Lemma 15.105.2. 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

\[ \mathop{\mathrm{Spec}}(A') \leftarrow \mathop{\mathrm{Spec}}((A')^ h) \rightarrow \mathop{\mathrm{Spec}}(A^ h) \]

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

\[ (A')^ h \otimes _{A^ h} \kappa ^ h = A' \otimes _ A A^ h \otimes _{A^ h} \kappa ^ h = A' \otimes _ A \kappa \]

because the residue field extension $\kappa \subset \kappa ^ h$ 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

\[ (A')^ h \otimes _ A Q(A) = A^ h \otimes _ A A' \otimes _ A Q(A) = A^ h \otimes _ A Q(A) \]

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.
$\square$

## Comments (2)

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