Definition 15.106.1. Let $A$ be a local ring. We say $A$ is *unibranch* if the reduction $A_{red}$ is a domain and if the integral closure $A'$ of $A_{red}$ in its field of fractions is local. We say $A$ is *geometrically unibranch* if $A$ is unibranch and moreover the residue field of $A'$ is purely inseparable over the residue field of $A$.

## 15.106 Local irreducibility

The following definition seems to be the generally accepted one. To parse it, observe that if $A \subset B$ is an integral extension of local domains, then $A \to B$ is a local ring homomorphism by going up (Algebra, Lemma 10.36.22).

Let $A$ be a local ring. Here is an equivalent formulation

$A$ is unibranch if $A$ has a unique minimal prime $\mathfrak p$ and the integral closure of $A/\mathfrak p$ in its fraction field is a local ring, and

$A$ is geometrically unibranch if $A$ has a unique minimal prime $\mathfrak p$ and the integral closure of $A/\mathfrak p$ in its fraction field is a local ring whose residue field is purely inseparable over the residue field of $A$.

A local ring which is normal is geometrically unibranch (follows from Definition 15.106.1 and Algebra, Definition 10.37.11). Lemmas 15.106.3 and 15.106.5 suggest that being (geometrically) unibranch is a reasonable property to look at.

Lemma 15.106.2. Let $A$ be a local ring. Assume $A$ has finitely many minimal prime ideals. Let $A'$ be the integral closure of $A$ in the total ring of fractions of $A_{red}$. Let $A^ h$ be the henselization of $A$. Consider the maps

where $(A')^ h = A' \otimes _ A A^ h$. Then

the left arrow is bijective on maximal ideals,

the right arrow is bijective on minimal primes,

every minimal prime of $(A')^ h$ is contained in a unique maximal ideal and every maximal ideal contains exactly one minimal prime.

**Proof.**
Let $I \subset A$ be the ideal of nilpotents. We have $(A/I)^ h = A^ h/IA^ h$ by (Algebra, Lemma 10.156.2). The spectra of $A$, $A^ h$, $A'$, and $(A')^ h$ are the same as the spectra of $A/I$, $A^ h/IA^ h$, $A'$, and $(A')^ h = A' \otimes _{A/I} A^ h/IA^ h$. Thus we may replace $A$ by $A_{red} = A/I$ and assume $A$ is reduced. Then $A \subset A'$ which we will use below without further mention.

Proof of (1). As $A'$ is integral over $A$ we see that $(A')^ h$ is integral over $A^ h$. By going up (Algebra, Lemma 10.36.22) every maximal ideal of $A'$, resp. $(A')^ h$ lies over the maximal ideal $\mathfrak m$, resp. $\mathfrak m^ h$ of $A$, resp. $A^ h$. Thus (1) follows from the isomorphism

because the residue field extension $\kappa ^ h/\kappa $ induced by $A \to A^ h$ is trivial. We will use below that the displayed ring is integral over a field hence spectrum of this ring is a profinite space, see Algebra, Lemmas 10.36.19 and 10.26.5.

Proof of (3). The ring $A'$ is a normal ring and in fact a finite product of normal domains, see Algebra, Lemma 10.37.16. Since $A^ h$ is a filtered colimit of étale $A$-algebras, $(A')^ h$ is filtered colimit of étale $A'$-algebras hence $(A')^ h$ is a normal ring by Algebra, Lemmas 10.163.9 and 10.37.17. Thus every local ring of $(A')^ h$ is a normal domain and we see that every maximal ideal contains a unique minimal prime. By Lemma 15.11.8 applied to $A^ h \to (A')^ h$ we see that $((A')^ h, \mathfrak m(A')^ h)$ is a henselian pair. If $\mathfrak q \subset (A')^ h$ is a minimal prime (or any prime), then the intersection of $V(\mathfrak q)$ with $V(\mathfrak m (A')^ h)$ is connected by Lemma 15.11.16 Since $V(\mathfrak m (A')^ h) = \mathop{\mathrm{Spec}}((A')^ h \otimes \kappa ^ h)$ is a profinite space by we see there is a unique maximal ideal containing $\mathfrak q$.

