The following type of statements is often bandied about around the mathematical watering hole: "etale close is closer than formally close". Namely, if one imagines some sort of 'absolute topology' encompassing all the usual Grothendieck topologies on schemes (fppf, fpqc, etale, Nisnevich,...) that zooming in 'formally close' is a coarser zoom than zooming in 'etale close'.

Can someone give meat to these statements? An actual comparison statement of some form? Also, perhaps more pressingly, are they 'intuitively true' (I've never really seen a rigorous formalization of these statements made)?

Here is a sampler of the sort of things I'm curious about:

  1. Is there a natural Grothendieck topology on schemes whose 'local rings' are completions? Namely, in the etale topology the local ring of a scheme $X$ at a point $x$ is $\mathcal{O}_{X,x}^\text{sh}$. Is there a natural Grothendieck topology whose local ring of $X$ at $x$ is $\widehat{\mathcal{O}_{X,x}}$?
  2. Does this topology have a natural notion of fundamental groups? Is there some category of 'finite ____ covers" which forms a reasonable Galois category? How do the fundamental groups in this category compare?
  3. A literal question: Let $R$ be any local ring. How does, $\pi_1^{\acute{et}}(\text{Spec}(\widehat{R}))$ relate to $\pi_1^{\acute{e}t}(\text{Spec}(R^\ast))$ for $\ast\in\{\text{h},\text{sh}\}$? Similarly, how does $G_{\text{Frac}(\widehat{R})}$ (if $R$ is a domain) relate to $G_{\text{Frac}(R^\text{h})}$ and $G_{\text{Frac}(R^{\text{sh}})}$? What if $R$ is a DVR?
  4. I'm almost positive that Artin approximation enters this picture, but not totally sure how so.

One reason I'm curious is that it's often easy to make nice intuitive relationships between geometry and the formal local geometry of a scheme (say over $\mathbb{C}$) but less obvious to me how to make analogies using henselizations.

For example, when one wants to intuit $G_{\mathbb{Q}_p}$ as a 'group classifying the geometry of a punctured $\text{Spec}(\mathbb{Z}$) near $p$ ' one often times looks to $G_{\mathbb{C}(T)}$. There one sees that the analogue of $G_{\mathbb{Q}_p}$ is $G_{\mathbb{C}((T))}$ which one can imagine as being the (profinite) fundamental group of a small punctured disk around a point $p$ which makes total sense, and is beautiful. That said, it's a small punctured formal disk around the point whereas, for me, it would have made more sense to consider small etale disks (whatever this means) around the point.

This is further backed up by the point that if $D^\ast$ is a small punctured disk around $p$ then $\widehat{\pi_1(D^\ast)}=\widehat{\mathbb{Z}}=G_{\mathbb{C}((T))}$. This should still be true if $\mathbb{C}((T))$ is replaced by $\mathbb{C}[T]_{(t)}^{\text{sh}}$, but it's then less clear that this equivalence of fundamental groups is not a mistake.


  • $\begingroup$ The "formal topology" should be finer than Nisnevich's. Given a point $x \in X$ we have the chain of local rings $$O_{X,x} \to O_{X,x}^{h} \to \widehat{O_{X,x}}$$ corresponding to the Zariski and Nisnevich topologies, both with common completion. As for the étale topology, its local ring is $O_{X,x}^{sh} $ whose completion has as residue field the separable closure of $k(x)$, the residue filed at $x$ that , in principle, may be obtained from $O_{X,x}$ by an inverse limit procedure. This would be correspond to a non-existent "formal-étale" topology. Is this related to your question? $\endgroup$
    – Leo Alonso
    Sep 19, 2016 at 15:51

1 Answer 1


For local noetherian rings, the henselization has much more direct algebro-geometric meaning than the completion since it is built from local-etale algebras. Hence, your statement that the completion provides more intuition than the henselization is just a matter of having less experience with henselizations. See Example 2 below for why there is no need in practice for anything like a "formal topology", which likely doesn't exist anyway. In particular, the answer #1 is likely "no", so #2 is then moot.

