$(x)$ is prime but not maximal in $\mathbb{Z}[x]$

Question

How prove that the ideal $(x)=x\mathbb{Z}[x]$ in $\mathbb{Z}[x]$ is prime but not maximal?

Answer 1

$p(x).q(x)\in (x)\Rightarrow p(0).q(0)=0\Rightarrow p(0)=0\mbox{ or }q(0)=0\Rightarrow x|p(x)\mbox{ or }x|q(x)$ hence prime ideal

As $(x)\subset(x)+(2)\subset\mathbb{Z}[x]$ hence not maximal

Answer 2

Show that $\mathbb{Z}[x]/(x)$ is an integral domain but not a field.

Answer 3

Structurally: use $\rm\:P\:$ is prime $\rm\iff R/P\:$ is a domain; $\rm\:P\:$ is maximal $\rm\iff R/P\:$ is a field.

Elementwise: $\rm\ \ f,g\not\in (x)\iff f(0),g(0)\neq 0\iff f(0)g(0)\neq 0 \iff fg\not\in (x)$

$\rm (x)$ max $\rm\iff [f\not\in(x)\, \Rightarrow\, (f,x)=1]\iff [f(0)\ne 0\:\Rightarrow (f(0)) = 1]\,$ by evaluation at $\rm\,x=0.$

It is very instructive to analyze how the elementwise proofs correspond to the structural proofs.

Answer 4

Zooming out for a moment: the following general facts should help. Let $A, B$ be commutative rings.

1. An ideal $\mathfrak a$ of $A$ is prime (resp. maximal) if and only if $A/\mathfrak a$ is an integral domain (resp. a field).
2. There is a canonical isomorphism from $B[x]/(x)$ to another ring. You probably have a guess for what the ring and the map are and it's likely correct; to prove it, try to define an inverse.

Related, but possibly overwhelming at this point: Mumford's treasure map.