Absolute value (algebra, $|\cdot |$)

An absolute value on a field is a nonnegative multiplicative function that vanishes only at zero and satisfies the triangle inequality.

For finitely many pairwise inequivalent nontrivial absolute values on a field, an element of the field can approximate independently prescribed targets in the associated completion of a valued field at each place to arbitrary positive accuracies. Equivalently the diagonal map has dense image in the finite product. Requiring finitely many inequivalent places is essential; the conclusion controls finitely many inequalities, not infinitely many independently prescribed bounds at once, or incompatible targets at one place. Weak approximation for number fields is the class-field-theoretic instance.

If two multiplicative absolute values on a field satisfy a comparison with fixed positive constants $c,C$, apply it to $x^n$ and take $n$th roots. Letting $n\to \infty$ eliminates both constants. This stronger bounded comparison differs from equivalence merely as topologies: replacing an absolute value by its square generally preserves topology but not equality.

The metric completion of a field for an absolute value on a field is formed from Cauchy sequences modulo sequences tending to zero. Addition and multiplication extend continuously, and every nonzero completed element is invertible, so the result remains a field. The absolute value on a field extends uniquely. A place of a number field determines the corresponding completion $K_v$.

Choose compatible primitive roots of unity ${\zeta }_{p^{n+1}}$. The partial sums $A_N={\sum }_{n=0}^N{p}^{2n}{\zeta }_{p^{n+1}}$ form a Cauchy sequence. Each $A_N$ generates ${\mathbb{Q}}_p({\zeta }_{p^{N+1}})$: for a nonidentity cyclotomic automorphism with $r=v_p(a-1)$, the first changed term has valuation $2r+1/(p-1)$, strictly below every later changed term. Thus the conjugate distance is at least $p^{-2N-1/(p-1)}$. The tail bound is $p^{-2N-2}$, strictly smaller. If the limit were algebraic, Krasner's lemma would place every one of these unbounded-degree cyclotomic fields in one finite extension, a contradiction.

The trivial absolute value on a field is zero at zero and one elsewhere. Its metric topology is discrete. Its valuation ring and residue field are both the field itself. This is why rational absolute-value classification and nontrivial discrete-value compactness criteria need a nontriviality hypothesis.

A non-Archimedean absolute value satisfies the ultrametric inequality $|x+y|\le \max (|x|,|y|)$.

An absolute value on a field is non-Archimedean if and only if $|n\cdot 1|\le 1$ for every positive integer $n$. Necessity follows from the ultrametric inequality. Conversely the binomial theorem gives $|x+y{|}^m\le (m+1)\max (|x|,|y|{)}^m$, since every binomial coefficient has absolute value at most one. Take $m$th roots and let $m\to \infty$. In particular an extension of an absolute value that is non-Archimedean stays non-Archimedean.

An absolute value on a field is non-Archimedean exactly when its values on the image of $\mathbb{Z}$ are bounded. The reverse implication follows by applying the ordinary triangle inequality to the binomial expansion of $(x+y{)}^n$ and taking $n$th roots as $n\to \infty$.

The ultrametric inequality, also called the strong triangle inequality, bounds a direct distance by the larger of the two distances through an intermediate point:A metric space satisfying this rule is an ultrametric space. For a Non-Archimedean absolute value, the corresponding inequality is $|a+b|\le \max \{|a|,|b|\}$, and the associated metric is $d(x,y)=|x-y|$. The ordinary triangle inequality follows because the maximum of two nonnegative real numbers is at most their sum.

For example, the 2-adic absolute value gives the three points $0,1,8$ distancesEach distance is bounded by the maximum of the other two. The two largest distances are equal: every triangle in an ultrametric space has this property. Otherwise its unique longest side would exceed the maximum of the other two, contradicting the ultrametric inequality. The drawn lengths of a triangle on paper need not represent these distances.

The ultrametric inequality gives a useful exact rule when $|a|\ne |b|$:To see why, suppose $|a|>|b|$. The upper bound is $|a+b|\le |a|$; applying the ultrametric inequality to $a=(a+b)-b$ forces the reverse bound. Equal sizes can instead cancel: $|1{|}_2=|1{|}_2=1$, but $|1+1{|}_2=1/2$. This makes a uniquely largest term decisive in a sum, a principle used in the Newton polygon and in arguments about roots of a polynomial.

An ultrametric space is a metric space whose metric satisfies the ultrametric inequality. The rational numbers with $d_p(x,y)=|x-y{|}_p$, and their P-adic number completion, are examples. Their geometry organizes points into nested clusters rather than measuring separation along a number line.

For integers $a,b$ and $k\ge 0$, the definition of the P-adic valuation givesThus a closed ball of radius $p^{-k}$ in the integers is a congruence class modulo $p^k$. In the p-adic integers the same closed ball is $a+p^k{\mathbb{Z}}_p$. More shared low-order digits mean a smaller p-adic absolute value of the difference.