But the real reason this is all somewhat of a wild goose chase is that you are correct about #4: for excellent local noetherian rings, the deep Artin-Popescu approximation is the ultimate answer to nearly all questions about passing between their completions and their henselizations. This theorem implies that for any such ring $R$ and any finite system of polynomial equations in several variables over $R$, any solution in the completion can be approximated arbitrarily closely (for the max-adic topology) by a solution in $R^{\rm{h}}$. (The real content, upon varying the system of equations, is that there is even a single solution in $R^{\rm{h}}$, let alone one that is "close" to the given solution in the completion.) However, that raw statement about systems of equations does not do justice to the significance, so let's illustrate with a couple of applications; see section 3 of Artin's IHES paper (http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1969__36_/PMIHES_1969__36__23_0/PMIHES_1969__36__23_0.pdf) for many more striking applications.

Example 1. A pretty algebraic application of this approximation theorem is that if $R$ is an excellent local normal noetherian domain then $R^{\rm{h}}$ is the "algebraic closure" of $R$ in the normal local noetherian completion $\widehat{R}$. Even for $R$ the local ring at the origin of an affine space over a field, I am not aware of a proof of this application without invoking Artin approximation.

Example 2. A geometric reason that Artin-Popescu approximation is so relevant is that it implies the following extremely useful fact: if $R$ is an exellent local noetherian ring and $A, B$ are two local $R$-algebras (with local structure map) essentially of finite type such that there exists an isomorphism $f:\widehat{A} \simeq \widehat{B}$ as $\widehat{R}$-algebras then $A$ and $B$ admit a common residually trivial local-etale neighborhood. (In particular, $A^{\rm{h}} \simeq B^{\rm{h}}$ over $R$; we definitely do not claim that such an isomorphism of henselizations can be found which induces $f$ on completions.) Even for $R$ a field, this is not at all obvious but is super-useful; e.g., it comes up when rigorously justifying the sufficiency of proving some facts about the geometry of specific singularities by studying analogous special cases with completions, such as in deJong's paper on alternations or various papers on semistable curves, etc.

It remains to address #3, under reasonable hypotheses on $R$. We will see that the first question in #3 is quite easy, and the second generally has a negative answer when $\dim R > 1$ (whereas the case $\dim R = 1$ is affirmative under reasonable assumptions on $R$, a fact used all the time in number theory; more on this below).

Now we come to the hypotheses that should have been made on $R$ for posing any questions of the sort being asked. Presumably you meant for $R$ to be noetherian (max-adic completion would not be appropriate otherwise), and also excellent (or else the completion could fail to be reduced when $R$ is reduced, etc.). Moreover, let's forget about strict henselization and only consider henselization, since the latter has the same completion and is the correct "algebraic" substitute for the completion. And since henselization of an excellent local ring is excellent (see the Remark after Prop. 1.21 in the book "Etale Cohomology and the Weil Conjectures" for a proof, which was overlooked in EGA even though preservation under strict henselization is done in EGA), we may as well assume $R$ is henselian.

Since the category of finite etale algebras over a henselian local ring is naturally equivalent to that over its residue field, the invariance of etale fundamental group under passage to the completion for any henselian local noetherian ring is obvious. That is, the first question in #3 has an easy affirmative answer.

I do not know what is the precise intended meaning of the 2nd to last paragraph in the posting (since equicharacteristic 0 is a poor guide to the rich arithmetic structure of mixed characteristic $(0,p)$), but one might contemplate trying to use the Artin-Popescu approximation theorem to affirmatively settle:

Question. Let $R$ be an excellent henselian local domain that is normal, with fraction field $K$. Let $\widehat{K}$ denote the fraction field of the completion $\widehat{R}$. Do the Galois groups of $K$ and $\widehat{K}$ naturally coincide?

Despite whatever intuition may be derived from Artin-Popescu approximation, the possibly surprising answer to the Question is generally negative when $\dim R > 1$, even when $R$ is regular. (I only realized that after failing to prove an affirmative answer beyond dimension 1 in this way.) Before discussing counterexamples, a few Remarks will be helpful to put things in context:

Remark 1: That the local noetherian ring $\widehat{R}$ is actually a normal domain requires a proof. One way to see it is to use Serre's homological characterization of normality and the fact that the flat map ${\rm{Spec}}(\widehat{R}) \rightarrow {\rm{Spec}}(R)$ has regular fibers (since $R$ is excellent); see the Corollary to 23.9 in Matsumura's "Commutative Ring Theory". Note also that when $\dim R > 1$, there is no "topological field" structure on $K$ and so $\widehat{K}$ is not a "completion" for $K$ in some intrinsic sense not referring to $R$; it is just notation for the fraction field of the completion of $R$.

