Valuation (algebra)

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In algebra (in particular in algebraic geometry or algebraic number theory), a valuation is a measure of size or multiplicity. They generalize to commutative algebra the notion of size inherent in consideration of the degree of a pole or multiplicity of a zero in complex analysis, the degree divisibility of a number by a prime number in number theory, and the geometrical concept of contact between two algebraic or analytic varieties in algebraic geometry.

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[edit] Algebraic definition

To define the algebraic concept of valuation, the following objects are needed:

and also an element \scriptstyle\infty such that

\begin{array}{cr} \infty \geq \mathfrak{{a}} & \forall\mathfrak{{a}}\in\mathfrak{G}\cup\{\infty\} \\ \infty+\mathfrak{a} = \mathfrak{a} + \infty = \infty & \forall\mathfrak{a}\in\mathfrak{G}\cup\{\infty\} \end{array}

Then a valuation is any map

v:\mathbb{K} \rightarrow {\mathfrak{G} \cup \{\infty\}}

which satisfies the following properties

\begin{array}{ll} v(a)=\infty & \mbox{iff}~a=0 \\ v(ab)=v(a)+v(b) & \forall a,b\in\mathbb{K}^* \\ v(a+b)\geq\min\{v(a),v(b)\} & \forall a,b\in\mathbb{K} \end{array}

Note that some authors use the term exponential valuation rather than "valuation". In this case the term "valuation" means "absolute value".

For valuations used in geometric applications, the first property implies that any non-empty germ of an analytic variety near a point contains that point. The second property assert that any valuation is a group homomorphism, while the third property is a translation of the triangle inequality from metric spaces to ordered groups.

It is possible to give a dual definition of the same concept: if, instead of \scriptstyle\infty , an element \scriptstyle\mathfrak{0} is given such that

\begin{array}{cr} \mathfrak{0} \leq \mathfrak{{a}} & \forall\mathfrak{{a}}\in\mathfrak{G}\cup\{\mathfrak{0}\} \\ \mathfrak{0}\mathfrak{a} = \mathfrak{a}\mathfrak{0} = \mathfrak{0} & \forall\mathfrak{a}\in\mathfrak{G}\cup\{\mathfrak{0}\} \end{array}

then a valuation is any map

v:\mathbb{K} \rightarrow \mathfrak{G \cup \{0\}}

satisfying the following properties (written using the multiplicative notation for group operation)

\begin{array}{ll} v(a)=\mathfrak{0} & \mbox{iff}~a=0 \\ v(ab)=v(a)v(b) & \forall a,b\in\mathbb{{K}}^* \\ v(a+b)\leq\max\{v(a),v(b)\} & \forall a,b\in\mathbb{K} \end{array}

A valuation is commonly required to be surjective, since many arguments used in ordinary mathematical research involving those objects use preimages of unspecified elements of the ordered group contained in its codomain. Also, the first definition of valuation given is more frequently encountered in ordinary mathematical research, thus it is the only one used in the following considerations and examples: then, in what follows, \scriptstyle\mathfrak{0} is the identity element the ordered group, or the zero element of the ring considered. See Jacobson 1989 for further details.

[edit] Equivalence of valuations

Two valuations are said to be equivalent if they have the same domain, codomain and are proportional i.e. they differ by a fixed element belonging to the ordered group in their codomain: using a symbolic notation

v_1 \propto v_2 \iff \exists\mathfrak{c}\in\mathfrak{G}\;|\; v_1(x) = v_2(x) + \mathfrak{c}\quad\forall x\in\mathbb{K}.

