That's the second part of my course becoming acquainted with Henselizations of fields and local rings. (in this question we focus on local rings as it is more algebro geometric motivated). So let $(R, \mathfrak m, \kappa= R/\mathfrak m )$ be a local ring with max ideal $m$.

We can obtain two new rings $R^h$ (the Henselization) and $\widehat{R}_m$ the completion wrt $m$. Consider $R$ as a stalk of a nice enough scheme $S$ we can use these two constructions to obtain new new objects stalkwise: $S^h$ (here we have to differ between strict and "weak" Henselization) and the completion $\widehat{S}$. (recall $\widehat{S}$ is not more a scheme but a ringed space: localizations and completions not behave well to each other).

I would like to compare the main differences & (dis)advantages of completions & Henselizations from viewpoint of commutative algebra and (as well possible) geometric intuition.

The main motivation is that I often read comments like "in practice it's often nicer to work with Henselizations than with completions" in order to study the ring $R$ itself.

Question: Could anybody point out what are the advantages making Henselizations from certain viewpoint nicer to handle with then with completions?

In many comments the hand wavy arguments appearing in this context are like $\widehat{R}_m$ is much "bigger" that $R^h$ making it not "so easy handable like $R^h$". Could anybody bring more light in this formulation? When is mean by "bigger" (the added limits of Cauchy sequences I guess) but much more interesting what makes $R^h$ more "handable"?

The only point that I found out is that $Frac(R)=K \subset K^h$ stays algebraic and in many situations even finite. Is $R \to R^h$ also a finite $R$-module. In general that's not true for completions $ R \to \widehat{R}_m$.

Is this the only point making $R^h$ more handable than $ R \to \widehat{R}_m$?

What can we say about the geometrical part? The completion $\widehat{S}$ gives in certain way "analytic structure" to an (algebraic) scheme $S$ (very hand wavy; I know). About what kind of "geometry" one can think when one consider a henselization of a scheme (as for completion: local ring wise)? Some sources refer to "etale topology". It's a starting point of a huge machinery cumulating in stack theory.

Is there a geometric intuition how one can draw comparisons between endowings of $S$ "analytical structure" (as for completions) and with "etale topology" for $S^h$?

I know that there are a couple of questions here with similar titles (eg https://mathoverflow.net/questions/105381/henselization-and-completion , https://mathoverflow.net/questions/133499/completion-versus-henselization , https://mathoverflow.net/questions/217540/comparison-of-completion-and-henselization-in-class-field-theory ) but none of them deal with question of pure comparison of two constructions in the way I explained above.

Rmk: This is exactly the same question I asked some days ago in MO.

  • $\begingroup$ Again look at $\Bbb{C}[t]_{(t)}$ as I did in your other questions. math.stackexchange.com/questions/3565898/… At first $R$ is a DVR, $K=Frac(R)$, $K_v$ is the completion, $\overline{K}$ is the algebraic closure, and $K_h=K_v\cap \overline{K}$. The next step is to generalize to $\Bbb{C}[x,y]_{(x,y)}$ a local ring of dimension $2$. The completion is a ring of formal power series whereas the Hensalization is a subring which is in the same time a ring of algebraic functions with formal power series expansion at almost every primes. $\endgroup$
    – reuns
    Mar 17, 2020 at 21:28
  • $\begingroup$ In some cases, there is no "the completion" at all, in the sense that two maximal immediate extensions may not be isomorphic over the valued field. (if by completion you mean the maximal dense extension in the valuation topology, then it does not contain the henselization in general). The henselization has the initial property among henselian extensions which makes it easy to use also when dealing with various embeddings. $\endgroup$
    – nombre
    Mar 17, 2020 at 22:56
  • $\begingroup$ @reuns: In your argument you use the identity $K_h=K_v\cap \overline{K}$. Do you know if there exist a quick argument why this identity holds or is it really a "deep" result. Meanwhile nombre gave here mathoverflow.net/questions/354591/… an excellent reference where the proof of this identity is explained but I saw in a couple of your answers that you using frequently this identity and I'm asking if you possibly know an "ad hoc" argument why this holds? $\endgroup$
    – user741314
    Mar 18, 2020 at 17:49
  • $\begingroup$ @nombre: I was talking indeed about this completion with resp the maximal ideal $m$. And you claim that in general the henselization is not contained in the completion. Do you have a conterexample? But at least there always exist a map $R^h \to \widehat{R}_m$ since $\widehat{R}_m$ has Hensel property and $R^h$ has universal property that every map $R \to T$ where $T$ has Hensel property, factorizes through $R^h$, right? $\endgroup$
    – user741314
    Mar 18, 2020 at 17:50
  • $\begingroup$ Additionally, you wrote "... henselization has the initial property among henselian extensions which makes it easy to use..." By "initial property" I guess you mean exactly the universal property I explaned above? Could you loose a couple of works on what you concretly mean by "makes it easy to use"? I think this is exactly the core of my question: it is often mentioned that the henselization is "easier to handle" but I not understand in concrete which way. $\endgroup$
    – user741314
    Mar 18, 2020 at 17:50

1 Answer 1


I can try to give a basic answer regarding the notion of "size". Henselization is a separable-algebraic extension, whereas completion need not be algebraic. For example, the completion of the rational function field $k(t)$ with respect to the $t$-adic valuation is the power series field $k((t))$. The transcendence degree of $k((t))/k(t)$ is uncountable.


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