Remark 2. In case $\dim R = 1$ the Question has an affirmative answer (even valid for rank-1 henselian valuation rings and valuation-theoretic completions thereof, without noetherian hypotheses; see 2.4.1--2.4.3 in Berkovich's paper in Publ. IHES 78). This does not require the approximation theorem, and is an instructive exercise in the use of Krasner's Lemma. The main task when $\dim R > 1$ is to try to replace Krasner's Lemma with an argument using the approximation theorem, bypassing the fact that $K$ (let alone its finite separable extensions) doesn't have a natural topological field structure when $\dim R > 1$; all such attempts are doomed to fail, as we'll see below.

Remark 3. Equality of the Galois groups (through the evident natural map) would imply via Galois theory that every finite separable extension of $\widehat{K}$ arises by "completing" a finite separable extension of $K$ and more generally that $E \rightsquigarrow E \otimes_K \widehat{K}$ is an equivalence of categories of finite etale algebraic over $K$ and $\widehat{K}$ respectively. This equivalence is very useful in number theory for the case $\dim R = 1$ when the answer is affirmative.

To see what goes wrong beyond dimension 1, consider a general $R$ as above, allowing $\dim R$ arbitrary for now and not yet assuming regularity. Let $K'/K$ be a finite separable extension, so by normality of $R$ the integral closure $R'$ of $R$ in $K'$ is module-finite over $R$. Since $R$ is henselian and $R'$ is a domain, it follows that $R'$ must be local (and henselian and excellent). Thus, $R' \otimes_R \widehat{R} = \widehat{R'}$ is a normal domain, and its fraction field $\widehat{K'}$ is clearly finite separable over $\widehat{K}$ with degree $[K':K]$.

In case $K'/K$ is Galois with Galois group $G$, it follows that $G \subset {\rm{Aut}}(\widehat{K'}/\widehat{K})$ with $\#G = [\widehat{K'}:\widehat{K}]$, so $\widehat{K'}/\widehat{K}$ is Galois with Galois group $G$. In this way, we see (by exhausting a separable algebraic extension by finite separable subextensions) that $K_s \otimes_K \widehat{K}$ is a Galois extension of $\widehat{K}$ with Galois group ${\rm{Gal}}(K_s/K)$. The Question is exactly asking if this is a separable closure of $\widehat{K}$.

By Galois theory over $\widehat{K}$, it is the same to ask that any given finite Galois extension $E/\widehat{K}$ is the "completion" (in the above sense based on integral closures of $R$) of some finite separable extension $K'/K$, necessarily of the same degree.

Now let's see that already for quadratic extensions one has counterexamples with $R$ any countable henselian excellent regular local ring with $\dim R > 1$ and ${\rm{char}}(K) \ne 2$, such as the henselization at any closed point on a smooth connected scheme of dimension $>1$ over a field not of characteristic 2; e.g., $R = \mathbf{Q}[x_1,\dots,x_n]_{(x_1,\dots,x_n)}^{\rm{h}}$ is a counterexample for any $n > 1$.

It is the same to show for such $R$ that the natural map $K^{\times}/(K^{\times})^2 \rightarrow \widehat{K}^{\times}/(\widehat{K}^{\times})^2$ is not surjective. Since $K^{\times}$ is countable, it suffices to show that the target is uncountable. This uncountability feels "obvious" when $\dim R > 1$, but we want to give a rigorous proof (e.g., to understand exactly how the condition $\dim R > 1$ and the completeness of $\widehat{R}$ are used).

The regularity of $R$ implies that of $\widehat{R}$, so $\widehat{R}$ is a UFD. Thus, the height-1 primes of $\widehat{R}$ are principal, corresponding to irreducibles in $\widehat{R}$ up to unit multiple. Any two irreducibles that are not $\widehat{R}^{\times}$-multiples of each other give rise to distinct elements in $\widehat{K}^{\times}/(\widehat{K}^{\times})^2$, by the UFD property. Thus, it suffices to show that $\widehat{R}$ contains uncountably many height-1 primes.