The binary tree first splits the sixteen integers into even and odd congruence classes, then refines modulo $4$, $8$, and $16$. Each node fixes one more low-order digit. The leaves $5$ and $13$ share the class $5$ modulo $8$, then separate modulo $16$, so their distance is exactly $1/8$. The leaves are deliberately ordered by their low-order digits; horizontal spacing and edge lengths are layout choices, not the metric. The singleton leaves describe this finite sample; a congruence class modulo $16$ in all p-adic integers remains infinite.

The ultrametric inequality also explains the unusual behavior of balls. Every point of an open ball can serve as its center: if $d(a,b)<r$, then $d(a,x)<r$ if and only if $d(b,x)<r$. Two open balls that intersect are nested, since a point of their intersection serves as a common center and their radii can be compared. The same nesting rule holds for closed balls and for a pair consisting of one open ball and one closed ball.

Every open ball is a clopen set. For a point $y$ outside $B(a,r)$, the open ball $B(y,d(a,y))$ stays outside $B(a,r)$: if $d(y,z)<d(a,y)$, the ultrametric inequality forces $d(a,z)=d(a,y)\ge r$. Hence the complement is an open set. Also every closed ball of positive radius is an open set, because $B(b,r)$ lies in the closed ball of radius $r$ around $a$ whenever $d(a,b)\le r$. A radius-zero closed ball is a singleton and need not be an open set. Distinct points can therefore be separated by a clopen set, proving that an ultrametric space is totally disconnected. These ball properties turn distance thresholds into a hierarchy of compatible clusters. ^[ref]

A topological space is totally disconnected when its only connected subsets are singletons.

Two nontrivial absolute values on a field are equivalent when they induce the same topology. Equivalently, one is a positive real power of the other.

For a field extension $L/K$, an extension of an absolute value $|\cdot |$ on $K$ is an absolute value on $L$ whose restriction to $K$ is $|\cdot |$.

If $K$ is complete and $L/K$ is finite, an absolute value on $K$ extends uniquely to $L$. With the normalization that it restricts exactly to the given absolute value,Every $K$-automorphism of $L$ preserves this extended absolute value.

Let $K$ be complete for a Non-Archimedean absolute value and let $V$ be a finite-dimensional vector space with a norm satisfying $\Vert av\Vert =|a|\Vert v\Vert$ and the ultrametric inequality. All such norms are equivalent to the maximum of coordinates in any basis. The upper bound is immediate. For the lower bound, induct on dimension. The span $W$ of all but the last basis vector is complete by induction, hence closed for the given norm. Thus $\delta ={\inf }_{w\in W}\Vert e_n-w\Vert >0$. For $v=w+a{e}_n$, $\Vert v\Vert \ge |a|\delta$; also $\Vert w\Vert \le \max (\Vert v\Vert ,|a|\Vert e_n\Vert )$. The induction bound on coordinates of $w$ now bounds all coordinates by a constant times $\Vert v\Vert$. This proof needs no local compactness.

On a finite-dimensional vector space over ${\mathbb{Q}}_p$, the triangle inequality gives an upper bound by a coordinate maximum norm. The coordinate unit sphere is compact. The other norm is continuous and nonzero there, so has a positive minimum. Scaling by powers of $p$ gives the lower bound. The resulting equivalence of norms in finite dimensions proves completeness of finite extensions, and applying the comparison to powers of a field element proves equality of two extending multiplicative absolute values on a field.

Suppose a Non-Archimedean absolute value extends uniquely to every finite extension. If $f\in {\mathcal{O}}_K[X]$ is monic and irreducible, every root has absolute value at most one: otherwise its leading power dominates all lower terms, contradicting $f(\alpha )=0$ by the ultrametric inequality. In a splitting field, let $\phi$ be the minimal polynomial of an algebraic element of the residue of one root. Lift its coefficients to ${\mathcal{O}}_K$. The lift has absolute value less than one at that root and hence, by equal absolute values of algebraic conjugates, at every root. Consequently every root residue is a zero of $\phi$. Since reduction preserves the product of the linear factors with multiplicities, every irreducible factor of $\overline{f}$ is $\phi$, proving the displayed identity for some $m\ge 1$.

Suppose a Non-Archimedean absolute value on $K$ has a unique extension to every finite field extension. Any K-isomorphism $K({\alpha }_i)\to K({\alpha }_j)$ between conjugate roots pulls the absolute value back to an extension on the source field, which must be the given one. Thus $|h({\alpha }_i)|=|h({\alpha }_j)|$ for every $h\in K[X]$. This argument includes inseparable irreducible polynomials: no transitive Galois action or completeness assumption is needed.

