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

$\mathop{\mathrm{Spec}}(A') \xleftarrow {c} \mathop{\mathrm{Spec}}((A')^{sh}) \xrightarrow {e} \mathop{\mathrm{Spec}}(A^{sh})$

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

1. 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 $\mathop{\mathrm{Hom}}\nolimits _\kappa (\kappa ', \overline{\kappa })$,

2. the right arrow is bijective on minimal primes,

3. 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

$(A')^{sh} \otimes _{A^{sh}} \kappa ^{sep} = A' \otimes _ A A^ h \otimes _{A^ h} \kappa ^{sep} = (A' \otimes _ A \kappa ) \otimes _{\kappa } \kappa ^{sep}$

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

$\mathop{\mathrm{Hom}}\nolimits _\kappa (\kappa ', \overline{\kappa }) \to \mathop{\mathrm{Spec}}(\kappa ' \otimes _\kappa \kappa ^{sep}),\quad \sigma \mapsto \mathop{\mathrm{Ker}}( \sigma \otimes 1 : \kappa ' \otimes _\kappa \kappa ^{sep} \to \overline{\kappa } )$

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

$(A')^{sh} \otimes _ A Q(A) = A^{sh} \otimes _ A A' \otimes _ A Q(A) = A^{sh} \otimes _ A Q(A)$

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. $\square$

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