## 15.45 Permanence of properties under henselization

Given a local ring $R$ we denote $R^ h$, resp. $R^{sh}$ the henselization, resp. strict henselization of $R$, see Algebra, Definition 10.155.3. Many of the properties of $R$ are reflected in $R^ h$ and $R^{sh}$ as we will show in this section.

Lemma 15.45.1. Let $(R, \mathfrak m, \kappa )$ be a local ring. Then we have the following

$R \to R^ h \to R^{sh}$ are faithfully flat ring maps,

$\mathfrak m R^ h = \mathfrak m^ h$ and $\mathfrak m R^{sh} = \mathfrak m^ h R^{sh} = \mathfrak m^{sh}$,

$R/\mathfrak m^ n = R^ h/\mathfrak m^ nR^ h$ for all $n$,

there exist elements $x_ i \in R^{sh}$ such that $R^{sh}/\mathfrak m^ nR^{sh}$ is a free $R/\mathfrak m^ n$-module on $x_ i \bmod \mathfrak m^ nR^{sh}$.

**Proof.**
By construction $R^ h$ is a colimit of étale $R$-algebras, see Algebra, Lemma 10.155.1. Since étale ring maps are flat (Algebra, Lemma 10.143.3) we see that $R^ h$ is flat over $R$ by Algebra, Lemma 10.39.3. As a flat local ring homomorphism is faithfully flat (Algebra, Lemma 10.39.17) we see that $R \to R^ h$ is faithfully flat. The ring map $R^ h \to R^{sh}$ is a colimit of finite étale ring maps, see proof of Algebra, Lemma 10.155.2. Hence the same arguments as above show that $R^ h \to R^{sh}$ is faithfully flat.

Part (2) follows from Algebra, Lemmas 10.155.1 and 10.155.2. Part (3) follows from Algebra, Lemma 10.101.1 because $R/\mathfrak m \to R^ h/\mathfrak mR^ h$ is an isomorphism and $R/\mathfrak m^ n \to R^ h/\mathfrak m^ nR^ h$ is flat as a base change of the flat ring map $R \to R^ h$ (Algebra, Lemma 10.39.7). Let $\kappa ^{sep}$ be the residue field of $R^{sh}$ (it is a separable algebraic closure of $\kappa $). Choose $x_ i \in R^{sh}$ mapping to a basis of $\kappa ^{sep}$ as a $\kappa $-vector space. Then (4) follows from Algebra, Lemma 10.101.1 in exactly the same way as above.
$\square$

Lemma 15.45.2. Let $(R, \mathfrak m, \kappa )$ be a local ring. Then

$R \to R^ h$, $R^ h \to R^{sh}$, and $R \to R^{sh}$ are formally étale,

$R \to R^ h$, $R^ h \to R^{sh}$, resp. $R \to R^{sh}$ are formally smooth in the $\mathfrak m^ h$, $\mathfrak m^{sh}$, resp. $\mathfrak m^{sh}$-topology.

**Proof.**
Part (1) follows from the fact that $R^ h$ and $R^{sh}$ are directed colimits of étale algebras (by construction), that étale algebras are formally étale (Algebra, Lemma 10.150.2), and that colimits of formally étale algebras are formally étale (Algebra, Lemma 10.150.3). Part (2) follows from the fact that a formally étale ring map is formally smooth and Lemma 15.37.2.
$\square$

reference
Lemma 15.45.3. Let $R$ be a local ring. The following are equivalent

$R$ is Noetherian,

$R^ h$ is Noetherian, and

$R^{sh}$ is Noetherian.

In this case we have

$(R^ h)^\wedge $ and $(R^{sh})^\wedge $ are Noetherian complete local rings,

$R^\wedge \to (R^ h)^\wedge $ is an isomorphism,

$R^ h \to (R^ h)^\wedge $ and $R^{sh} \to (R^{sh})^\wedge $ are flat,

$R^\wedge \to (R^{sh})^\wedge $ is formally smooth in the $\mathfrak m_{(R^{sh})^\wedge }$-adic topology,

$(R^\wedge )^{sh} = R^\wedge \otimes _{R^ h} R^{sh}$, and

$((R^\wedge )^{sh})^\wedge = (R^{sh})^\wedge $.

**Proof.**
Since $R \to R^ h \to R^{sh}$ are faithfully flat (Lemma 15.45.1), we see that $R^ h$ or $R^{sh}$ being Noetherian implies that $R$ is Noetherian, see Algebra, Lemma 10.164.1. In the rest of the proof we assume $R$ is Noetherian.