Proof of (2). The minimal primes of $A'$ are exactly the primes lying over a minimal prime of $A$ (by construction). Since $A' \to (A')^ h$ is flat by going down (Algebra, Lemma 10.39.19) every minimal prime of $(A')^ h$ lies over a minimal prime of $A'$. Conversely, any prime of $(A')^ h$ lying over a minimal prime of $A'$ is minimal because $(A')^ h$ is a filtered colimit of étale hence quasi-finite algebras over $A'$ (small detail omitted). We conclude that the minimal primes of $(A')^ h$ are exactly the primes which lie over a minimal prime of $A$. Similarly, the minimal primes of $A^ h$ are exactly the primes lying over minimal primes of $A$. By construction we have $A' \otimes _ A Q(A) = Q(A)$ where $Q(A)$ is the total fraction ring of our reduced local ring $A$. Of course $Q(A)$ is the finite product of residue fields of the minimal primes of $A$. It follows that

Our discussion above shows the spectrum of the ring on the left is the set of minimal primes of $(A')^ h$ and the spectrum of the ring on the right is the is the set of minimal primes of $A^ h$. This finishes the proof. $\square$

Lemma 15.106.3. Let $A$ be a local ring. Let $A^ h$ be the henselization of $A$. The following are equivalent

$A$ is unibranch, and

$A^ h$ has a unique minimal prime.

**Proof.**
This follows from Lemma 15.106.2 but we will also give a direct proof. Denote $\mathfrak m$ the maximal ideal of the ring $A$. Recall that the residue field $\kappa = A/\mathfrak m$ is the same as the residue field of $A^ h$.

Assume (2). Let $\mathfrak p^ h$ be the unique minimal prime of $A^ h$. The flatness of $A \to A^ h$ implies that $\mathfrak p = A \cap \mathfrak p^ h$ is the unique minimal prime of $A$ (by going down, see Algebra, Lemma 10.39.19). Also, since $A^ h/\mathfrak pA^ h = (A/\mathfrak p)^ h$ (see Algebra, Lemma 10.156.2) is reduced by Lemma 15.45.4 we see that $\mathfrak p^ h = \mathfrak pA^ h$. Let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. We have to show that $A'$ is local. Since $A \to A'$ is integral, every maximal ideal of $A'$ lies over $\mathfrak m$ (by going up for integral ring maps, see Algebra, Lemma 10.36.22). If $A'$ is not local, then we can find distinct maximal ideals $\mathfrak m_1$, $\mathfrak m_2$. Choose elements $f_1, f_2 \in A'$ with $f_ i \in \mathfrak m_ i$ and $f_ i \not\in \mathfrak m_{3 - i}$. We find a finite subalgebra $B = A[f_1, f_2] \subset A'$ with distinct maximal ideals $B \cap \mathfrak m_ i$, $i = 1, 2$. Note that the inclusions

give, on tensoring with the flat ring map $A \to A^ h$ the inclusions

the last inclusion because $\kappa (\mathfrak p) \otimes _ A A^ h = \kappa (\mathfrak p) \otimes _{A/\mathfrak p} A^ h/\mathfrak p^ h$ is a localization of the domain $A^ h/\mathfrak p^ h$. Note that $B \otimes _ A \kappa $ has at least two maximal ideals because $B/\mathfrak mB$ has two maximal ideals. Hence, as $A^ h$ is henselian we see that $B \otimes _ A A^ h$ is a product of $\geq 2$ local rings, see Algebra, Lemma 10.153.5. But we've just seen that $B \otimes _ A A^ h$ is a subring of a domain and we get a contradiction.

