## 10.49 Valuation rings

Here are some definitions.

Definition 10.49.1. Valuation rings.

1. Let $K$ be a field. Let $A$, $B$ be local rings contained in $K$. We say that $B$ dominates $A$ if $A \subset B$ and $\mathfrak m_ A = A \cap \mathfrak m_ B$.

2. Let $A$ be a ring. We say $A$ is a valuation ring if $A$ is a local domain and if $A$ is maximal for the relation of domination among local rings contained in the fraction field of $A$.

3. Let $A$ be a valuation ring with fraction field $K$. If $R \subset K$ is a subring of $K$, then we say $A$ is centered on $R$ if $R \subset A$.

With this definition a field is a valuation ring.

Lemma 10.49.2. Let $K$ be a field. Let $A \subset K$ be a local subring. Then there exists a valuation ring with fraction field $K$ dominating $A$.

Proof. We consider the collection of local subrings of $K$ as a partially ordered set using the relation of domination. Suppose that $\{ A_ i\} _{i \in I}$ is a totally ordered collection of local subrings of $K$. Then $B = \bigcup A_ i$ is a local subring which dominates all of the $A_ i$. Hence by Zorn's Lemma, it suffices to show that if $A \subset K$ is a local ring whose fraction field is not $K$, then there exists a local ring $B \subset K$, $B \not= A$ dominating $A$.

Pick $t \in K$ which is not in the fraction field of $A$. If $t$ is transcendental over $A$, then $A[t] \subset K$ and hence $A[t]_{(t, \mathfrak m)} \subset K$ is a local ring distinct from $A$ dominating $A$. Suppose $t$ is algebraic over $A$. Then for some $a \in A$ the element $at$ is integral over $A$. In this case the subring $A' \subset K$ generated by $A$ and $ta$ is finite over $A$. By Lemma 10.35.17 there exists a prime ideal $\mathfrak m' \subset A'$ lying over $\mathfrak m$. Then $A'_{\mathfrak m'}$ dominates $A$. If $A = A'_{\mathfrak m'}$, then $t$ is in the fraction field of $A$ which we assumed not to be the case. Thus $A \not= A'_{\mathfrak m'}$ as desired. $\square$

Lemma 10.49.3. Let $A$ be a valuation ring with maximal ideal $\mathfrak m$ and fraction field $K$. Let $x \in K$. Then either $x \in A$ or $x^{-1} \in A$ or both.

Proof. Assume that $x$ is not in $A$. Let $A'$ denote the subring of $K$ generated by $A$ and $x$. Since $A$ is a valuation ring we see that there is no prime of $A'$ lying over $\mathfrak m$. Since $\mathfrak m$ is maximal we see that $V(\mathfrak m A') = \emptyset$. Then $\mathfrak m A' = A'$ by Lemma 10.16.2. Hence we can write $1 = \sum _{i = 0}^ d t_ i x^ i$ with $t_ i \in \mathfrak m$. This implies that $(1 - t_0) (x^{-1})^ d - \sum t_ i (x^{-1})^{d - i} = 0$. In particular we see that $x^{-1}$ is integral over $A$. Thus the subring $A''$ of $K$ generated by $A$ and $x^{-1}$ is finite over $A$ and we see there exists a prime ideal $\mathfrak m'' \subset A''$ lying over $\mathfrak m$ by Lemma 10.35.17. Since $A$ is a valuation ring we conclude that $A = (A'')_{\mathfrak m''}$ and hence $x^{-1} \in A$. $\square$

Lemma 10.49.4. Let $A \subset K$ be a subring of a field $K$ such that for all $x \in K$ either $x \in A$ or $x^{-1} \in A$ or both. Then $A$ is a valuation ring with fraction field $K$.

