nLab valuation ring

Definition and Properties

A valuation ring is an integral domain $O$ which satisfies any of the following equivalent conditions:

1. for every nonzero element $x$ in its field of fractions $K$, either $x \in O$ or $x^{-1} \in O$;

2. the ideals of $O$ are totally ordered by inclusion;

3. the principal ideals of $O$ are totally ordered by inclusion.

These equivalences are not difficult to establish. For example, if 1. holds and $I, J$ are distinct ideals, then some element $x$ belongs to one but not the other, say $x \in J$, $x \notin I$. Then for each $y \in I$, the condition $x/y \in O$ leads to $x = y(x/y) \in I$ which is false; therefore $y/x \in O$, whence $y = x(y/x) \in J$, and we conclude $I \subset J$. That 2. implies 3. is trivial, and if 3. holds and $x \in K$, write $x = a/b$ where $a, b \in O$, and conclude either $(a) \subseteq (b)$ or $(b) \subseteq (a)$, where $x \in O$ in the former case and $x^{-1} \in O$ in the latter.

Proposition

A valuation ring $O$ is a local ring. Its maximal ideal is said to be the valuation ideal?.

Proof

Suppose $x$, $y$ are nonzero non-invertible elements of $O$. Either $x/y$ or $y/x$ belongs to $O$, say $x/y$. Then $(x+y)/y$ belongs to $O$ as well. If $x+y$ were a unit of $O$, it would follow that $1/y$ belongs to $O$, i.e., both $y$ and $1/y$ belong to $O$, so that $y$ is a unit of $O$, contradiction. It follows that non-unit elements of $O$ are closed under addition. It is also clear that if $x$ is a non-unit element of $O$ and $y$ is any element of $O$, then $x y$ is a non-unit element of $O$. Therefore the non-units form an ideal of $O$, clearly the unique maximal ideal of $O$.

Proposition

A valuation ring $O$ is integrally closed in its field of fractions $F$.

Proof

Suppose $x \in F$ satisfies an equation $x^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0$ where the $a_i$ belong to $O$. Either $x$ or $1/x$ belongs to $O$, and if $1/x$ belongs to $O$, then so does $x$ because

$$-x = a_{n-1} + a_{n-1}x^{-1} + \ldots + a_0 x^{-n+1}$$

and this completes the proof.

Theorem

A valuation ring is a Prüfer domain?.

Proof

There are many characterizations of Prüfer domains, but one is that the lattice of ideals is distributive, which is obviously the case if the lattice is linearly ordered.

Example: germs of definable functions

Local rings often arise as stalks of sheaves of real or complex-valued functions, and similarly, some of the more interesting examples of valuation rings arise by considering germs of functions at an “ideal” point, for example at an ultrafilter or infinite point where the functions are not formally defined. Examples such as these are often rich sources of rings and fields with infinitesimal elements. The following example should give the flavor of this phenomenon.

Consider the class of all functions $\mathbb{R} \to \mathbb{R}$ which can be defined by a first-order formula, starting with the basic operations $+, -, \cdot, \exp$, any constant $a \in \mathbb{R}$, and the relations $\lt$ and $=$. This is an enormous class of functions: it includes the logarithm function and any function which can be built from polynomials, exponentials, and logarithms using the four basic arithmetic operations and composition, and also implicitly defined functions such as:

$$f(x) = max \{y: y^5 - (e^{e^x - (\log x)^2 + 1/x})y^2 + \log(1 + x^{\sqrt{2}}))y - 17 = 0\}$$

and many, many more. Of interest are rates of growth (cf. O notation?) of these functions, or, at an even more refined level, the precise ordering of these functions $f(x) \leq g(x)$ when $x$ is large. A quite remarkable and deep fact is the following:

Theorem

Any such definable function $f(x)$ is either positive for all sufficiently large $x$, $0$ for all sufficiently large $x$, or negative for all sufficiently large $x$.

This theorem implies that the germs at infinity of such functions ($\sim$-equivalence classes of functions where $f \sim g$ if $f(x) = g(x)$ for all sufficiently large $x$) form a totally ordered field, in fact a real closed field. The real numbers are embedded in this field as germs of constant functions, but lying between ordinary real numbers are other “numbers” infinitesimally close to reals, such as $2 - [1/x]$, $1 + [1/\log \log(x)]$, as well as infinite numbers such as $[e^x]$.

Sitting inside this field $Germ(\mathbb{R}_{exp})$ is the valuation ring of germs of bounded definable functions, in other words the ring of finite “numbers”, which contains infinitesimals of incredibly rich variety.

Such examples are close in spirit to hyperreal numbers, which form a considerably larger real closed field. In this case, the procedure is similar, except that one takes germs of all functions $\mathbb{N} \to \mathbb{R}$ in the neighborhood of a non-principal ultrafilter on $\mathbb{N}$, which can be considered an ideal point at “infinity”. This is called an ultrapower of the standard real numbers. Again the finite hyperreals form a valuation ring sitting inside.