Assume (1). Let $\mathfrak p \subset A$ be the unique minimal prime and let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. Let $A \to B$ be a local map of local rings inducing an isomorphism of residue fields which is a localization of an étale $A$-algebra. In particular $\mathfrak m_ B$ is the unique prime containing $\mathfrak m B$. Then $B' = A' \otimes _ A B$ is integral over $B$ and the assumption that $A \to A'$ is local implies that $B'$ is local (Algebra, Lemma 10.156.5). On the other hand, $A' \to B'$ is the localization of an étale ring map, hence $B'$ is normal, see Algebra, Lemma 10.163.9. Thus $B'$ is a (local) normal domain. Finally, we have

Hence $B/\mathfrak pB$ is a domain, which implies that $B$ has a unique minimal prime (since by flatness of $A \to B$ these all have to lie over $\mathfrak p$). Since $A^ h$ is a filtered colimit of the local rings $B$ it follows that $A^ h$ has a unique minimal prime. Namely, if $fg = 0$ in $A^ h$ for some non-nilpotent elements $f, g$, then we can find a $B$ as above containing both $f$ and $g$ which leads to a contradiction. $\square$

Lemma 15.106.4. Let $(A, \mathfrak m, \kappa )$ be a local ring. Assume $A$ has finitely many minimal prime ideals. Let $A'$ be the integral closure of $A$ in the total ring of fractions of $A_{red}$. Choose an algebraic closure $\overline{\kappa }$ of $\kappa $ and denote $\kappa ^{sep} \subset \overline{\kappa }$ the separable algebraic closure of $\kappa $. Let $A^{sh}$ be the strict henselization of $A$ with respect to $\kappa ^{sep}$. Consider the maps

where $(A')^{sh} = A' \otimes _ A A^{sh}$. Then

for $\mathfrak m' \subset A'$ maximal the residue field $\kappa '$ is algebraic over $\kappa $ and the fibre of $c$ over $\mathfrak m'$ can be canonically identified with $\mathop{\mathrm{Hom}}\nolimits _\kappa (\kappa ', \overline{\kappa })$,

the right arrow is bijective on minimal primes,

every minimal prime of $(A')^{sh}$ is contained in a unique maximal ideal and every maximal ideal contains a unique minimal prime.

**Proof.**
The proof is almost exactly the same as for Lemma 15.106.2. Let $I \subset A$ be the ideal of nilpotents. We have $(A/I)^{sh} = A^{sh}/IA^{sh}$ by (Algebra, Lemma 10.156.2). The spectra of $A$, $A^{sh}$, $A'$, and $(A')^ h$ are the same as the spectra of $A/I$, $A^{sh}/IA^{sh}$, $A'$, and $(A')^{sh} = A' \otimes _{A/I} A^{sh}/IA^{sh}$. Thus we may replace $A$ by $A_{red} = A/I$ and assume $A$ is reduced. Then $A \subset A'$ which we will use below without further mention.

Proof of (1). The field extension $\kappa '/\kappa $ is algebraic because $A'$ is integral over $A$. Since $A'$ is integral over $A$, we see that $(A')^{sh}$ is integral over $A^{sh}$. By going up (Algebra, Lemma 10.36.22) every maximal ideal of $A'$, resp. $(A')^{sh}$ lies over the maximal ideal $\mathfrak m$, resp. $\mathfrak m^{sh}$ of $A$, resp. $A^ h$. We have

because the residue field of $A^{sh}$ is $\kappa ^{sep}$. Thus the fibre of $c$ over $\mathfrak m'$ is the spectrum of $\kappa ' \otimes _\kappa \kappa ^{sep}$. We conclude (1) is true because there is a bijection

We will use below that the displayed ring is integral over a field hence spectrum of this ring is a profinite space, see Algebra, Lemmas 10.36.19 and 10.26.5.

Proof of (3). The ring $A'$ is a normal ring and in fact a finite product of normal domains, see Algebra, Lemma 10.37.16. Since $A^{sh}$ is a filtered colimit of étale $A$-algebras, $(A')^{sh}$ is filtered colimit of étale $A'$-algebras hence $(A')^{sh}$ is a normal ring by Algebra, Lemmas 10.163.9 and 10.37.17. Thus every local ring of $(A')^{sh}$ is a normal domain and we see that every maximal ideal contains a unique minimal prime. By Lemma 15.11.8 applied to $A^{sh} \to (A')^{sh}$ to see that $((A')^{sh}, \mathfrak m(A')^{sh})$ is a henselian pair. If $\mathfrak q \subset (A')^{sh}$ is a minimal prime (or any prime), then the intersection of $V(\mathfrak q)$ with $V(\mathfrak m (A')^{sh})$ is connected by Lemma 15.11.16 Since $V(\mathfrak m (A')^{sh}) = \mathop{\mathrm{Spec}}((A')^{sh} \otimes \kappa ^{sh})$ is a profinite space by we see there is a unique maximal ideal containing $\mathfrak q$.

