Number field places classification

A nontrivial absolute value on K cannot restrict trivially to the rationals. If it did, for each algebraic x and all n, reduction by its fixed minimal equation would express $x^n$ in a fixed finite list of powers of x with rational coefficients of value at most one. Thus $\mathrm{\mid }x{\mathrm{\mid }}^n$ is bounded independently of n and $\mathrm{\mid }x\mathrm{\mid }\le 1$. Apply the same argument to $x^{-1}$ to obtain $\mathrm{\mid }x\mathrm{\mid }=1$ for nonzero x. Ostrowski and the equivalence characterization therefore leave precisely a p-adic restriction up to positive power, or the usual archimedean restriction up to positive power.

If the restriction is a positive power of the $p$-adic value, then all integers have value at most one. The binomial theorem gives $\mathrm{\mid }x+y{\mathrm{\mid }}^n\le (n+1)\max (\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }{)}^n$; taking $n$th roots and letting $n\to \mathrm{\infty }$ proves the ultrametric inequality on $K$. For this nonarchimedean value, every algebraic integer has value at most one: otherwise its leading term in a monic integral equation would have strictly greater value than the sum of the others. Then $P=\{a\in {\mathcal{O}}_K:\mathrm{\mid }a\mathrm{\mid }<1\}$ is a proper prime ideal, and it is nonzero since it contains the rational prime p. By [F10], ${\mathcal{O}}_K$ is Dedekind, so [F6] applies and $({\mathcal{O}}_K{)}_P$ is a DVR. Elements outside P have value one; every nonzero element of K is $u{\varpi }^m$ with u a local unit, so its value is $\mathrm{\mid }\varpi {\mathrm{\mid }}^m$, $0<\mathrm{\mid }\varpi \mathrm{\mid }<1$. Hence its place is exactly P's valuation place. The elements of ${\mathcal{O}}_K$ with value less than one recover P, proving uniqueness. Conversely a DVR valuation defines that nontrivial absolute-value place.

In the archimedean case write the restriction as the ordinary value to a power $0<a\le 1$; a>1 is excluded by the triangle inequality on positive integers. The completion contains the complete real field with this powered value. The image V of $\mathbb{R}{\otimes }_{\mathbb{Q}}K$ in the completion is a finite-dimensional normed real vector space over that valued real field. It is complete, hence closed, and contains the dense K, so equals the completion. V is a finite-dimensional real domain and thus a field (multiplication by any nonzero element is injective and hence surjective). By the fundamental theorem of algebra, real irreducible polynomials have degree at most two: after adjoining one nonreal element one gets $\mathbb{C}$, which has no proper finite algebraic extensions. Therefore V is $\mathbb{R}$ or $\mathbb{C}$. Uniqueness of extending values identifies its value with the ordinary modulus to the a-th power.

Each embedding into the reals or complexes gives such a place. If two embeddings give equivalent places, normalize their restriction to the same real power. The equivalence exponent must then be one on the rationals; their completions are isometrically isomorphic fixing the dense K and hence the reals. A real automorphism of the complexes sends i to i or -i, so these embeddings are equal or conjugate. Conversely conjugation preserves modulus. Finally $NP=p^f$ and ${\mathrm{ord}}_P(x)=e{\mathrm{ord}}_p(x)$ for rational x (rational units at p are local units) give the stated exponent ef.