As $\mathfrak m \subset R$ is finitely generated it follows that $\mathfrak m^ h = \mathfrak m R^ h$ and $\mathfrak m^{sh} = \mathfrak mR^{sh}$ are finitely generated, see Lemma 15.45.1. Hence $(R^ h)^\wedge $ and $(R^{sh})^\wedge $ are Noetherian by Algebra, Lemma 10.160.3. This proves (a).

Note that (b) is immediate from Lemma 15.45.1. In particular we see that $(R^ h)^\wedge $ is flat over $R$, see Algebra, Lemma 10.97.3.

Next, we show that $R^ h \to (R^ h)^\wedge $ is flat. Write $R^ h = \mathop{\mathrm{colim}}\nolimits _ i R_ i$ as a directed colimit of localizations of étale $R$-algebras. By Algebra, Lemma 10.39.6 if $(R^ h)^\wedge $ is flat over each $R_ i$, then $R^ h \to (R^ h)^\wedge $ is flat. Note that $R^ h = R_ i^ h$ (by construction). Hence $R_ i^\wedge = (R^ h)^\wedge $ by part (b) is flat over $R_ i$ as desired. To finish the proof of (c) we show that $R^{sh} \to (R^{sh})^\wedge $ is flat. To do this, by a limit argument as above, it suffices to show that $(R^{sh})^\wedge $ is flat over $R$. Note that it follows from Lemma 15.45.1 that $(R^{sh})^\wedge $ is the completion of a free $R$-module. By Lemma 15.27.2 we see this is flat over $R$ as desired. This finishes the proof of (c).

At this point we know (c) is true and that $(R^ h)^\wedge $ and $(R^{sh})^\wedge $ are Noetherian. It follows from Algebra, Lemma 10.164.1 that $R^ h$ and $R^{sh}$ are Noetherian.

Part (d) follows from Lemma 15.45.2 and Lemma 15.37.4.

Part (e) follows from Algebra, Lemma 10.155.13 and the fact that $R^\wedge $ is henselian by Algebra, Lemma 10.153.9.

Proof of (f). Using (e) there is a map $R^{sh} \to (R^\wedge )^{sh}$ which induces a map $(R^{sh})^\wedge \to ((R^\wedge )^{sh})^\wedge $ upon completion. Using (e) there is a map $R^\wedge \to (R^{sh})^\wedge $. Since $(R^{sh})^\wedge $ is strictly henselian (see above) this map induces a map $(R^\wedge )^{sh} \to (R^{sh})^\wedge $ by Algebra, Lemma 10.155.10. Completing we obtain a map $((R^\wedge )^{sh})^\wedge \to (R^{sh})^\wedge $. We omit the verification that these two maps are mutually inverse.
$\square$

slogan
Lemma 15.45.4. Let $R$ be a local ring. The following are equivalent: $R$ is reduced, the henselization $R^ h$ of $R$ is reduced, and the strict henselization $R^{sh}$ of $R$ is reduced.

**Proof.**
The ring maps $R \to R^ h \to R^{sh}$ are faithfully flat. Hence one direction of the implications follows from Algebra, Lemma 10.164.2. Conversely, assume $R$ is reduced. Since $R^ h$ and $R^{sh}$ are filtered colimits of étale, hence smooth $R$-algebras, the result follows from Algebra, Lemma 10.163.7.
$\square$

Lemma 15.45.5. Let $R$ be a local ring. Let $nil(R)$ denote the ideal of nilpotent elements of $R$. Then $nil(R)R^ h = nil(R^ h)$ and $nil(R)R^{sh} = nil(R^{sh})$.

**Proof.**
Note that $nil(R)$ is the biggest ideal consisting of nilpotent elements such that the quotient $R/nil(R)$ is reduced. Note that $nil(R)R^ h$ consists of nilpotent elements by Algebra, Lemma 10.32.3. Also, note that $R^ h/nil(R) R^ h$ is the henselization of $R/nil(R)$ by Algebra, Lemma 10.156.2. Hence $R^ h/nil(R)R^ h$ is reduced by Lemma 15.45.4. We conclude that $nil(R) R^ h = nil(R^ h)$ as desired. Similarly for the strict henselization but using Algebra, Lemma 10.156.4.
$\square$

Lemma 15.45.6. Let $R$ be a local ring. The following are equivalent: $R$ is a normal domain, the henselization $R^ h$ of $R$ is a normal domain, and the strict henselization $R^{sh}$ of $R$ is a normal domain.

