The Stacks project

Lemma 15.123.4. Let $A$ be a valuation ring. Let $A \to B$ be an étale ring map and let $\mathfrak m \subset B$ be a prime lying over the maximal ideal of $A$. Then $A \subset B_\mathfrak m$ is an extension of valuation rings which is weakly unramified.

Proof. The ring $A$ has weak dimension $\leq 1$ by Lemma 15.104.18. Then $B$ has weak dimension $\leq 1$ by Lemmas 15.104.4 and 15.104.14. hence the local ring $B_\mathfrak m$ is a valuation ring by Lemma 15.104.18. Since the extension $A \subset B_\mathfrak m$ induces a finite extension of fraction fields, we see that the $\Gamma _ A$ has finite index in the value group of $B_{\mathfrak m}$. Thus for every $h \in B_\mathfrak m$ there exists an $n > 0$, an element $f \in A$, and a unit $w \in B_\mathfrak m$ such that $f = w h^ n$ in $B_\mathfrak m$. We will show that this implies $f = ug^ n$ for some $g \in A$ and unit $u \in A$; this will show that the value groups of $A$ and $B_\mathfrak m$ agree, as claimed in the lemma.

Write $A = \mathop{\mathrm{colim}}\nolimits A_ i$ as the colimit of its local subrings which are essentially of finite type over $\mathbf{Z}$. Since $A$ is a normal domain (Algebra, Lemma 10.50.3), we may assume that each $A_ i$ is normal (here we use that taking normalizations the local rings remain essentially of finite type over $\mathbf{Z}$ by Algebra, Proposition 10.162.16). For some $i$ we can find an étale extension $A_ i \to B_ i$ such that $B = A \otimes _{A_ i} B_ i$, see Algebra, Lemma 10.143.3. Let $\mathfrak m_ i$ be the intersection of $B_ i$ with $\mathfrak m$. Then we may apply Lemma 15.123.3 to the ring map $A_ i \to (B_ i)_{\mathfrak m_ i}$ to conclude. The hypotheses of the lemma are satisfied because:

  1. $A_ i$ and $(B_ i)_{\mathfrak m_ i}$ are Noetherian as they are essentially of finite type over $\mathbf{Z}$,

  2. $A_ i \to (B_ i)_{\mathfrak m_ i}$ is flat as $A_ i \to B_ i$ is étale,

  3. $B_ i$ is normal as $A_ i \to B_ i$ is étale, see Algebra, Lemma 10.163.9,

  4. for every height $1$ prime of $A_ i$ there exists a height $1$ prime of $(B_ i)_{\mathfrak m_ i}$ lying over it by Algebra, Lemma 10.113.2 and the fact that $\mathop{\mathrm{Spec}}((B_ i)_{\mathfrak m_ i}) \to \mathop{\mathrm{Spec}}(A_ i)$ is surjective,

  5. the induced extensions $(A_ i)_\mathfrak p \to (B_ i)_\mathfrak q$ are unramified for every prime $\mathfrak q$ lying over a prime $\mathfrak p$ as $A_ i \to B_ i$ is étale.

This concludes the proof of the lemma. $\square$

Comments (0)

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 0ASJ. Beware of the difference between the letter 'O' and the digit '0'.