Proof. If $A$ is not $K$, then $A$ is not a field and there is a nonzero maximal ideal $\mathfrak m$. If $\mathfrak m'$ is a second maximal ideal, then choose $x, y \in A$ with $x \in \mathfrak m$, $y \not\in \mathfrak m$, $x \not\in \mathfrak m'$, and $y \in \mathfrak m'$ (see Lemma 10.14.2). Then neither $x/y \in A$ nor $y/x \in A$ contradicting the assumption of the lemma. Thus we see that $A$ is a local ring. Suppose that $A'$ is a local ring contained in $K$ which dominates $A$. Let $x \in A'$. We have to show that $x \in A$. If not, then $x^{-1} \in A$, and of course $x^{-1} \in \mathfrak m_ A$. But then $x^{-1} \in \mathfrak m_{A'}$ which contradicts $x \in A'$. $\square$

Lemma 10.49.5. Let $I$ be a directed set. Let $(A_ i, \varphi _{ij})$ be a system of valuation rings over $I$. Then $A = \mathop{\mathrm{colim}}\nolimits A_ i$ is a valuation ring.

Proof. It is clear that $A$ is a domain. Let $a, b \in A$. Lemma 10.49.4 tells us we have to show that either $a | b$ or $b | a$ in $A$. Choose $i$ so large that there exist $a_ i, b_ i \in A_ i$ mapping to $a, b$. Then Lemma 10.49.3 applied to $a_ i, b_ i$ in $A_ i$ implies the result for $a, b$ in $A$. $\square$

Lemma 10.49.6. Let $K \subset L$ be an extension of fields. If $B \subset L$ is a valuation ring, then $A = K \cap B$ is a valuation ring.

Proof. We can replace $L$ by the fraction field $F$ of $B$ and $K$ by $K \cap F$. Then the lemma follows from a combination of Lemmas 10.49.3 and 10.49.4. $\square$

Lemma 10.49.7. Let $K \subset L$ be an algebraic extension of fields. If $B \subset L$ is a valuation ring with fraction field $L$ and not a field, then $A = K \cap B$ is a valuation ring and not a field.

Proof. By Lemma 10.49.6 the ring $A$ is a valuation ring. If $A$ is a field, then $A = K$. Then $A = K \subset B$ is an integral extension, hence there are no proper inclusions among the primes of $B$ (Lemma 10.35.20). This contradicts the assumption that $B$ is a local domain and not a field. $\square$

Lemma 10.49.8. Let $A$ be a valuation ring. For any prime ideal $\mathfrak p \subset A$ the quotient $A/\mathfrak p$ is a valuation ring. The same is true for the localization $A_\mathfrak p$ and in fact any localization of $A$.

Proof. Use the characterization of valuation rings given in Lemma 10.49.4. $\square$

Lemma 10.49.9. Let $A'$ be a valuation ring with residue field $K$. Let $A$ be a valuation ring with fraction field $K$. Then $C = \{ \lambda \in A' \mid \lambda \bmod \mathfrak m_{A'} \in A\}$ is a valuation ring.

Proof. Note that $\mathfrak m_{A'} \subset C$ and $C/\mathfrak m_{A'} = A$. In particular, the fraction field of $C$ is equal to the fraction field of $A'$. We will use the criterion of Lemma 10.49.4 to prove the lemma. Let $x$ be an element of the fraction field of $C$. By the lemma we may assume $x \in A'$. If $x \in \mathfrak m_{A'}$, then we see $x \in C$. If not, then $x$ is a unit of $A'$ and we also have $x^{-1} \in A'$. Hence either $x$ or $x^{-1}$ maps to an element of $A$ by the lemma again. $\square$

Lemma 10.49.10. Let $A$ be a valuation ring. Then $A$ is a normal domain.

Proof. Suppose $x$ is in the field of fractions of $A$ and integral over $A$, say $x^{d + 1} + \sum _{i \leq d} a_ i x^ i = 0$. By Lemma 10.49.4 either $x \in A$ (and we're done) or $x^{-1} \in A$. In the second case we see that $x = - \sum a_ i x^{i - d} \in A$ as well. $\square$

Lemma 10.49.11. Let $A$ be a normal domain with fraction field $K$.

1. For every $x \in K$, $x \not\in A$ there exists a valuation ring $A \subset V \subset K$ with fraction field $K$ such that $x \not\in V$.

