# The maximal ideals of $\Bbb Z[X]$

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

• 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). May 26, 2019 at 3:27
• 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 $\in \Bbb{F}_p[X]$. May 26, 2019 at 3:59
• 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 $M=(p,f)$. May 26, 2019 at 20:23
• 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/…. May 26, 2019 at 20:38
• I think your question is a duplicate of math.stackexchange.com/questions/174595/…. At least all you asked is answered there. May 26, 2019 at 20:39

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.

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.