**Proof.**
A preliminary remark is that a local ring is normal if and only if it is a normal domain (see Algebra, Definition 10.37.11). The ring maps $R \to R^ h \to R^{sh}$ are faithfully flat. Hence one direction of the implications follows from Algebra, Lemma 10.164.3. Conversely, assume $R$ is normal. Since $R^ h$ and $R^{sh}$ are filtered colimits of étale hence smooth $R$-algebras, the result follows from Algebra, Lemmas 10.163.9 and 10.37.17.
$\square$

Lemma 15.45.7. Given any local ring $R$ we have $\dim (R) = \dim (R^ h) = \dim (R^{sh})$.

**Proof.**
Since $R \to R^{sh}$ is faithfully flat (Lemma 15.45.1) we see that $\dim (R^{sh}) \geq \dim (R)$ by going down, see Algebra, Lemma 10.112.1. For the converse, we write $R^{sh} = \mathop{\mathrm{colim}}\nolimits R_ i$ as a directed colimit of local rings $R_ i$ each of which is a localization of an étale $R$-algebra. Now if $\mathfrak q_0 \subset \mathfrak q_1 \subset \ldots \subset \mathfrak q_ n$ is a chain of prime ideals in $R^{sh}$, then for some sufficiently large $i$ the sequence

\[ R_ i \cap \mathfrak q_0 \subset R_ i \cap \mathfrak q_1 \subset \ldots \subset R_ i \cap \mathfrak q_ n \]

is a chain of primes in $R_ i$. Thus we see that $\dim (R^{sh}) \leq \sup _ i \dim (R_ i)$. But by the result of Lemma 15.44.2 we have $\dim (R_ i) = \dim (R)$ for each $i$ and we win.
$\square$

Lemma 15.45.8. Given a Noetherian local ring $R$ we have $\text{depth}(R) = \text{depth}(R^ h) = \text{depth}(R^{sh})$.

**Proof.**
By Lemma 15.45.3 we know that $R^ h$ and $R^{sh}$ are Noetherian. Hence the lemma follows from Algebra, Lemma 10.163.2.
$\square$

Lemma 15.45.9. Let $R$ be a Noetherian local ring. The following are equivalent: $R$ is Cohen-Macaulay, the henselization $R^ h$ of $R$ is Cohen-Macaulay, and the strict henselization $R^{sh}$ of $R$ is Cohen-Macaulay.

**Proof.**
By Lemma 15.45.3 we know that $R^ h$ and $R^{sh}$ are Noetherian, hence the lemma makes sense. Since we have $\text{depth}(R) = \text{depth}(R^ h) = \text{depth}(R^{sh})$ and $\dim (R) = \dim (R^ h) = \dim (R^{sh})$ by Lemmas 15.45.8 and 15.45.7 we conclude.
$\square$

Lemma 15.45.10. Let $R$ be a Noetherian local ring. The following are equivalent: $R$ is a regular local ring, the henselization $R^ h$ of $R$ is a regular local ring, and the strict henselization $R^{sh}$ of $R$ is a regular local ring.

**Proof.**
By Lemma 15.45.3 we know that $R^ h$ and $R^{sh}$ are Noetherian, hence the lemma makes sense. Let $\mathfrak m$ be the maximal ideal of $R$. Let $x_1, \ldots , x_ t \in \mathfrak m$ be a minimal system of generators of $\mathfrak m$, i.e., such that the images in $\mathfrak m/\mathfrak m^2$ form a basis over $\kappa = R/\mathfrak m$. Because $R \to R^ h$ and $R \to R^{sh}$ are faithfully flat, it follows that the images $x_1^ h, \ldots , x_ t^ h$ in $R^ h$, resp. $x_1^{sh}, \ldots , x_ t^{sh}$ in $R^{sh}$ are a minimal system of generators for $\mathfrak m^ h = \mathfrak mR^ h$, resp. $\mathfrak m^{sh} = \mathfrak mR^{sh}$. Regularity of $R$ by definition means $t = \dim (R)$ and similarly for $R^ h$ and $R^{sh}$. Hence the lemma follows from the equality of dimensions $\dim (R) = \dim (R^ h) = \dim (R^{sh})$ of Lemma 15.45.7
$\square$

Lemma 15.45.11. Let $R$ be a Noetherian local ring. Then $R$ is a discrete valuation ring if and only if $R^ h$ is a discrete valuation ring if and only if $R^{sh}$ is a discrete valuation ring.

