# Classification of prime ideals of $\mathbb{Z}[X]$

Let $$\mathbb{Z}[X]$$ be the ring of polynomials in one variable over $$\Bbb Z$$.

My question: Is every prime ideal of $$\mathbb{Z}[X]$$ one of following types? If yes, how would you prove this?

1. $$(0)$$.

2. $$(f(X))$$, where $$f(X)$$ is an irreducible polynomial.

3. $$(p)$$, where $$p$$ is a prime number.

4. $$(p, f(X))$$, where $$p$$ is a prime number and $$f(X)$$ is an irreducible polynomial modulo $$p$$.

• Yes. There is a famous picture due to Mumford of $\operatorname{Spec} \mathbb{Z}[x]$, which can be found in [The red book of varieties and schemes, 2nd ed., p. 75]. Jul 24, 2012 at 9:49
• You can also find a proof (or at least, a detailed discussion) of this in Hartshorne's book, just a few pages into the chapter on schemes. Jul 24, 2012 at 16:58

Let $\mathfrak{P}$ be a prime ideal of $\mathbb{Z}[x]$. Then $\mathfrak{P}\cap\mathbb{Z}$ is a prime ideal of $\mathbb{Z}$: this holds whenever $R\subseteq S$ are commutative rings. Indeed, if $a,b\in R$, $ab\in R\cap P$, then $a\in P$ or $b\in P$ (since $P$ is prime). (More generally, the contraction of a prime ideal is always a prime ideal, and $\mathfrak{P}\cap\mathbb{Z}$ is the contraction of $\mathfrak{P}$ along the embedding $\mathbb{Z}\hookrightarrow\mathbb{Z}[x]$).

Thus, we have two possibilities: $\mathfrak{P}\cap\mathbb{Z}=(0)$, or $\mathfrak{P}\cap\mathbb{Z}=(p)$ for some prime integer $p$.

Case 1. $\mathfrak{P}\cap\mathbb{Z}=(0)$. If $\mathfrak{P}=(0)$, we are done; otherwise, let $S=\mathbb{Z}-\{0\}$. Then $S\cap \mathfrak{P}=\varnothing$, $S$ is a multiplicative set, so we can localize $\mathbb{Z}[x]$ at $S$ to obtain $\mathbb{Q}[x]$; the ideal $S^{-1}\mathfrak{P}$ is prime in $\mathbb{Q}[x]$, and so is of the form $(q(x))$ for some irreducible polynomial $q(x)$. Clearing denominators and factoring out content we may assume that $q(x)$ has integer coefficients and the gcd of the coefficients is $1$.

I claim that $\mathfrak{P}=(q(x))$. Indeed, from the theory of localizations, we know that $\mathfrak{P}$ consists precisely of the elements of $\mathbb{Z}[x]$ which, when considered to be elements of $\mathbb{Q}[x]$, lie in $S^{-1}\mathfrak{P}$. That is, $\mathfrak{P}$ consists precisely of the rational multiples of $q(x)$ that have integer coefficients. In particular, every integer multiple of $q(x)$ lies in $\mathfrak{P}$, so $(q(x))\subseteq \mathfrak{P}$. But, moreover, if $f(x)=\frac{r}{s}q(x)\in\mathbb{Z}[x]$, then $s$ divides all coefficients of $q(x)$; since $q(x)$ is primitive, it follows that $s\in\{1,-1\}$, so $f(x)$ is actually an integer multiple of $q(x)$. Thus, $\mathfrak{P}\subseteq (q(x))$, proving equality.

Thus, if $\mathfrak{P}\cap\mathbb{Z}=(0)$, then either $\mathfrak{P}=(0)$, or $\mathfrak{P}=(q(x))$ where $q(x)\in\mathbb{Z}[x]$ is irreducible.

Case 2. $\mathfrak{P}\cap\mathbb{Z}=(p)$.

We can then consider the image of $\mathfrak{P}$ in $\mathbb{Z}[x]/(p)\cong\mathbb{F}_p[x]$. The image is prime, since the map is onto; the prime ideals of $\mathbb{F}_p[x]$ are $(0)$ and ideals of the form $(q(x))$ with $q(x)$ monic irreducible over $\mathbb{F}_p[x]$. If the image is $(0)$, then $\mathfrak{P}=(p)$, and we are done.

Otherwise, let $p(x)$ be a polynomial in $\mathbb{Z}[x]$ that reduces to $q(x)$ modulo $p$ and that is monic. Note that $p(x)$ must be irreducible in $\mathbb{Z}[x]$, since any nontrivial factorization in $\mathbb{Z}[x]$ would induce a nontrivial factorization in $\mathbb{F}_p[x]$ (since $p(x)$ and $q(x)$ are both monic).

I claim that $\mathfrak{P}=(p,p(x))$. Indeed, the isomorphism theorems guarantee that $(p,p(x))\subseteq \mathfrak{P}$. Conversely, let $r(x)\in\mathfrak{P}(x)$. Then there exists a polynomial $s(x)\in\mathbb{F}_p[x]$ such that $s(x)q(x) = \overline{r}(x)$. If $t(x)$ is any polynomial that reduces to $s(x)$ modulo $p$, then $t(x)p(x)-r(x)\in (p)$, hence there exists a polynomial $u(x)\in\mathbb{Z}[x]$ such that $r(x) = t(x)p(x)+pu(x)$. Therefore, $r(x)\in (p,p(x))$, hence $\mathfrak{P}\subseteq (p,p(x))$, giving equality.

Thus, if $\mathfrak{P}\cap\mathbb{Z}[x]=(p)$ with $p$ a prime, then either $\mathfrak{P}=(p)$ or $\mathfrak{P}=(p,p(x))$ with $p(x)\in\mathbb{Z}[x]$ irreducible.

This proves the desired classification.

• This generalizes easily to any PID; for an arbitrary UFD it gets a bit more complicated. Jul 24, 2012 at 17:04
• Is $S^{-1}\mathfrak{P}$ equal to the extension of $\mathfrak{P}$ under the natural inclusion $\pi: \mathbb{Z}[x] \rightarrow \mathbb{Q}[x]$ given by $f \mapsto f/1$?
– Sam
Dec 3, 2014 at 7:48
• @ArturoMagidin Do you mean that the result continues to hold for UFDs but the proof is harder, or that the result is no longer true in its full extent? A reference is welcomed, thanks! Apr 26, 2016 at 21:55
• "for an arbitrary UFD it gets a bit more complicated" looks like a mysterious remark. For instance, it's not clear at all how to describe the prime ideals in polynomial rings over fields when we have at least three indeterminates. Mar 2, 2019 at 9:53