The valuation function

(also called multiplicative or exponential valuation)

The nonzero elements of $K$ may be partially ordered as follows: write $x \leq y$ if $x/y$ belongs to $O$. For any two nonzero elements $x$, $y$ of $K$, exactly one of the following conditions holds:

• $x/y$ is a non-unit of $O$;

• $x/y$ is a unit of $O$: $x/y$ and $y/x$ belong to $O$;

• $y/x$ is a non-unit of $O$.

We call this the trichotomy law. Thus, if we write $O^*$ for the units of $O$, it follows from trichotomy that the partial order on $K^*$ descends to a total order on the quotient group $G = K^*/O^*$. This totally ordered group is called the value group of the valuation ring $O$. When the value group is isomorphic to $\mathbb{Z}$, the ring $O$ is called a discrete valuation ring. Many local rings which arise in practice, for example localizations of rings of algebraic integers $O$ with respect to a prime ideal $\mathcal{p}$, or their completions as rings of $\mathcal{p}$-adic integers $O_{(\mathcal{p})}$, are discrete valuation rings.

We may then define a valuation function

$$v: K \to K^*/O^* \cup \{0\}$$

where $v(x)$ is the coset $x O^*$ if $x \in K^*$, and $v(0) = 0$. The codomain becomes a totally ordered monoid if $0$ is regarded as the bottom element and is absorbent ($0 x = 0$ for all $x$). This function satisfies the following conditions:

1. $v(x) = 0$ if and only if $x = 0$;

2. $v(x y) = v(x)v(y)$;

3. $v(x + y) \leq \max \{v(x), v(y)\}$

4. $v$ is surjective.

Example: rates of growth

Let us return to our earlier example of a valuation ring $O$, consisting of germs of bounded functions which are first-order definable in the theory of the reals as ordered field with exponentiation. Two “numbers” $[f]$, $[g]$ have the same coset, $[f]O^* = [g]O^*$, if both $[f/g]$ and $[g/f]$ are bounded. But this is precisely to say $f$ is $O(g)$ and $g$ is $O(f)$.

Thus, the elements of the value group in this case can be described as the various “rates of growth” of definable functions. The order relation is that $[f]O^* \leq [g]O^*$ if $f$ is $O(g)$. Thus the classical analysis notion of ‘O notation’ fits within the theory of valuation rings.

Rates of growth, as elements of the value group $K^*/O^*$, can also be regarded as “numbers” containing infinitesimal and infinite quantities. Thus, ordinary real numbers $a$ would correspond to rates of growth of power functions $x^a$, whereas the rate of growth $\log x$ is infinitesimal and the rate of growth of $e^x$ is infinite. The fact that rates of growth of definable functions are totally ordered is essentially due to G.H. Hardy, and in his honor, fields of germs of definable functions are frequently called “Hardy fields”.

• Hardy actually studied the class of functions definable from polynomials, exponential, and logarithmic functions using the four arithmetic operations and composition. This of course is a small subclass of the definable functions considered above, but contains all functions likely to be of interest in analytic number theory (for example).

Valuation rings from valuation functions

Quite generally, we may define a valuation on a field $K$ to be a function

$$v: K \to G \cup \{0\}$$

(where $G$ is a totally ordered group, extended to a totally ordered monoid $G \cup \{0\}$ as above), satisfying conditions 1 - 4 listed above. Two valuations $v$, $v'$ are equivalent if there is an isomorphism

$$\phi: G \cup \{0\} \to G' \cup \{0\}$$

of totally ordered monoids such that $v' = \phi \circ v$. In fact, valuations $v$ may be preordered: if we regard $v$ as a special sort of group homomorphism $v: K^* \to G$, then define $v \leq v'$ if there is a surjective homomorphism of ordered groups $\phi: G \to G'$ such that $v' = \phi \circ v$.

From the data of a valuation on $K$, we may construct a valuation ring $O_v$ inside $K$:

$$O_v = \{x \in K: v(x) \leq 1\}$$

where $1$ is the identity element of $G$. (The fact that $O_v$ is a ring follows straightforwardly from the axioms 1 - 3, and that $O_v$ is a valuation ring follows from the fact that $G$ is totally ordered.)

In summary, we have the following result whose proof is straightforward:

Proposition

There is a natural bijective correspondence between equivalence classes of valuations on $K$ and valuation rings in $K$.

To express the naturality more precisely: there are two functors

$$V: Field^{op} \to Pos, \qquad V': Field^{op} \to Pos$$

from fields to posets. Here $V$ assigns to a field $K$ the poset of equivalence classes of valuations on $K$, where the partial order is inherited from the preorder described above; $V'$ assigns to a field $K$ the poset of valuation rings inside $K$, ordered by inclusion. If $i: K \to K'$ is a field homomorphism, then $V'(i): V'(K') \to V'(K)$ takes a valuation ring $O' \subseteq K$ to the valuation ring $i^{-1}(O') \subseteq K$, and $V(i)$ is defined similarly. The proposition asserts that the functors $V$, $V'$ are naturally isomorphic. We freely conflate them, denoting either functor as $Val: Field^{op} \to Pos$.