Proportionality in this sense is an equivalence relation:

v = v + \mathfrak{0} \iff v \propto v ,
v_1 \propto v_2 \Rightarrow v_1 = v_2 + \mathfrak{c} \iff v_2 = v_1 + \mathfrak{c}^{-1} \Rightarrow v_2 \propto v_1
  • It is transitive since, given three valuation \scriptstyle v_1,v_2,v_3\, such that \scriptstyle v_1\, is equivalent to \scriptstyle v_2\, which is in turn equivalent to \scriptstyle v_3\,, then
\begin{array}{l} v_1 \propto v_2 \\ v_2 \propto v_3 \end{array} \Rightarrow \begin{array}{l} v_1 = v_2 + \mathfrak{c} \\ v_2 = v_3 + \mathfrak{k} \end{array} \Rightarrow\; v_1 = v_2 + \mathfrak{c} = (v_3 + \mathfrak{k}) + \mathfrak{c} = v_3 + (\mathfrak{k} + \mathfrak{c}) \;\Rightarrow\; v_1 \propto v_3

Every equivalence class of valuations over a field with respect to this equivalence relation is called a place. Ostrowski's theorem gives a complete classification of places of the field of rational numbers \scriptstyle\mathbb{Q} : these are precisely equivalence classes of valuations for the p-adic completions of \scriptstyle\mathbb{{Q}} .

[edit] Dedekind valuation

A Dedekind valuation is a valuation for which the ordered abelian group \scriptstyle\mathfrak{G} in its codomain is the additive group of the integers, i.e.

(\mathfrak{G},+,\leq) = (\mathbb{Z},+,\leq)

Dedekind valuations are also known under the name of discrete valuations, even if some authors consider a discrete valuation as a valuation where the group \scriptstyle\mathfrak{G} is a subgroup of the real numbers isomorphic to the integers.

[edit] Examples

[edit] p-adic valuation

Let \scriptstyle\mathfrak{{R}} be a principal ideal domain, \scriptstyle\mathbb{{K}} be its field of fractions, \scriptstyle\mathfrak{{p}} \in \mathfrak{{R}} be one of its irreducible elements. Then, if the ideal \scriptstyle(\mathfrak{p}) is prime,

\mathfrak{g} \in (\mathfrak{{p}})^k \quad \forall \mathfrak{{g\in R}}, k\in\mathbb{N}

i.e. any element belongs to its k-th power, for a proper natural number k: this can be easily seen since

  • if \scriptstyle\mathfrak{{g = 0}} , then \scriptstyle\mathfrak{g} belongs to \scriptstyle(\mathfrak{p})^k for any natural number k,
  • if \scriptstyle\mathfrak{g} and \scriptstyle\mathfrak{p} share non trivial common factors, then \scriptstyle\mathfrak{g} belongs to \scriptstyle( \mathfrak{p} ) , i.e. k = 1,
  • if \scriptstyle\mathfrak{g} is coprime respect to \scriptstyle\mathfrak{p} , it is sufficient to choose k = 0: then
\mathfrak{p^0=R} \iff \mathfrak{g} \in \mathfrak{p^0}

Therefore, any element s of the field \scriptstyle\mathbb{K} can be written as follows

s = \mathfrak{q}/\mathfrak{r} \cdot \mathfrak{p}^k \quad \mathfrak{q,r\in R}\,,\, k\in\mathbb{Z}

where \scriptstyle\mathfrak{q,r} are coprime respect to \scriptstyle\mathfrak{p} , and k is now an integer. Then the map \scriptstyle v:\mathbb{K} \rightarrow \mathbb{Z} defined as

v(s) = \begin{cases} k & \forall s \in \mathbb{K}^* \\ \infty & s=0 \in \mathbb{K} \end{cases}

is easily proven to be a valuation. When the principal ideal domain considered is the ring of integers, \scriptstyle\mathfrak{p} is a prime number p, and this valuation is called the p-adic valuation on the set \scriptstyle\mathbb{Q} of rational numbers.

