## 15.123 Extensions of valuation rings

This section is the analogue of Section 15.111 for general valuation rings.

Definition 15.123.1. We say that $A \to B$ or $A \subset B$ is an *extension of valuation rings* if $A$ and $B$ are valuation rings and $A \to B$ is injective and local. Such an extension induces a commutative diagram

\[ \xymatrix{ A \setminus \{ 0\} \ar[r] \ar[d]_ v & B \setminus \{ 0\} \ar[d]^ v \\ \Gamma _ A \ar[r] & \Gamma _ B } \]

where $\Gamma _ A$ and $\Gamma _ B$ are the value groups. We say that $B$ is *weakly unramified* over $A$ if the lower horizontal arrow is a bijection. If the extension of residue fields $\kappa _ A = A/\mathfrak m_ A \subset \kappa _ B = B/\mathfrak m_ B$ is finite, then we set $f = [\kappa _ B : \kappa _ A]$ and we call it the *residual degree* or *residue degree* of the extension $A \subset B$.

Note that $\Gamma _ A \to \Gamma _ B$ is injective, because the units of $A$ are the inverse of the units of $B$ under the map $A \to B$. Note also, that we do not require the extension of fraction fields to be finite.

Lemma 15.123.2. Let $A \subset B$ be an extension of valuation rings with fraction fields $K \subset L$. If the extension $L/K$ is finite, then the residue field extension is finite, the index of $\Gamma _ A$ in $\Gamma _ B$ is finite, and

\[ [\Gamma _ B : \Gamma _ A] [\kappa _ B : \kappa _ A] \leq [L : K]. \]

**Proof.**
Let $b_1, \ldots , b_ n \in B$ be units whose images in $\kappa _ B$ are linearly independent over $\kappa _ A$. Let $c_1, \ldots , c_ m \in B$ be nonzero elements whose images in $\Gamma _ B/\Gamma _ A$ are pairwise distinct. We claim that $b_ i c_ j$ are $K$-linearly independent in $L$. Namely, we claim a sum

\[ \sum a_{ij} b_ i c_ j \]

with $a_{ij} \in K$ not all zero cannot be zero. Choose $(i_0, j_0)$ with $v(a_{i_0j_0}b_{i_0}c_{j_0})$ minimal. Replace $a_{ij}$ by $a_{ij}/a_{i_0j_0}$, so that $a_{i_0 j_0} = 1$. Let

\[ P = \{ (i, j) \mid v(a_{ij}b_ ic_ j) = v(a_{i_0j_0}b_{i_0}c_{j_0}) \} \]

By our choice of $c_1, \ldots , c_ m$ we see that $(i, j) \in P$ implies $j = j_0$. Hence if $(i, j) \in P$, then $v(a_{ij}) = v(a_{i_0j_0}) = 0$, i.e., $a_{ij}$ is a unit. By our choice of $b_1, \ldots , b_ n$ we see that

\[ \sum \nolimits _{(i, j) \in P} a_{ij}b_ i \]

is a unit in $B$. Thus the valuation of $\sum \nolimits _{(i, j) \in P} a_{ij}b_ ic_ j$ is $v(c_{j_0}) = v(a_{i_0j_0}b_{i_0}c_{j_0})$. Since the terms with $(i, j) \not\in P$ in the first displayed sum have strictly bigger valuation, we conclude that this sum cannot be zero, thereby proving the lemma.
$\square$

Lemma 15.123.3. Let $A$ be a valuation ring with fraction field $K$ of characteristic $p > 0$. Let $L/K$ be a purely inseparable extension. Then the integral closure $B$ of $A$ in $L$ is a valuation ring with fraction field $L$ and $A \subset B$ is an extension of valuation rings.

**Proof.**
Omitted. Hints: use Algebra, Lemmas 10.50.5 and 10.36.17 for example.
$\square$

Lemma 15.123.4. Let $A \to B$ be a flat local homomorphism of Noetherian local normal domains. Let $f \in A$ and $h \in B$ such that $f = w h^ n$ for some $n > 1$ and some unit $w$ of $B$. Assume that for every height $1$ prime $\mathfrak p \subset A$ there is a height $1$ prime $\mathfrak q \subset B$ lying over $\mathfrak p$ such that the extension $A_\mathfrak p \subset B_\mathfrak q$ is weakly unramified. Then $f = u g^ n$ for some $g \in A$ and unit $u$ of $A$.

**Proof.**
The local rings of $A$ and $B$ at height $1$ primes are discrete valuation rings (Algebra, Lemma 10.119.7). Thus the assumption makes sense (via Definition 15.111.1). Let $\mathfrak p_1, \ldots , \mathfrak p_ r$ be the primes of $A$ minimal over $f$. These have height $1$ by Algebra, Lemma 10.60.11. For each $i$ let $\mathfrak q_{i, j} \subset B$, $j = 1, \ldots , r_ i$ be the height $1$ primes of $B$ lying over $\mathfrak p_ i$. Say we number them so that $A_{\mathfrak p_ i} \to B_{\mathfrak q_{i, 1}}$ is weakly unramified. Since $f$ maps to an $n$th power times a unit in $B_{\mathfrak q_{i, 1}}$ we see that the valuation $v_ i$ of $f$ in $A_{\mathfrak p_ i}$ is divisible by $n$. Say $v_ i = n w_ i$ for some $w_ i \geq 0$. Consider the exact sequence

\[ 0 \to I \to A \to \prod \nolimits _{i = 1, \ldots , r} A_{\mathfrak p_ i}/\mathfrak p_ i^{w_ i}A_{\mathfrak p_ i} \]

defining the ideal $I$. Applying the exact functor $- \otimes _ A B$ we obtain an exact sequence

\[ 0 \to I \otimes _ A B \to B \to \prod \nolimits _{i = 1, \ldots , r} (A_{\mathfrak p_ i}/\mathfrak p_ i^{w_ i}A_{\mathfrak p_ i}) \otimes _ A B \]