Examples

1. The algebraic notion of Riemann surface from 19th century is constructed in a way in which valuations are used to construct points of Riemann surfaces over various fields. See Riemann surface via valuations.

2. There is a very general construction which takes as input an arbitrary field $k$ and a totally ordered abelian group $G$, and produces as output a valuation ring whose value group is naturally identified with $G$. This is the ring of Hahn series, see there.

Equivalent characterisations

Proposition

(Homological characterisation) For a commutative local ring $V$, the following are equivalent :

1. $V$ is a valuation ring ;
2. all torsion-free $V$-modules are flat ;
3. the weak global dimension? of $V$ satisfies $wgldim(V) \leq 1$ ;
4. $V$ is semi-hereditary?.

Proof

$1 \Rightarrow 2$. Let $M$ be a torsion-free $V$-module and let $\sum_{0 \leq i\leq n} v_i m^i = 0$ be a linear relation in $M$. One can assume that $v_0 \in V$ is the element with the minimal valuation. It follows that for every $i \gt 0$ , $v_i$ is divisible by $v_0$ in V. Now since $M$ is torsion-free, it means that we can divide the whole linear equation by $v_0$ and reduce to the case where $v_0 = 1$, that is $m^0 + \sum_{1 \leq i \leq n} v_i m^i = 0$. Let $b_i \coloneqq m_i$ for $i \gt 0$ and define $a^i_j$ for $0 \leq i \leq n$ and $0 \lt j \leq n$ with $a^{j+1}_j \coloneqq 1$, $a^0_j \coloneqq - v_j$ for every $0 \lt j$ and $a^i_j \coloneqq 0$ otherwise. Then $m^i = \sum_j a^i_j b^j$ for every $0 \leq i$ and $\sum_i v_i a^i_j = 0$ for every $0 \lt j$.

$2 \Rightarrow 3$. We must show that $Tor_2^V(M, N) = 0$ for every $V$-modules $M, N$. Since $Tor_2$ commutes with filtered colimits, it is enough to show this for every finitely generated $M$. Using the long exact sequence for Tor, one can further reduce to the case where $M$ is a cyclic module $V/I$. By assumption, since all ideals $I \subset V$ are torsion-free, they are flat, that is $Tor_1^V(I, N) = 0$ for every $V$-module $N$. Tensoring the short exact sequence $0 \to I \to V \to V/I \to 0$ with $N$, one gets $Tor_2^V(V/I, N) = Tor_1^V(I, N) = 0$ for every $N$.

$3 \Rightarrow 4$. Let $I \subset V$ be a finitely generated ideal. By assumption $Tor_2^V(V/I, M) = 0$ for every $V$-module $M$. So $Tor_1^V(I, M) = 0$ for every $M$ ; $I$ is flat. Finitely generated flat modules over a local ring are free.

$4 \Rightarrow 1$. Let $I \subset V$ be a finitely generated ideal. Since $V$ is semi-hereditary, $I$ is projective. Because $V$ is local this means that $I$ is free ; this is possible only if $I$ is principal and generated by a regular element of $V$ or the ideal $(0)$. It follows that $V$ must be an integral domain and that every finitely generated ideal is principal. Let $x, y \in V$ be distinct, since $(x, y)$ is finitely generated, there is a regular $z \in V$ such that $(x) + (y) = (z)$. Then there exists $\alpha, \beta, \lambda, \mu$ such that $x = \alpha z$, $y = \beta z$ and $z = \lambda x + \mu y$. Then $z (1 - \alpha \lambda - \beta \mu) = 0$ and since $z$ is regular $\alpha \lambda + \beta \mu = 1$. Now because $V$ is a local ring, one of these two terms must be invertible. If $\alpha \lambda$ is invertible, then again since $V$ is local, $\alpha$ must be invertible and $(y) \subset (x)$. In the other case one has $(x) \subset (y)$. We have thus proved that principal ideals in $V$ are totally ordered by inclusion and thus $V$ is a valuation ring.

Related concepts

• valuation

• discrete valuation

• class field theory

Literature

• Albrecht Fröhlich, J. W. S. Cassels (eds.), Algebraic number theory, Acad. Press 1967, with many reprints; Fröhlich, Cassels, Birch, Atiyah, Wall, Gruenberg, Serre, Tate, Heilbronn, Rouqette, Kneser, Hasse, Swinerton-Dyer, Hoechsmann, systematic lecture notes from the instructional conference at Univ. of Sussex, Brighton, Sep. 1-17, 1965 (ISBN:9780950273426, pdf, errata pdf by Kevin Buzzard)

• Serge Lang, Algebraic number theory. GTM 110, Springer 1970, 2000

• O. Zariski, Samuel, Commutative algebra