2. If $A$ is local, we can moreover choose $V$ which dominates $A$.

In other words, $A$ is the intersection of all valuation rings in $K$ containing $A$ and if $A$ is local, then $A$ is the intersection of all valuation rings in $K$ dominating $A$.

Proof. Suppose $x \in K$, $x \not\in A$. Consider $B = A[x^{-1}]$. Then $x \not\in B$. Namely, if $x = a_0 + a_1x^{-1} + \ldots + a_ d x^{-d}$ then $x^{d + 1} - a_0x^ d - \ldots - a_ d = 0$ and $x$ is integral over $A$ in contradiction with the fact that $A$ is normal. Thus $x^{-1}$ is not a unit in $B$. Thus $V(x^{-1}) \subset \mathop{\mathrm{Spec}}(B)$ is not empty (Lemma 10.16.2), and we can choose a prime $\mathfrak p \subset B$ with $x^{-1} \in \mathfrak p$. Choose a valuation ring $V \subset K$ dominating $B_\mathfrak p$ (Lemma 10.49.2). Then $x \not\in V$ as $x^{-1} \in \mathfrak m_ V$.

If $A$ is local, then we claim that $x^{-1} B + \mathfrak m_ A B \not= B$. Namely, if $1 = (a_0 + a_1x^{-1} + \ldots + a_ d x^{-d})x^{-1} + a'_0 + \ldots + a'_ d x^{-d}$ with $a_ i \in A$ and $a'_ i \in \mathfrak m_ A$, then we'd get

$(1 - a'_0) x^{d + 1} - (a_0 + a'_1) x^ d - \ldots - a_ d = 0$

Since $a'_0 \in \mathfrak m_ A$ we see that $1 - a'_0$ is a unit in $A$ and we conclude that $x$ would be integral over $A$, a contradiction as before. Then choose the prime $\mathfrak p \supset x^{-1} B + \mathfrak m_ A B$ we find $V$ dominating $A$. $\square$

An totally ordered abelian group is a pair $(\Gamma , \geq )$ consisting of an abelian group $\Gamma$ endowed with a total ordering $\geq$ such that $\gamma \geq \gamma ' \Rightarrow \gamma + \gamma '' \geq \gamma ' + \gamma ''$ for all $\gamma , \gamma ', \gamma '' \in \Gamma$.

Lemma 10.49.12. Let $A$ be a valuation ring with field of fractions $K$. Set $\Gamma = K^*/A^*$ (with group law written additively). For $\gamma , \gamma ' \in \Gamma$ define $\gamma \geq \gamma '$ if and only if $\gamma - \gamma '$ is in the image of $A - \{ 0\} \to \Gamma$. Then $(\Gamma , \geq )$ is a totally ordered abelian group.

Proof. Omitted, but follows easily from Lemma 10.49.3. Note that in case $A = K$ we obtain the zero group $\Gamma = \{ 0\}$ endowed with its unique total ordering. $\square$

Definition 10.49.13. Let $A$ be a valuation ring.

1. The totally ordered abelian group $(\Gamma , \geq )$ of Lemma 10.49.12 is called the value group of the valuation ring $A$.

2. The map $v : A - \{ 0\} \to \Gamma$ and also $v : K^* \to \Gamma$ is called the valuation associated to $A$.

3. The valuation ring $A$ is called a discrete valuation ring if $\Gamma \cong \mathbf{Z}$.

Note that if $\Gamma \cong \mathbf{Z}$ then there is a unique such isomorphism such that $1 \geq 0$. If the isomorphism is chosen in this way, then the ordering becomes the usual ordering of the integers.

Lemma 10.49.14. Let $A$ be a valuation ring. The valuation $v : A -\{ 0\} \to \Gamma _{\geq 0}$ has the following properties:

1. $v(a) = 0 \Leftrightarrow a \in A^*$,

2. $v(ab) = v(a) + v(b)$,

3. $v(a + b) \geq \min (v(a), v(b))$.

