The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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 $\mathop{\mathrm{Im}}(v) \subset \mathbf{Z}_{\geq 0}$. 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. $\square$


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 00I8. Beware of the difference between the letter 'O' and the digit '0'.