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.

  • $\begingroup$ 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 $P=p$. $\endgroup$ – KReiser Jul 24 '12 at 10:30
  • $\begingroup$ I meant $P$ contains a prime number $p$. I'll edit it later. $\endgroup$ – Makoto Kato Jul 24 '12 at 10:32

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.

  • $\begingroup$ This is the best, most succinct treatment I've see. Thanks. $\endgroup$ – user12802 Oct 19 '12 at 15:56
  • $\begingroup$ A wonderful proof!!! $\endgroup$ – Lao-tzu Jan 9 '14 at 7:58
  • $\begingroup$ 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. $\endgroup$ – Vincent J. Ruan Dec 6 '16 at 3:10
  • 1
    $\begingroup$ @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 $R$. $\endgroup$ – Jyrki Lahtonen Dec 6 '16 at 6:06
  • 1
    $\begingroup$ @VincentJ.Ruan If it holds for $A$ it holds for $A/P$, by the usual abuse of notation. $\endgroup$ – 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 $\,\left(\Bbb Z/p\Bbb Z\right)[x]\,$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.