Quotienting $\mathbb Z[x]$ by a maximal ideal gives a finite field.

Question

Let $R = \mathbb Z[x]$, a polynomial ring over $\mathbb Z$, and $\mathfrak m$ be any maximal ideal of $R$. How do we show that $R / \mathfrak m$ is a finite field? I know that the fact that it is a finite field directly follows, but not sure about the finite part.

Answer 1

Let $A=R/\mathfrak{m}$ and $F$ be its prime subfield. Assume $F$ is finite, then $A$ is generated over $\mathbb{Z}$ by a single element, so the same holds for $A$ over $F$. In other words, $A$ is a quotient of $F[x]$ by a maximal ideal (as $A$ is a field) so is finite.

So we only need to show that $F$ cannot be $\mathbb{Q}$. Indeed, assume for the sake of contradiction that $F=\mathbb{Q}$. Then $\mathfrak{m}\mathbb{Q}[x]$ is a maximal ideal of $\mathbb{Q}[x]$ (because it’s a localization of $\mathbb{Z}[x]$ at a multiplicative subset not meeting $\mathfrak{m}$), and the quotients are equal.

There is an irreducible polynomial $f(x) \in \mathfrak{m}$, and $(f) \subset \mathbb{Q}[x]$ is a maximal ideal, so $\mathfrak{m}\mathbb{Q}[x]=(f)$, so $\mathfrak{m}=f(x)\mathbb{Q}[x] \cap \mathbb{Z}[x]$.

Now, let $g(x)=h(x)/D$ be a rational polynomial with $D > 1$ integer, $h(x) \in \mathbb{Z}[x]$ primitive (ie its coefficients do not have a nontrivial common divisor – note that $f$ is also primitive, since it’s irreducible) and assume $f(x)g(x) \in \mathbb{Z}[x]$. Then, if $p$ prime divides $D$, $f(x)h(x)=0$ mod $p$, so $f(x)=0$ mod $p$ or $h(x)=0$ mod $p$, both impossible. So $\mathfrak{m}=f\mathbb{Q}[x] \cap \mathbb{Z}[x]=f\mathbb{Z}[x]$.

But then, let $n \geq 1$ be an integer, then $n$ is invertible mod $\mathfrak{m}$, so there is a polynomial $p$ such that $f|np-1$ in $\mathbb{Z}[x]$. Take $n=|f(N)|$ where $N$ is a large integer: $f(x)|np(x)-1$ so in particular $f(N)|np(N)-1$, thus $f(N)|1$, hence a contradiction.