Classification of prime ideals of $\mathbb{Z}[X]/(f(X))$ Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable.
Let $f(X) \in \mathbb{Z}[X]$ be a monic irreducible polynomial.
Let $A = \mathbb{Z}[X]/(f(X))$.
Let $\theta$ = $X$ (mod $f(X)$).
My question: Is the following proposition correct? If yes, how would you prove this?
Proposition
Let $P$ be a non-zero prime ideal of $A$.
Then the following assertions hold.
(1) $P$ contains a prime number $p$.
(2) One of the following two cases occurs.   
a. If $f(X)$ is irreducible mod $p$, then $P = (p)$.
b. If $f(X)$ is not irreducible mod $p$, then $P = (p, g(\theta))$, where $g(X)$ is an irreducible factor of $f(X)$ mod $p$.
(3) $P$ is a maximal ideal and $A/P$ is a finite field of characteristic $p$.
This is a related question.
 A: Let us denote by $P$ the corresponding ideal of $\mathbb{Z}[X]$ as well. It will contain polynomials that are not multiples of $f(X)$. Let $g(X)$ be one such. Because $f(X)$ is irreducible, the greatest common divisor (in the Euclidean domain $\mathbb{Q}[X]$) must be equal to 1. By Bezout's identity there exist polynomials $u(X),v(X)\in \mathbb{Q}[X]$ such that
$$
u(X)f(X)+v(X)g(X)=1.
$$
Multiplying this by the least common multiple $m$ of the denominators of the coefficients of $u(X)$ and $v(X)$ we see that there exists polynomials $U(X)=mu(X),V(X)=mv(X)\in \mathbb{Z}[X]$ such that
$$
m=U(x)f(X)+V(X)g(X)\in P.
$$
So the ideal $P$ contains non-zero integer constants. Because $P$ is a prime ideal, the intersection $P\cap\mathbb{Z}$ is also a (non-zero) prime ideal, and hence it contains a prime number $p$. Because $P$ is non-trivial, $p$ is the only prime number in
$P$. Part (1) is settled.
From  (1) we deduce that the quotient ring $A/P$ is finite. As $A$ is an integral domain, and $P$ is a prime ideal, the ring $A/P$ is also an integral domain. A finite integral domain is always a field, so (3) is proven.
Clearly $A/P$ is generated (as a ring) by $\theta$. So $A/P=\mathbb{F}_p[\theta]$, where I again abuse notation and denote $X+P$ also with $\theta$. Let $g(X)\in\mathbb{F}_p[X]$ be the minimal polynomial of $\theta$ over the prime field. Obviously $g(X)\mid \overline{f}(X)$, where $\overline{f}(X)$ stands for the reduction of $f(X)$ modulo $p$. The claim (2) follows from this relatively easily.
A: You may want to read this , or  also this . 
Basically and bottom line, a maximal ideal in $\,\Bbb Z[x]\,$ is of the form $\,(p,f(x))\,$ , with $\,p\,$ a prime number and $\,f(x)\,$ an irreducible polynomial when reduced modulo $\,p\,$, i.e. in $\,\left(\Bbb Z/p\Bbb Z\right)[x]\,$ 
