The Stacks project

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

  1. the left arrow is bijective on maximal ideals,

  2. the right arrow is bijective on minimal primes,

  3. 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)

Comment #5905 by Dario Weißmann on

typo in (3): .. contains exact one minimal prime -> exactly

There are also:

  • 2 comment(s) on Section 15.105: Local irreducibility

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0C24. Beware of the difference between the letter 'O' and the digit '0'.