Proof of (2). The minimal primes of $A'$ are exactly the primes lying over a minimal prime of $A$ (by construction). Since $A' \to (A')^{sh}$ is flat by going down (Algebra, Lemma 10.39.19) every minimal prime of $(A')^{sh}$ lies over a minimal prime of $A'$. Conversely, any prime of $(A')^{sh}$ lying over a minimal prime of $A'$ is minimal because $(A')^{sh}$ is a filtered colimit of étale hence quasi-finite algebras over $A'$ (small detail omitted). We conclude that the minimal primes of $(A')^{sh}$ are exactly the primes which lie over a minimal prime of $A$. Similarly, the minimal primes of $A^{sh}$ are exactly the primes lying over minimal primes of $A$. By construction we have $A' \otimes _ A Q(A) = Q(A)$ where $Q(A)$ is the total fraction ring of our reduced local ring $A$. Of course $Q(A)$ is the finite product of residue fields of the minimal primes of $A$. It follows that

Our discussion above shows the spectrum of the ring on the left is the set of minimal primes of $(A')^{sh}$ and the spectrum of the ring on the right is the is the set of minimal primes of $A^{sh}$. This finishes the proof. $\square$

Lemma 15.106.5. Let $A$ be a local ring. Let $A^{sh}$ be a strict henselization of $A$. The following are equivalent

$A$ is geometrically unibranch, and

$A^{sh}$ has a unique minimal prime.

**Proof.**
This follows from Lemma 15.106.4 but we will also give a direct proof; this direct proof is almost exactly the same as the direct proof of Lemma 15.106.3. Denote $\mathfrak m$ the maximal ideal of the ring $A$. Denote $\kappa $, $\kappa ^{sh}$ the residue field of $A$, $A^{sh}$.

Assume (2). Let $\mathfrak p^{sh}$ be the unique minimal prime of $A^{sh}$. The flatness of $A \to A^{sh}$ implies that $\mathfrak p = A \cap \mathfrak p^{sh}$ is the unique minimal prime of $A$ (by going down, see Algebra, Lemma 10.39.19). Also, since $A^{sh}/\mathfrak pA^{sh} = (A/\mathfrak p)^{sh}$ (see Algebra, Lemma 10.156.4) is reduced by Lemma 15.45.4 we see that $\mathfrak p^{sh} = \mathfrak pA^{sh}$. Let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. We have to show that $A'$ is local and that its residue field is purely inseparable over $\kappa $. Since $A \to A'$ is integral, every maximal ideal of $A'$ lies over $\mathfrak m$ (by going up for integral ring maps, see Algebra, Lemma 10.36.22). If $A'$ is not local, then we can find distinct maximal ideals $\mathfrak m_1$, $\mathfrak m_2$. Choosing elements $f_1, f_2 \in A'$ with $f_ i \in \mathfrak m_ i, f_ i \not\in \mathfrak m_{3 - i}$ we find a finite subalgebra $B = A[f_1, f_2] \subset A'$ with distinct maximal ideals $B \cap \mathfrak m_ i$, $i = 1, 2$. If $A'$ is local with maximal ideal $\mathfrak m'$, but $A/\mathfrak m \subset A'/\mathfrak m'$ is not purely inseparable, then we can find $f \in A'$ whose image in $A'/\mathfrak m'$ generates a finite, not purely inseparable extension of $A/\mathfrak m$ and we find a finite local subalgebra $B = A[f] \subset A'$ whose residue field is not a purely inseparable extension of $A/\mathfrak m$. Note that the inclusions