Let $K$ be complete for a discrete valuation and let $L/K$ be finite. The valuation ring ${\mathcal{O}}_L$ is exactly the integral closure of ${\mathcal{O}}_K$ in $L$. One direction follows from a monic integral equation; for the other, the coefficients of the minimal polynomial are elementary symmetric polynomials in conjugates of absolute value at most one.

If $L/K$ is a finite normal extension, the Galois group acts transitively on the absolute values of $L$ extending a fixed absolute value of $K$. For number fields, this is equivalent to transitivity on the prime ideals above a fixed prime.

Give $\mathbb{Q}$ its $5$-adic valuation, take $L=\mathbb{Q}(i)$, and choose the extension associated with the prime $(2+i)$. Then $(2+i)/(2-i)$ lies in the chosen valuation ring but has negative valuation at the conjugate prime. It is therefore not in the integral closure of ${\mathbb{Z}}_{(5)}$ in $L$.

If $L=K(\alpha )$ is a finite separable extension and $v$ is an absolute value on $K$, the extensions of $v$ to $L$ correspond to the irreducible factors of the minimal polynomial of $\alpha$ over the completion of $K$ at $v$.

This polynomial is Eisenstein at $2$. Over ${\mathbb{Q}}_3$, its Newton polygon has a single slope $-2/5$, so it is irreducible and a root generates a totally ramified extension of degree five. Over ${\mathbb{Q}}_7$, its factor degrees are $1,1,3$: the simple residue roots $1,-1$ lift, and the remaining roots have valuation $1/3$. Thus the degree-five number field generated by a root has one prime above $3$ and three above $7$.

For a finite extension of number fields $L/K$ and a nonzero prime ideal $\mathfrak{p}$ of $K$, the primes of $L$ above $\mathfrak{p}$ correspond to the orbits ofAn embedding determines a prime by pulling back the local absolute value on a field. Embeddings with the same prime extend to embeddings of the same completion and are conjugate. If $L=K(\alpha )$, these orbits are the irreducible factors of its minimal polynomial over $K_{\mathfrak{p}}$.

For a number field $K$ and a nonzero prime ideal $\mathfrak{p}\subset {\mathcal{O}}_K$, the completion $K_{\mathfrak{p}}$ is the completion for the $\mathfrak{p}$-adic absolute value. It is a finite extension of ${\mathbb{Q}}_p$, where $p\mathbb{Z}=\mathfrak{p}\cap \mathbb{Z}$.

For a number field $K$ and a rational prime $p$, the natural diagonal map gives an isomorphism of ${\mathbb{Q}}_p$-algebras

If $L/K$ is a finite Galois extension of number fields, its Galois group acts transitively on the prime ideals of ${\mathcal{O}}_L$ lying above a fixed prime ideal of ${\mathcal{O}}_K$.

For a prime $\mathcal{P}$ of a finite Galois extension $L/K$ above $\mathfrak{p}$, the decomposition group is the stabilizer

For a finite Galois extension of number fields, choose a nonzero residue field generator at $\mathfrak{q}$ and lift it to an integral element $a$ whose residues at the other primes above $\mathfrak{p}$ are zero, using the Chinese remainder theorem. The orbit polynomial ${\prod }_{\sigma \in G}(T-\sigma a)$ has coefficients in the base ring of integers. Modulo $\mathfrak{q}$, its nonzero roots are precisely the orbit of $\overline{a}$ under the decomposition group. Its coefficients belong to $k_{\mathfrak{p}}$, so finite-field Frobenius automorphism permutes these roots. Consequently some decomposition-group element acts as Frobenius on the generator, and therefore on the entire residue field. Frobenius generates the finite-field Galois group, proving surjectivity.

Let $M/K$ be a finite Galois extension, $L=M^H$, and $D$ the decomposition group of a prime $\mathfrak{P}$ above $\mathfrak{p}$. The bijectionfollows from transitivity on primes for both $M/K$ and $M/L$.

If $M/K$ is the Galois closure of a finite extension $L/K$, a prime splits completely in $L$ exactly when it splits completely in $M$. Complete splitting means every right action of $D$ on $H\backslash G$ fixes each coset, so $D$ lies in the subgroup core of $H$. That core is trivial because the conjugates of $L$ generate $M$.

Restriction to completions gives a canonical isomorphismConsequently the number of primes above $\mathfrak{p}$ is the index of the decomposition group.

For a prime $\mathfrak{q}$ of a finite separable extension $L/K$ above $\mathfrak{p}$, completion preserves the ramification index and residue-field extension:

The polynomialsplits into three linear factors over ${\mathbb{Q}}_2$. Over ${\mathbb{Q}}_5$ it is an Eisenstein polynomial, so adjoining a root gives a totally ramified cubic extension. Its splitting field over ${\mathbb{Q}}_5$ has decomposition group $S_3$ and inertia group $A_3\cong C_3$.