Complete local noetherian rings satisfy "countable prime avoidance": if an ideal $I$ is contained in a union of countably many prime ideals then it is contained in one of those primes. (This fact is originally a lemma of L. Burch. It can also be deduced easily from the Baire category theorem by using that such rings with their max-adic topology are complete metric spaces in which all ideals are closed, as noted in section 2 of the paper "Baire's category theorem and prime avoidance in complete local rings" by Sharp and Vamos in Arch. Math. 44 (1985), pp. 243-248.) By the UFD property, every non-unit in $\widehat{R}$ is divisible by an irreducible element and so lies in a height-1 prime. Thus, the maximal ideal $m$ of $\widehat{R}$ is contained in the union of all height-1 prime ideals. But $m$ is not contained in (equivalently, not equal to!) any of those height-1 primes since $\dim \widehat{R} = \dim R > 1$ (!), so by countable prime avoidance in $\widehat{R}$ it follows that $\widehat{R}$ must have uncountably many height-1 primes, as desired. (Related reasoning comes up in Example 2.4 in the paper of Sharp and Vamos mentioned above.)

Remark: For any $R$ as in the above counterexamples, and $B'$ the module-finite integral closure of $\widehat{R}$ in a quadratic extension of $\widehat{K}$ not arising from a quadratic extension of $K$, there is no finite $R$-algebra $B$ satisfying $B \otimes_R \widehat{R} \simeq B'$. This may seem to contradict Theorem 3.11 in Artin's IHES paper linked above, but it does not (since there is no reason that the proper closed non-etale locus in the base for the generically etale ${\rm{Spec}}(B') \rightarrow {\rm{Spec}}(\widehat{R})$ is contained in a proper closed subset of ${\rm{Spec}}(\widehat{R})$ that is the preimage of one in ${\rm{Spec}}(R)$).

  • $\begingroup$ Dear Dr. nfdC, thanks for your helpful comments--I think you've answered essentially all my MO questions... I'll study your answer intensely tonight and report back if I have any questions. One question I would greatly appreciate your help in understanding is the following (which I didn't see answered on my first read through of your answer). Suppose that you want to intuit something like I mentioned above: $G_{\mathbb{Q}_p}$ should be like studying the local geometry 'at $p$' of something like a punctured $\text{Spec}(\mathbb{Z})$. Here's one way which I can make this semi-precise $\endgroup$
    – SomeGuy
    Sep 20, 2016 at 20:04
  • $\begingroup$ in my head (or, rather, semi-convincing). If we pretend we're studying something like a punctured (compact) Riemann surface instead of a punctured $\text{Spec}(\mathbb{Z})$ then studying the local geometry of this surface at $p$ is like studying a small punctured disk at $p$. Now, while this should correspond in the intuitive 'etale topology=replacement for standard topology' to studying something like the fraction field of the strict Henselization of the local ring of our (algebraic) Riemann surface at $p$, I DON'T know how to think of this as the type of disk I'd like $\endgroup$
    – SomeGuy
    Sep 20, 2016 at 20:06
  • $\begingroup$ (the disk I'd get if I thought of this in the standard topology). But, of course, one can think of $\mathbb{C}((T))$ as being the functions on a small punctured (formal) disk at $p$. I mean, this is super intuitively clear to me since, for example, its functor of points on complete Huber pairs is what one would expect for a formal open disk (the topological nilpotents). Is there an analagous way in which I can justify intuitively to myself that the fraction field of the strictly local ring of our (algebraic) Riemann surface at $p$ is 'like a disk'. Thanks again! $\endgroup$
    – SomeGuy
    Sep 20, 2016 at 20:08
  • $\begingroup$ Also, just a comment, I discussed this with someone I highly trust, and I think under some reasonable hypotheses one needn't use Krasner's lemma or any baby form of Artin approximation to show that $\text{Frac}(R^h)$ and $\text{Frac}(\widehat{R})$ have the same Galois group. That said, the proof that I/we ultimately came up with only works for $R$ a Henselian DVR (maybe with perfect residue field) which I suppose is a MUCH weaker version than the case you were trying to explain the solution to. $\endgroup$
    – SomeGuy
    Sep 20, 2016 at 20:19
  • 1
    $\begingroup$ I understand why it's not overly indicative: as you point out, really the local picture of the fundamental group of a disk is really only telling about $\text{Gal}(K^\text{tame}/K^\text{ur})$. I disagree that it's not saying anything--it explains what sort of 'local' one means when discussing how Galois groups of local fields inform the Galois groups of global fields. I think most people 'think topologically' or try to intuit 'topologically' when thinking about arithmetic problems, which is (at least I think) OK with the obvious caveat that it's just a cartoon. Thanks again for your $\endgroup$
    – SomeGuy
    Sep 21, 2016 at 3:17

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