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Let $\mathbb{Z}[X]$ be the polynomial ring in $X$ over the ring $\mathbb{Z}$ of integers. Its profinite completion is the inverse limit over all ideals of finite index. So, what is the profinite completion $\widehat{\mathbb{Z}[X]}$ of $\mathbb{Z}[X]$? Is it $\widehat{\mathbb{Z}}[[X]]$? Is $\widehat{\mathbb{Z}[X]}$ a Noetherian ring?

lovemathguy
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1 Answers1

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They idea is to do similar computations as in the case of the profinite completion of $\mathbb{Z}$. If ideals of finite index of $\mathbb{Z}[X]$ where products of prime ideals of finite index, then by the chinese reminder theorem we would have an isomorphism $$(\ast) \ \ \ \ \widehat{\mathbb{Z}[X]} \rightarrow \prod_{\wp} \ \lim_{\leftarrow} \mathbb{Z}[X]/\wp^n,$$ with the product being indexed by primes of finite index. Unfortnately, there are ideals of finite index that are not product of primes, as for example $(2,x^2)$. However, for every ideal of finite rank $I$, there are some prime ideals of finite index $\wp_1,...,\wp_r$ such that $\wp_1^{e_1} \dots \wp_r^{e_r} \subset I$. Indeed, let $\wp_1,\dots,\wp_r$ be the prime ideals containing $I$ (they come in finite number as $I$ is of finite index). Then $\sqrt{I} = \wp_1 \cap \dots \cap \wp_r = \wp_1 \dots \wp_r$ as they are coprimes. As $\mathbb{Z}[X]$ is noetherian, there is some $n$ such that $\sqrt{I}^n \subset I$, thus $(\wp_1 \dots \wp_r)^n \subset I$.
This is sufficient to establish the isomorphism $(\ast)$. More generally, the isomorphism $(\ast)$ holds as product of primes form a final subposet of all ideals of finite index, hence the limit might be reduced to this final subposet. Now we study primes of finite index.

If $\wp$ is a prime ideal of finite index, then $\mathbb{Z} \cap \wp$ is a non zero prime ideal of $\mathbb{Z}$, thus of the form $(p)$, and $\wp$ contains $p$. Moreover, the quotient $\mathbb{Z}[X]/\wp$ is a finite integral domain, thus is a finite field. Hence $\wp$ is of the form $(p,P)$, such that $P$ is irreducible mod $p$, and $\mathbb{Z}[X]/\wp \cong \mathbb{F}_p[X]/(\overline{P})$.


We are ready to compute $\widehat{\mathbb{Z}[X]}$. Write $\mathbb{Z}[X]^{\wedge,\wp}$ for the completion for the $\wp$-adic topology, that is $\underset{\leftarrow}{\textrm{lim}} \ \mathbb{Z}[X] / \wp^n$. We thus have $\widehat{\mathbb{Z}[X]} \cong \prod_{\wp} \mathbb{Z}[X]^{\wedge,\wp}$.
If $\wp=(p,P)$, by general computation of the completion of finite type ideals, see for example Completion of a polynomial ring w.r.t. a maximal ideal, we have an isomorphism $$\mathbb{Z}[X]^{\wedge,\wp} \cong \mathbb{Z}[X] [\![X_1,X_2]\!]/(p-X_1,P-X_2) \cong \mathbb{Z_p}[X][\![P]\!]. $$ Thus, you find the isomorphism $$\widehat{\mathbb{Z}[X]} \cong \prod_p \prod_{P} \mathbb{Z}_p[X][\![P]\!], $$ where $p$ ranges over all primes, and $P$ ranges over all polynomials that are irreducible mod $p$.


Every finite field is a residue field of this ring, hence it is not isomorphic to $\widehat{\mathbb{Z}}[\![X]\!]$, for which only prime finite fields are residue fields. It is not a noetherian ring, as it is an infinite product of non trivial rings.

Enguerrand Moulinier
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  • Therefore, even its pro-p completion (i.e. the inverse limit over all ideals of finite index of $p$-power) is not Noetherian? – lovemathguy Oct 31 '24 at 11:48
  • Indeed, its pro-$p$ completion is not noetherian – Enguerrand Moulinier Oct 31 '24 at 12:58
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    @EnguerrandMoulinier. Can you elaborate about the existence of a decomposition of a nonzero ideal into the product of prime ideals and how it relates to factoriality ? – GreginGre Oct 31 '24 at 17:43
  • @GreginGre You are right to ask the question, as it is a mistake ; it is not true that every ideal in a UFD has a decomposition into primes. However, I have been able to correct my error. Here is my statement ; Every ideal of finite index in $\mathbb{Z}[X]$ is a product of primes.

    More generally, I believe that in a noetherian ring $A$ that is integrally closed in its field of fractions, then any finite index ideal has a decomposition into primes, but I still need to work out some details. I will add those corrections

     – Enguerrand Moulinier Oct 31 '24 at 19:11
  • How is $(2,x^2)$ a product of primes? – Aphelli Nov 11 '24 at 11:55
  • @Aphelli Hello, no it's not. I need to edit my post ; the final isomorphism still holds, but instead of writing ideals as a product of primes, I need to write them as intersections of primary ideals. Things almost do not change, as if $Q$ is a primary ideal of $\mathbb{Z}[X]$ of finite index, then $\sqrt{Q}$ is prime of finite index, thus maximal, of the form $\wp$ as in my post. Then I may show that $I$ contains a product of primes, and things will pass to the limit – Enguerrand Moulinier Nov 11 '24 at 12:17
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    @Aphelli The key observation is that in the ring $\mathbb Z [X]$ products of primes form a final subposet of all finite index ideals, so computing the completion can be reduced to taking limit of quotients by only those ideals which are products of primes. The relevant notion is of cofiltered limit — limit of a diagram where every finite subdiagram already has a cone. – xsnl Nov 11 '24 at 16:36
  • @xsnl: you’re right, good point! – Aphelli Nov 11 '24 at 18:19