Bbb Z=\langle p \rangle.$$ Let's take the map $$\phi:\Bbb Z[X] \longrightarrow \Bbb Z_p[X],\ f(X)\longmapsto \phi(f(X)):=\overline{f(X)}.$$ It is easy to verify that this is a ring epimorphism. So, $\phi(M)\trianglelefteq\Bbb Z_p[X]$.

Also, let's take the map $$\psi:\Bbb Z[X] \longrightarrow \Bbb Z_p[X]/\phi(M),$$ which is an empimorphism. Then, if we apply the 1st Isomorphism Theorem, we take that $\ker\psi=M$ and hence $$\Bbb Z[X]/M\cong \Bbb Z_p[X]/\phi(M).$$ We have $\Bbb Z[X]/M$ is an integral domain $\iff$ $\Bbb Z_p[X]/\phi(M)$ is an integral domain, but it is also finite. So, $\Bbb Z_p[X]/\phi(M)$ is a field. And since $\Bbb Z_p[X]$ is a PID, there is an irreducible polynomial $\overline{f_0(X)}\in \Bbb Z_p[X]$, s.t. $\phi(M)=\langle \overline{f_0(X)} \rangle $.

I have done up until here.

Questions. 1) Are all these thoughts in the right way?

2) How can we proceed to take $M=\langle p,f(X)\rangle $ and how should we reject the case $M\cap \Bbb Z=\{0\}$?

Of course any other possible ways are welcome!

Thank you

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    $\begingroup$ What is $B$ in $B\cap\mathbb{Z}[X]=\{0\}$? Use the fact that $\mathbb{Z}[X]/M$ is a field (because $M$ is maximal) instead of just an integral domain makes your proof shorter and can reject $M\cap\mathbb{Z}=\{0\}$ (Hint: $\mathbb{Q}$ is not a finitely generated $\mathbb{Z}$-algebra). $\endgroup$ May 26, 2019 at 3:27
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    $\begingroup$ The integral domain $\Bbb{Z}[X]/M$ is a finitely generated $\Bbb{Z}$-module, it won't contain $\Bbb{Q}$ which is not finitely generated. Since it is also a field $\Bbb{Z}[X]/M$ is not of characteristic $0$, so it is of characteristic $p$ prime and $M = (p,m)$ for some maximal ideal $m$ of $\Bbb{F}_p[X]$ ie. $ m =(f(x)),M = (p,f(x))$ for some irreducible polynomial $\in \Bbb{F}_p[X]$. $\endgroup$
    – reuns
    May 26, 2019 at 3:59
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    $\begingroup$ If $M\cap\mathbb Z=p\mathbb Z$ then $p\in M$ and thus $p\mathbb Z[X]\subseteq M$. Then $M/p\mathbb Z[X]$ is a maximal ideal in $\mathbb Z[X]/p\mathbb Z[X]\simeq(\mathbb Z/p\mathbb Z)[X]$ which is a principal ideal domain. This means that $M/p\mathbb Z[X]$ is generated by a polynomial $f\in\mathbb Z[X]$ which is irreducible modulo $p$, and therefore $M=(p,f)$. $\endgroup$
    – user26857
    May 26, 2019 at 20:23
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    $\begingroup$ If $M\cap\mathbb Z=(0)$ then, for $S=\mathbb Z\setminus\{0\}$, $S^{-1}M$ is a maximal ideal of $S^{-1}\mathbb Z[X]=\mathbb Q[X]$, and therefore it is principal. By using Gauss Lemma one can show that $M$ is also principal, a contradiction with math.stackexchange.com/questions/1591558/…. $\endgroup$
    – user26857
    May 26, 2019 at 20:38
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    $\begingroup$ I think your question is a duplicate of math.stackexchange.com/questions/174595/…. At least all you asked is answered there. $\endgroup$
    – user26857
    May 26, 2019 at 20:39

2 Answers 2

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Your reasoning looks good in the case $\mathbb Z \cap M \neq 0$. As you say, $M \cap \mathbb Z$ is an ideal of $\mathbb Z$ generated by a prime number $p$, and the image of $M$ under the reduction map $\phi: \mathbb Z[X] \rightarrow \mathbb Z_p[X]$ is a maximal ideal of $\mathbb Z_p[X]$, generated by a an irreducible polynomial $\overline{f_0}$ of $\mathbb Z_p[X]$.