[edit] μ-adic valuation

Let \scriptstyle(\mathfrak{{R}},\mu) be a local integral ring with maximal ideal μ: then

\mathfrak{f} \in (\mu)^k \quad \forall \mathfrak{f}\in\mathfrak{{R}}\,,\,k\in\mathbb{N}

i.e. every element of the local ring belongs to the k-th power of its maximal ideal, for a proper natural number k. Now define the map \scriptstyle v:\mathfrak{R}\rightarrow\mathbb{Z} as

v(\mathfrak{f}) = k\,\Longleftrightarrow\,\mathfrak{{f}}\in\mu^k\ \mbox{but}~\mathfrak{{f}}\notin\mu^{{k+ 1}}\quad \forall \mathfrak{{f}} \in \mathfrak{{R}}

and extend it to the field of fractions \scriptstyle\mathbb{{K}} of \scriptstyle\mathfrak{{R}} as follows:

v\mathfrak{(f/g)}= \begin{cases} v(\mathfrak{{f}}) - v(\mathfrak{{g}}) & \forall \mathfrak{f/g} \in \mathbb{K}^* \\ \infty & \mathfrak{f}=\mathfrak{0} \in \mathbb{K} \end{cases}

It is easy to prove that this map is a well-defined valuation: it is called μ-adic valuation on \scriptstyle\mathbb{K} . If, for example, the local integral ring considered is the ring of formal power series in two variables over the complex field i.e. \scriptstyle\mathfrak{R} = \mathbb{C}[[x,y]] , then its maximal ideal is \scriptstyle\mu = (x;y) \, and its μ-adic valuation is given by the difference of the orders of the power series in the numerator and the denominator: as examples, computation of μ-valuation for some fractions is reported

v(x^2 + y^2 + x^3y^2)=2 \,
v(x^3/y^2) = 3 - 2 = 1 \,

[edit] Geometric notion of contact

Let \scriptstyle\mathfrak{{R}} = \mathbb{{C}}[x,y] be the ring of polynomials of two variables over the complex field, \scriptstyle\mathbb{{K}} = \mathbb{{C}}(x,y) be the field of rational functions over the same field, and consider the (convergent) power series

f(x,y) = y - \sum_{n=3}^{\infty} \frac{x^n}{n!} \in \mathbb{{C}}\{x,y\}

whose zero set, the analytic variety \scriptstyle V_f\,, can be parametrized by one coordinate t as follows

V_f = \{(x,y)\in\mathbb{C}^2\,|\, f(x,y) = 0\} = \left\{ (x,y)\in\mathbb{C}^2\,|\,(x,y) = \left(t,\sum_{n=3}^{\infty}t^i\right)\right\}

It is possible to define a map \scriptstyle v: \mathbb{{C}}[x,y] \rightarrow \mathbb{Z} as the value of the order of the formal power series in the variable t obtained by restriction of any polynomial P in \scriptstyle\mathbb{C}[x,y] to the points of the set \scriptstyle V_f\,

v(P) = \mathrm{ord}_t\left(P|_{V_f}\right) = {\mathrm{ord}}_t \left(P\left(t,\sum_{n=3}^{+\infty}t^i\right)\right) \quad \forall P\in \mathbb{C}[x,y]

It is also possible to extend the map v from its original ring of definition to the whole field \scriptstyle\mathbb{{C}}(x,y) as follows

v(P/Q) = \begin{cases} v(P) - v(Q) & \forall P/Q \in {\mathbb{C}(x,y)}^* \\ \infty & P \equiv 0 \in \mathbb{C}(x,y) \end{cases}

As the power series f is not a polynomial, it is easy to prove that the extended map v is a valuation: the value \scriptstyle v(P) is called intersection number between the curves (1-dimensional analytic varieties) \scriptstyle V_P\, and \scriptstyle V_f\,. As an example, the computation of some intersection numbers follows

\begin{array}{l} v(x) = \mathrm{ord}_t(t) = 1 \\ v(x^6-y^2)=\mathrm{ord}_t(t^6-t^6-2t^7-3t^8-\cdots)=\mathrm{ord}_t (-2t^7-3t^8-\cdots)=7 \\ v\left(\frac{x^6 - y^2}{x}\right)= \mathrm{ord}_t (-2t^7-3t^8-\cdots) - \mathrm{ord}_t(t) = 7 - 1 = 6 \end{array}

[edit] See also

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