# 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.

• When you say "(1) $P$ contains a prime ideal $p$", what do you mean? This statement has no content-it is always trivially true with $P=p$. – KReiser Jul 24 '12 at 10:30
• I meant $P$ contains a prime number $p$. I'll edit it later. – Makoto Kato Jul 24 '12 at 10:32

## 2 Answers

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.

• This is the best, most succinct treatment I've see. Thanks. – user12802 Oct 19 '12 at 15:56
• A wonderful proof!!! – Lao-tzu Jan 9 '14 at 7:58
• I don't understand this part of your proof "Clearly $A/P$ is generated (as a ring) by $X \; \text{mod} \; f$. So $A/P = \mathbb{F}_p[(X \; \text{mod} \; f) + P]$". Please elaborate. Thank you. – Vincent J. Ruan Dec 6 '16 at 3:10
• @VincentJ.Ruan: $A$ is a quotient ring of $R=\Bbb{Z}[X]$. The ring $R$ is generated by $X$, so $A$ is generated by the image of $X$ under the projection $R\to A$. Ditto for the quotient ring $A/P$ that is a homomorphic image of $R$. – Jyrki Lahtonen Dec 6 '16 at 6:06
• @VincentJ.Ruan If it holds for $A$ it holds for $A/P$, by the usual abuse of notation. – Jyrki Lahtonen Dec 6 '16 at 13:25

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 $\,\left(\Bbb Z/p\Bbb Z\right)[x]\,$