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

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.

• @EricWofsey, this may be a duplicate but I think there are simpler proofs for one variable. I suggest reopening.
– lhf
Feb 14 at 0:49
• See math.stackexchange.com/questions/148745/… for a stronger result, but as lhf commented, there is probably a simpler proof in your special case. Feb 14 at 0:52
• Feb 14 at 1:06

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.