Proof. Omitted. $\square$

Lemma 10.49.15. Let $A$ be a ring. The following are equivalent

1. $A$ is a valuation ring,

2. $A$ is a local domain and every finitely generated ideal of $A$ is principal.

Proof. Assume $A$ is a valuation ring and let $f_1, \ldots , f_ n \in A$. Choose $i$ such that $v(f_ i)$ is minimal among $v(f_ j)$. Then $(f_ i) = (f_1, \ldots , f_ n)$. Conversely, assume $A$ is a local domain and every finitely generated ideal of $A$ is principal. Pick $f, g \in A$ and write $(f, g) = (h)$. Then $f = ah$ and $g = bh$ and $h = cf + dg$ for some $a, b, c, d \in A$. Thus $ac + bd = 1$ and we see that either $a$ or $b$ is a unit, i.e., either $g/f$ or $f/g$ is an element of $A$. This shows $A$ is a valuation ring by Lemma 10.49.4. $\square$

Lemma 10.49.16. Let $(\Gamma , \geq )$ be a totally ordered abelian group. Let $K$ be a field. Let $v : K^* \to \Gamma$ be a homomorphism of abelian groups such that $v(a + b) \geq \min (v(a), v(b))$ for $a, b \in K$ with $a, b, a + b$ not zero. Then

$A = \{ x \in K \mid x = 0 \text{ or } v(x) \geq 0 \}$

is a valuation ring with value group $\mathop{\mathrm{Im}}(v) \subset \Gamma$, with maximal ideal

$\mathfrak m = \{ x \in K \mid x = 0 \text{ or } v(x) > 0 \}$

and with group of units

$A^* = \{ x \in K^* \mid v(x) = 0 \} .$

Proof. Omitted. $\square$

Let $(\Gamma , \geq )$ be a totally ordered abelian group. An ideal of $\Gamma$ is a subset $I \subset \Gamma$ such that all elements of $I$ are $\geq 0$ and $\gamma \in I$, $\gamma ' \geq \gamma$ implies $\gamma ' \in I$. We say that such an ideal is prime if $\gamma + \gamma ' \in I, \gamma , \gamma ' \geq 0 \Rightarrow \gamma \in I \text{ or } \gamma ' \in I$.

Lemma 10.49.17. Let $A$ be a valuation ring. Ideals in $A$ correspond $1 - 1$ with ideals of $\Gamma$. This bijection is inclusion preserving, and maps prime ideals to prime ideals.

Proof. Omitted. $\square$

Lemma 10.49.18. A valuation ring is Noetherian if and only if it is a discrete valuation ring or a field.

Proof. Suppose $A$ is a discrete valuation ring with valuation $v : A \setminus \{ 0\} \to \mathbf{Z}$ normalized so that $\mathop{\mathrm{Im}}(v) = \mathbf{Z}_{\geq 0}$. By Lemma 10.49.17 the ideals of $A$ are the subsets $I_ n = \{ 0\} \cup v^{-1}(\mathbf{Z}_{\geq n})$. It is clear that any element $x \in A$ with $v(x) = n$ generates $I_ n$. Hence $A$ is a PID so certainly Noetherian.

Suppose $A$ is a Noetherian valuation ring with value group $\Gamma$. By Lemma 10.49.17 we see the ascending chain condition holds for ideals in $\Gamma$. We may assume $A$ is not a field, i.e., there is a $\gamma \in \Gamma$ with $\gamma > 0$. Applying the ascending chain condition to the subsets $\gamma + \Gamma _{\geq 0}$ with $\gamma > 0$ we see there exists a smallest element $\gamma _0$ which is bigger than $0$. Let $\gamma \in \Gamma$ be an element $\gamma > 0$. Consider the sequence of elements $\gamma$, $\gamma - \gamma _0$, $\gamma - 2\gamma _0$, etc. By the ascending chain condition these cannot all be $> 0$. Let $\gamma - n \gamma _0$ be the last one $\geq 0$. By minimality of $\gamma _0$ we see that $0 = \gamma - n \gamma _0$. Hence $\Gamma$ is a cyclic group as desired. $\square$

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