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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 $p$.

Community
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Makoto Kato
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    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]. – Zhen Lin Jul 24 '12 at 09:49
  • 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. – Derek Allums Jul 24 '12 at 16:58
  • For future visitors, here are some more links/references for Mumford's famous picture of $$\text{Spec}(\mathbb Z[x])$ and similar pictures: http://www.neverendingbooks.org/mumfords-treasure-map, https://pbelmans.ncag.info/blog/2011/05/25/a-latex-version-of-mumfords-impression-of-spec-zx-or-some-tikz-tricks/, and Eisenbud&Harris pg. 85-86, pg. 90 – D.R. Mar 28 '23 at 05:28

1 Answers1

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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.

Xam
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Arturo Magidin
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    This generalizes easily to any PID; for an arbitrary UFD it gets a bit more complicated. – Arturo Magidin Jul 24 '12 at 17:04
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    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 03 '14 at 07:48
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    @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! – Matemáticos Chibchas Apr 26 '16 at 21:55
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    "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. – user26857 Mar 02 '19 at 09:53
  • @Auturo: In case one can we use the following argument instead of localization? If $g$ is another primitive irreducible polynomial, then they are coprime in $\mathbb Q[x]$ and there is $a,b\in\mathbb Z[x]$ s.t. $a(x)f(x)+b(x)g(x)=d\in \mathbb Z\cap \mathfrak P$, contradiction. Thus any $h(x)\in\mathfrak P$ must have a factor $f(x)$ – William Leynoid Aug 01 '23 at 02:39
  • @WilliamLeynoid what "other"? Who is $f$? If your $f$ is supposed to be my $q$, then how do you get $q$ in the first place without localization? PS careful with spelling. – Arturo Magidin Aug 01 '23 at 02:53
  • @Arturo Magidin: If $\mathfrak P\cap \mathbb Z=(0)$ but $\mathfrak P\neq (0)$, then there is a positive degree polynomial $q_1(x)\in \mathfrak P$. Factorize $q_1(x)$ in $\mathbb Z[x]$, we must have an irreducible primitive factor $q(x)\in \mathfrak P$. If $p(x)\neq q(x)$ is another irreducible primitive polynomial in $\mathfrak P$, $p,q$ must be coprime in $\mathbb Q[x]$, so there is $a,b\in \mathbb Z[x]$ s.t. $a(x)p(x)+b(x)q(x)=d\in \mathbb Z\cap \mathfrak P=(0)$, contradiction. Thus any $f(x)\in \mathfrak P$ must have a fractor $q(x)$. – William Leynoid Aug 01 '23 at 03:23
  • @WilliamLeynoid I read the answer (which unnecessarily bumped an old question with a partial answer). Why do you feel the need to repeat it? – Arturo Magidin Aug 01 '23 at 03:34
  • @Arturo: Sorry. I delete it. Just want to figure out if it's correct :) – William Leynoid Aug 01 '23 at 03:38
  • I know it is an old answer, but does someone knows why does it say that "$\mathfrak{P}$ consists precisely of the rational multiples of $q$ with integer coefficients"? It doesn't. It contains multiples of $q$ by a rational number and by a polynomial with integer coefficients, right? I'd say the argument is not complete if you don't acknowledge this. There must be something I'm not getting – Joel Aug 09 '23 at 15:12
  • @Joel How familiar are you with the theory of localizations, which is invoked to justify that? – Arturo Magidin Aug 09 '23 at 15:38
  • @Joel: It was inartly phrased, but you only need to worry about polynomials of least degree, i.e. of degree $\deg(q)$, and all of those are as described. – Arturo Magidin Aug 09 '23 at 20:34
  • @Arturo I am quite familiar with it, hence my confusion. But yeah, I think I didn't get the phrasing right, that's all. I mean, from what it says, "rational multiples" is products of q by rational numbers. That's simply not $\mathfrak{P}$, so I don't now if the phrasing is inartly or just wrongly expresing something correct. Anyway, now I see what you meant. Thanks! – Joel Aug 10 '23 at 01:45
  • @Joel It's not well written; "rational" and "integer" should be "rational polynomial" and "integer polynomial". But I don't think it's worth the edit, which will bump the post unnecessarily. The question about localizations wasn't a criticism, but on first blush I wanted to know if I needed to explain some of that stuff first. – Arturo Magidin Aug 10 '23 at 02:53
  • @ArturoMagidin Okay, now I can point more precisely my question, then. If you understand it as rational polynomial and integer polynomial, that's what I thought, then I don't see why can you conclude that $s$ must divide the coefficients of $q$ (and therefore be a unit). Because $r$ (which now I understand as $r(x)$) is also a polynomial. But here it might be that I am forgetting some basic fact about contents and polynomials. – Joel Aug 10 '23 at 15:33
  • @Joel: Because $\mathfrak{P}$ is prime and contains no nonzero integers, if $f(x)\in\mathfrak{P}$ is not primtive, then $f(x) = cF(x)$ with $c\in\mathbb{Z}$ and $F(x)$ primitive; but $c\notin\mathfrak{P}$, so $F(x)\in\mathfrak{P}$. So we may restrict our attention to primitive polynomials. By Gauss's Lemma, if $q(x)p(x)$ has integer coefficients, $p(x)$ is primitive with integer coefficients, then $q(x)$ has rational coefficients, then $q(x)$ must have integer coefficients and be primitive. So restricting to primitive $f$ forces $r$ to be a primitive polynomial with integer coefficients. – Arturo Magidin Aug 10 '23 at 15:56
  • @ArturoMagidin Thank you so much, now I get it!! – Joel Aug 10 '23 at 16:13
  • @ArturoMagidin Hi! "the ideal $S^{-1}P$ is prime in $Q[x]$ ,and so is of the form $(q(x))$ for some irreducible polynomial $q(x)$." Can you please tell me why $q(x)$ must be irreducible? if $S^{-1} P$ is a prime ideal. Which result you used? –  Oct 01 '23 at 14:09
  • @ArturoMagidin Similarly, in the line 2 of case 2 I am not able to understand why $q(x)$ must be irreducible? Can you please tell which theorem you used to deduce this? –  Oct 01 '23 at 14:26
  • @ArturoMagidin In the second last para of your solution , can you please tell me how the isomorphism theorem tells that $(p, p(x) )\subseteq \mathbb{P}$? –  Oct 01 '23 at 14:40