Fix $i$. We claim that the canonical map

\[ (A_{\mathfrak p_ i}/\mathfrak p_ i^{w_ i}A_{\mathfrak p_ i}) \otimes _ A B \to \prod \nolimits _{j = 1, \ldots , r_ i} B_{\mathfrak q_{i, j}}/\mathfrak q_{i, j}^{e_{i, j}w_ i}B_{\mathfrak q_{i, j}} \]

is injective. Here $e_{i, j}$ is the ramification index of $A_{\mathfrak p_ i} \to B_{\mathfrak q_{i, j}}$. The claim asserts that $\mathfrak p_ i^{w_ i}B_{\mathfrak p_ i}$ is equal to the set of elements $b$ of $B_{\mathfrak p_ i}$ whose valuation at $\mathfrak q_{i, j}$ is $\geq e_{i, j}w_ i$. Choose a generator $a \in A_{\mathfrak p_ i}$ of the principal ideal $\mathfrak p_ i^{w_ i}$. Then the valuation of $a$ at $\mathfrak q_{i, j}$ is equal to $e_{i, j}w_ i$. Hence, as $B_{\mathfrak p_ i}$ is a normal domain whose height one primes are the primes $\mathfrak q_{i, j}$, $j = 1, \ldots , r_ i$, we see that, for $b$ as above, we have $b/a \in B_{\mathfrak p_ i}$ by Algebra, Lemma 10.157.6. Thus the claim.

The claim combined with the second exact sequence above determines an exact sequence

\[ 0 \to I \otimes _ A B \to B \to \prod \nolimits _{i = 1, \ldots , r} \prod \nolimits _{j = 1, \ldots , r_ i} B_{\mathfrak q_{i, j}}/\mathfrak q_{i, j}^{e_{i, j}w_ i}B_{\mathfrak q_{i, j}} \]

It follows that $I \otimes _ A B$ is the set of elements $h'$ of $B$ which have valuation $\geq e_{i, j}w_ i$ at $\mathfrak q_{i, j}$. Since $f = wh^ n$ in $B$ we see that $h$ has valuation $e_{i, j}w_ i$ at $\mathfrak q_{i, j}$. Thus $h'/h \in B$ by Algebra, Lemma 10.157.6. It follows that $I \otimes _ A B$ is a free $B$-module of rank $1$ (generated by $h$). Therefore $I$ is a free $A$-module of rank $1$, see Algebra, Lemma 10.78.6. Let $g \in I$ be a generator. Then we see that $g$ and $h$ differ by a unit in $B$. Working backwards we conclude that the valuation of $g$ in $A_{\mathfrak p_ i}$ is $w_ i = v_ i/n$. Hence $g^ n$ and $f$ differ by a unit in $A$ (by Algebra, Lemma 10.157.6) as desired.
$\square$

Lemma 15.123.5. 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 15.123.4 to the ring map $A_ i \to (B_ i)_{\mathfrak m_ i}$ to conclude. The hypotheses of the lemma are satisfied because:

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

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

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

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,

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.
$\square$

Lemma 15.123.6. Let $A$ be a valuation ring. Let $A^ h$, resp. $A^{sh}$ be its henselization, resp. strict henselization. Then

\[ A \subset A^ h \subset A^{sh} \]

are extensions of valuation rings which induce bijections on value groups, i.e., which are weakly unramified.

**Proof.**
Write $A^ h = \mathop{\mathrm{colim}}\nolimits (B_ i)_{\mathfrak q_ i}$ where $A \to B_ i$ is étale and $\mathfrak q_ i \subset B_ i$ is a prime ideal lying over $\mathfrak m_ A$, see Algebra, Lemma 10.155.7. Then Lemma 15.123.5 tells us that $(B_ i)_{\mathfrak q_ i}$ is a valuation ring and that the induced map

\[ (A \setminus \{ 0\} )/A^* \longrightarrow ((B_ i)_{\mathfrak q_ i} \setminus \{ 0\} ) / (B_ i)_{\mathfrak q_ i}^* \]

is bijective. By Algebra, Lemma 10.50.6 we conclude that $A^ h$ is a valuation ring. It also follows that $(A \setminus \{ 0\} )/A^* \to (A^ h \setminus \{ 0\} )/(A^ h)^*$ is bijective. This proves the lemma for the inclusion $A \subset A^ h$. To prove it for $A \subset A^{sh}$ we can use exactly the same argument except we replace Algebra, Lemma 10.155.7 by Algebra, Lemma 10.155.11. Since $A^{sh} = (A^ h)^{sh}$ we see that this also proves the assertions of the lemma for the inclusion $A^ h \subset A^{sh}$.
$\square$

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