It is a simple matter to pull $\overline{f_0}$ back to a polynomial $f_0$ in $\mathbb Z[X]$, which is irreducible because $\overline{f_0}$ is. Clearly, $\langle p, f_0 \rangle \subseteq M$, and this must be an equality because $\langle p, f_0\rangle$ is a maximal ideal.

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Your thoughts are correct but I am gonna give a detailed proof and somehow clarify your original thoughts. We will characterize maximal ideals of $\mathbb{Z}[x]$ in the following two steps.

1.Let $\mathfrak{m}$ be a maximal ideal of $\mathbb{Z}[x]$, then $\mathfrak{m}$ contains some prime number $p$:

Consider the inclusion $i:\mathbb{Z}\rightarrow\mathbb{Z}[x]$. Note that $\mathfrak{m}$ is also a prime ideal of $\mathbb{Z}[x]$, then $i^{-1}(\mathfrak{m})=\mathfrak{m}\cap\mathbb{Z}$ is a prime ideal of $\mathbb{Z}$. Hence $\mathfrak{m}\cap\mathbb{Z}=p\mathbb{Z}$ for some prime $p$, or $\mathfrak{m}\cap\mathbb{Z}=(0)$. We will exempt the second case. Note that $i$ induces an injective homomorphism $\overline{i}:\mathbb{Z}/(\mathfrak{m}\cap\mathbb{Z})\rightarrow\mathbb{Z}[x]/\mathfrak{m}$. If $\mathfrak{m}\cap\mathbb{Z}=(0)$, then $\overline{i}:\mathbb{Z}\rightarrow\mathbb{Z}[x]/\mathfrak{m}$, which means that the field $\mathbb{Z}[x]/\mathfrak{m}$ contains $\mathbb{Z}$ and thus it contains $\mathbb{Q}$ as well. However, $\mathbb{Z}[x]/\mathfrak{m}$ is a finitely generated abelian group but $\mathbb{Q}$ is not. Hence we get the contradiction. This finishes the proof of 1.

To see why $\mathbb{Z}[x]/\mathfrak{m}$ is a finitely generated abelian group, we can pick $g(x)\in\mathfrak{m}$ and consider $\mathbb{Z}[x]/\mathfrak{m}\subset\mathbb{Z}[x]/(g(x))$, where $\mathbb{Z}[x]/(g(x))$ is finitely generated by $1,x,…,x^{n}$, $n=\deg(g)$.

2.There is a one-to-one correspondence between the maximal ideals of $\mathbb{Z}[x]$ that contain $p$ and the irreducible polynomials of $\mathbb{F}_p[x]$:

First note that $\mathbb{F}_p[x]/\overline{\mathfrak{m}}\cong\mathbb{Z}[x]/(\mathfrak{m},p)$, where $\overline{\mathfrak{m}}$ denotes the image of $\mathfrak{m}$ in $\mathbb{F}_p[x]$. To see this, consider the map $\varphi:\mathbb{Z}[x]\rightarrow(\mathbb{Z}/p)[x]\rightarrow(\mathbb{Z}/p)[x]/\overline{\mathfrak{m}}, f(x)\mapsto\overline{f}(x)\mapsto\overline{f}(x)+\overline{\mathfrak{m}}$, it is straightforward to check that $\ker\varphi=(\mathfrak{m},p)$. Hence $\mathfrak{m}$ is a maximal ideal of $\mathbb{Z}[x]$ containing $p$ if and only if $\mathbb{F}_{p}[x]/\mathfrak{m}$ is a field , if and only if $\mathfrak{m}$ is a maximal ideal of $\mathbb{F}_p[x]$, if and only if $\overline{\mathfrak{m}}=(\overline{f}(x))$, ${f}$ is irreducible in $\mathbb{F}_p[x]$, in which case, $m=(f(x),p)$, $f$ is a lift of $\overline{f}$ in $\mathbb{Z}[x]$. This completes the proof.

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