**Proof.**
This follows from Lemmas 15.45.7 and 15.45.10 and Algebra, Lemma 10.119.7.
$\square$

Lemma 15.45.12. Let $A$ be a ring. Let $B$ be a filtered colimit of étale $A$-algebras. Let $\mathfrak p$ be a prime of $A$. If $B$ is Noetherian, then there are finitely many primes $\mathfrak q_1, \ldots , \mathfrak q_ r$ lying over $\mathfrak p$, we have $B \otimes _ A \kappa (\mathfrak p) = \prod \kappa (\mathfrak q_ i)$, and each of the field extensions $\kappa (\mathfrak q_ i)/\kappa (\mathfrak p)$ is separable algebraic.

**Proof.**
Write $B$ as a filtered colimit $B = \mathop{\mathrm{colim}}\nolimits B_ i$ with $A \to B_ i$ étale. Then on the one hand $B \otimes _ A \kappa (\mathfrak p) = \mathop{\mathrm{colim}}\nolimits B_ i \otimes _ A \kappa (\mathfrak p)$ is a filtered colimit of étale $\kappa (\mathfrak p)$-algebras, and on the other hand it is Noetherian. An étale $\kappa (\mathfrak p)$-algebra is a finite product of finite separable field extensions (Algebra, Lemma 10.143.4). Hence there are no nontrivial specializations between the primes (which are all maximal and minimal primes) of the algebras $B_ i \otimes _ A \kappa (\mathfrak p)$ and hence there are no nontrivial specializations between the primes of $B \otimes _ A \kappa (\mathfrak p)$. Thus $B \otimes _ A \kappa (\mathfrak p)$ is reduced and has finitely many primes which all minimal. Thus it is a finite product of fields (use Algebra, Lemma 10.25.4 or Algebra, Proposition 10.60.7). Each of these fields is a colimit of finite separable extensions and hence the final statement of the lemma follows.
$\square$

Lemma 15.45.13. Let $R$ be a Noetherian local ring. Let $\mathfrak p \subset R$ be a prime. Then

\[ R^ h \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , t} \kappa (\mathfrak q_ i) \quad \text{resp.}\quad R^{sh} \otimes _ R \kappa (\mathfrak p) = \prod \nolimits _{i = 1, \ldots , s} \kappa (\mathfrak r_ i) \]

where $\mathfrak q_1, \ldots , \mathfrak q_ t$, resp. $\mathfrak r_1, \ldots , \mathfrak r_ s$ are the prime of $R^ h$, resp. $R^{sh}$ lying over $\mathfrak p$. Moreover, the field extensions $\kappa (\mathfrak q_ i)/\kappa (\mathfrak p)$ resp. $\kappa (\mathfrak r_ i)/\kappa (\mathfrak p)$ are separable algebraic.

**Proof.**
This can be deduced from the more general Lemma 15.45.12 using that the henselization and strict henselization are Noetherian (as we've seen above). But we also give a direct proof as follows.

We will use without further mention the results of Lemmas 15.45.1 and 15.45.3. Note that $R^ h/\mathfrak pR^ h$, resp. $R^{sh}/\mathfrak pR^{sh}$ is the henselization, resp. strict henselization of $R/\mathfrak p$, see Algebra, Lemma 10.156.2 resp. Algebra, Lemma 10.156.4. Hence we may replace $R$ by $R/\mathfrak p$ and assume that $R$ is a Noetherian local domain and that $\mathfrak p = (0)$. Since $R^ h$, resp. $R^{sh}$ is Noetherian, it has finitely many minimal primes $\mathfrak q_1, \ldots , \mathfrak q_ t$, resp. $\mathfrak r_1, \ldots , \mathfrak r_ s$. Since $R \to R^ h$, resp. $R \to R^{sh}$ is flat these are exactly the primes lying over $\mathfrak p = (0)$ (by going down). Finally, as $R$ is a domain, we see that $R^ h$, resp. $R^{sh}$ is reduced, see Lemma 15.45.4. Thus we see that $R^ h \otimes _ R \kappa (\mathfrak p)$ resp. $R^{sh} \otimes _ R \kappa (\mathfrak p)$ is a reduced Noetherian ring with finitely many primes, all of which are minimal (and hence maximal). Thus these rings are Artinian and are products of their localizations at maximal ideals, each necessarily a field (see Algebra, Proposition 10.60.7 and Algebra, Lemma 10.25.1).

The final statement follows from the fact that $R \to R^ h$, resp. $R \to R^{sh}$ is a colimit of étale ring maps and hence the induced residue field extensions are colimits of finite separable extensions, see Algebra, Lemma 10.143.5.
$\square$

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