give, on tensoring with the flat ring map $A \to A^{sh}$ the inclusions

the last inclusion because $\kappa (\mathfrak p) \otimes _ A A^{sh} = \kappa (\mathfrak p) \otimes _{A/\mathfrak p} A^{sh}/\mathfrak p^{sh}$ is a localization of the domain $A^{sh}/\mathfrak p^{sh}$. Note that $B \otimes _ A \kappa ^{sh}$ has at least two maximal ideals because $B/\mathfrak mB$ either has two maximal ideals or one whose residue field is not purely inseparable over $\kappa $, and because $\kappa ^{sh}$ is separably algebraically closed. Hence, as $A^{sh}$ is strictly henselian we see that $B \otimes _ A A^{sh}$ is a product of $\geq 2$ local rings, see Algebra, Lemma 10.153.6. But we've just seen that $B \otimes _ A A^{sh}$ is a subring of a domain and we get a contradiction.

Assume (1). Let $\mathfrak p \subset A$ be the unique minimal prime and let $A'$ be the integral closure of $A/\mathfrak p$ in its fraction field. Let $A \to B$ be a local map of local rings which is a localization of an étale $A$-algebra. In particular $\mathfrak m_ B$ is the unique prime containing $\mathfrak m_ AB$. Then $B' = A' \otimes _ A B$ is integral over $B$ and the assumption that $A \to A'$ is local with purely inseparable residue field extension implies that $B'$ is local (Algebra, Lemma 10.156.5). On the other hand, $A' \to B'$ is the localization of an étale ring map, hence $B'$ is normal, see Algebra, Lemma 10.163.9. Thus $B'$ is a (local) normal domain. Finally, we have

Hence $B/\mathfrak pB$ is a domain, which implies that $B$ has a unique minimal prime (since by flatness of $A \to B$ these all have to lie over $\mathfrak p$). Since $A^{sh}$ is a filtered colimit of the local rings $B$ it follows that $A^{sh}$ has a unique minimal prime. Namely, if $fg = 0$ in $A^{sh}$ for some non-nilpotent elements $f, g$, then we can find a $B$ as above containing both $f$ and $g$ which leads to a contradiction. $\square$

Definition 15.106.6. Let $A$ be a local ring with henselization $A^ h$ and strict henselization $A^{sh}$. The *number of branches of $A$* is the number of minimal primes of $A^ h$ if finite and $\infty $ otherwise. The *number of geometric branches of $A$* is the number of minimal primes of $A^{sh}$ if finite and $\infty $ otherwise.

We spell out the relationship with Definition 15.106.1.

Lemma 15.106.7. Let $(A, \mathfrak m, \kappa )$ be a local ring.

If $A$ has infinitely many minimal prime ideals, then the number of (geometric) branches of $A$ is $\infty $.

The number of branches of $A$ is $1$ if and only if $A$ is unibranch.

The number of geometric branches of $A$ is $1$ if and only if $A$ is geometrically unibranch.

Assume $A$ has finitely many minimal primes and let $A'$ be the integral closure of $A$ in the total ring of fractions of $A_{red}$. Then

the number of branches of $A$ is the number of maximal ideals $\mathfrak m'$ of $A'$,

to get the number of geometric branches of $A$ we have to count each maximal ideal $\mathfrak m'$ of $A'$ with multiplicity given by the separable degree of $\kappa (\mathfrak m')/\kappa $.

**Proof.**
This lemma follows immediately from the definitions, Lemma 15.106.2, Lemma 15.106.4, and Fields, Lemma 9.14.8.
$\square$

Lemma 15.106.8. Let $A \to B$ be a local homomorphism of local rings which is the localization of a smooth ring map.

The number of geometric branches of $A$ is equal to the number of geometric branches of $B$.

If $A \to B$ induces a purely inseparable extension of residue fields, then the number of branches of $A$ is the number of branches of $B$.

**Proof.**
We will use that smooth ring maps are flat (Algebra, Lemma 10.137.10), that localizations are flat (Algebra, Lemma 10.39.18), that compositions of flat ring maps are flat (Algebra, Lemma 10.39.4), that base change of a flat ring map is flat (Algebra, Lemma 10.39.7), that flat local homomorphisms are faithfully flat (Algebra, Lemma 10.39.17), that (strict) henselization is flat (Lemma 15.45.1), and Going down for flat ring maps (Algebra, Lemma 10.39.19).

Proof of (2). Let $A^ h$, $B^ h$ be the henselizations of $A$, $B$. Then $B^ h$ is the henselization of $A^ h \otimes _ A B$ at the unique maximal ideal lying over $\mathfrak m_ B$, see Algebra, Lemma 10.155.8. Thus we may and do assume $A$ is henselian. Since $A \to B \to B^ h$ is flat, every minimal prime of $B^ h$ lies over a minimal prime of $A$ and since $A \to B^ h$ is faithfully flat, every minimal prime of $A$ does lie under a minimal prime of $B^ h$; in both cases use going down for flat ring maps. Therefore it suffices to show that given a minimal prime $\mathfrak p \subset A$, there is at most one minimal prime of $B^ h$ lying over $\mathfrak p$. After replacing $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ we may assume that $A$ is a domain; the $A$ is still henselian by Algebra, Lemma 10.156.2. By Lemma 15.106.3 we see that the integral closure $A'$ of $A$ in its field of fractions is a local domain. Of course $A'$ is a normal domain. By Algebra, Lemma 10.163.9 we see that $A' \otimes _ A B^ h$ is a normal ring (the lemma just gives it for $A' \otimes _ A B$, to go up to $A' \otimes _ A B^ h$ use that $B^ h$ is a colimit of étale $B$-algebras and use Algebra, Lemma 10.37.17). By Algebra, Lemma 10.156.5 we see that $A' \otimes _ A B^ h$ is local (this is where we use the assumption on the residue fields of $A$ and $B$). Hence $A' \otimes _ A B^ h$ is a local normal ring, hence a local domain. Since $B^ h \subset A' \otimes _ A B^ h$ by flatness of $A \to B^ h$ we conclude that $B^ h$ is a domain as desired.

Proof of (1). Let $A^{sh}$, $B^{sh}$ be strict henselizations of $A$, $B$. Then $B^{sh}$ is a strict henselization of $A^ h \otimes _ A B$ at a maximal ideal lying over $\mathfrak m_ B$ and $\mathfrak m_{A^ h}$, see Algebra, Lemma 10.155.12. Thus we may and do assume $A$ is strictly henselian. Since $A \to B \to B^{sh}$ is flat, every minimal prime of $B^{sh}$ lies over a minimal prime of $A$ and since $A \to B^{sh}$ is faithfully flat, every minimal prime of $A$ does lie under a minimal prime of $B^{sh}$; in both cases use going down for flat ring maps. Therefore it suffices to show that given a minimal prime $\mathfrak p \subset A$, there is at most one minimal prime of $B^{sh}$ lying over $\mathfrak p$. After replacing $A$ by $A/\mathfrak p$ and $B$ by $B/\mathfrak p B$ we may assume that $A$ is a domain; then $A$ is still strictly henselian by Algebra, Lemma 10.156.4. By Lemma 15.106.5 we see that the integral closure $A'$ of $A$ in its field of fractions is a local domain whose residue field is a purely inseparable extension of the residue field of $A$. Of course $A'$ is a normal domain. By Algebra, Lemma 10.163.9 we see that $A' \otimes _ A B^{sh}$ is a normal ring (the lemma just gives it for $A' \otimes _ A B$, to go up to $A' \otimes _ A B^{sh}$ use that $B^{sh}$ is a colimit of étale $B$-algebras and use Algebra, Lemma 10.37.17). By Algebra, Lemma 10.156.5 we see that $A' \otimes _ A B^{sh}$ is local (since $A \subset A'$ induces a purely inseparable residue field extension). Hence $A' \otimes _ A B^{sh}$ is a local normal ring, hence a local domain. Since $B^{sh} \subset A' \otimes _ A B^{sh}$ by flatness of $A \to B^{sh}$ we conclude that $B^{sh}$ is a domain as desired. $\square$

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