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Such a field embedding pulls the p-adic absolute value back to a finite place of a number field above $p$. It extends to an embedding of the corresponding completion, whose image is a finite extension of the P-adic numbers.

In the absolute-value description, a place of a number field is an equivalence class of nontrivial absolute values, where equivalent absolute values give the same topology, or differ by a positive power. More generally this defines the rank-one absolute-value places used in weak approximation for a field. For a number field, finite places correspond to nonzero prime ideals of its ring of integers, while infinite places correspond to real embeddings and pairs of conjugate complex embeddings. The completion at $v$ is denoted $K_v$.

A valuation place can also be described by a valuation ring ${\mathcal{O}}_v\subset K$ and its residue map into $k_v$, extended by sending elements outside that ring to infinity. The finite places above are the discrete-valuation instances of this description.

The weak approximation theorem states that for finitely many pairwise inequivalent nontrivial absolute values $v_1,\dots ,v_r$, the diagonal image of $K$ is dense in ${\prod }_i{K}_{v_i}$. Equivalently, for arbitrary $a_i\in K_{v_i}$ and ${ϵ}_i>0$, there is $a\in K$ such thatThe places must be distinct: two requirements at the same completion need not be compatible. In the number-field setting this permits simultaneous prescribed finite congruences and archimedean approximations.

A place of a number field is an equivalence class of nontrivial absolute values on a field. Its finite places correspond to nonzero prime ideals of the ring of integers of a number field; its infinite places come from real embeddings and conjugate pairs of complex embeddings.

For a embedding of a number field into a p-adic algebraic closure $\iota :K\hookrightarrow {\overline{\mathbb{Q}}}_p$, pull back the p-adic absolute value. This gives a place above $p$. Every element of $\mathrm{Gal}({\overline{\mathbb{Q}}}_p/{\mathbb{Q}}_p)$ preserves that absolute value, by uniqueness of its extension to each finite local extension, so equivalent embeddings give the same place.

Conversely, a place $v$ above $p$ gives a completion of a number field at a prime ideal $K_v$, a finite extension of ${\mathbb{Q}}_p$. Embed it into ${\overline{\mathbb{Q}}}_p$ and restrict to $K$. If two embeddings give the same place, they extend to two ${\mathbb{Q}}_p$-embeddings of this same completion. They are conjugate under the absolute Galois group: their finite separable images lie in a finite normal closure, and an isomorphism of such images extends to an automorphism of the algebraic closure. The completion really is the closure of the embedded $K$, since rational coefficients are dense in the ${\mathbb{Q}}_p$-span of a primitive element. Thus we obtain the embedding orbit description of finite places:

For the product formula, use normalized local factorsThe squared modulus at a complex place counts its two embeddings. Equivalently one can use ordinary complex modulus and put exponent two in the product. These normalized local factors for the product formula must be specified; arbitrary representatives of place classes would not satisfy the unweighted printed formula.

The fractional principal ideal $(x)$ has only finitely many nonzero prime exponents. Its ideal norm is $|N_{K/\mathbb{Q}}(x)|$: for integral $x$, multiplication by $x$ on an integral basis has this determinant in absolute value, equal to the index of $(x)$ in ${\mathcal{O}}_K$; a quotient gives the fractional case. Hence the finite-place product is $|N_{K/\mathbb{Q}}(x){|}^{-1}$. The infinite-place product is $|N_{K/\mathbb{Q}}(x)|$, by the embedding expression for the field norm. Therefore all but finitely many local factors are one, and

Finally, let ${\Delta }_K\ne 0$ be the integer field discriminant, defined by the determinant of the trace pairing on an integral basis. If a rational prime $p$ ramifies, the algebra ${\mathcal{O}}_K/p{\mathcal{O}}_K$ has a nonzero nilpotent element: use prime ideal factorization and the Chinese remainder theorem to see a nonzero nilpotent in a factor ${\mathcal{O}}_K/{\mathfrak{p}}^e$ with $e>1$. If $y$ is nilpotent, multiplication by $yz$ is nilpotent for every $z$ in this commutative algebra, and therefore has trace zero. Thus $y$ is in the radical of its trace pairing, making the reduced discriminant zero. We have proved the discriminant obstruction to ramification:Only finitely many rational primes divide this nonzero integer, so only finitely many primes ramify.

Give ${\overline{K}}_{\mathfrak{p}}$ the unique extension of the absolute value of $K_{\mathfrak{p}}$. We use two standard facts: a complete non-Archimedean absolute value extends uniquely to every finite algebraic extension, and any two embeddings of a finite separable extension into an algebraic closure are conjugate by an automorphism of that algebraic closure.

A $K$-embedding $\tau :L\to {\overline{K}}_{\mathfrak{p}}$ pulls back this absolute value to a finite place of a number field on $L$ extending the given place of $K$. By the permitted equivalence of places and primes, it determines a prime ideal ${\mathfrak{q}}_{\tau }$ above $\mathfrak{p}$. Every element of $\mathrm{Gal}({\overline{K}}_{\mathfrak{p}}/K_{\mathfrak{p}})$ preserves the absolute value by uniqueness, so ${\mathfrak{q}}_{\tau }$ is constant on each orbit.

Every prime $\mathfrak{q}$ above $\mathfrak{p}$ occurs: its completion of a number field at a prime ideal $L_{\mathfrak{q}}$ is a finite extension of $K_{\mathfrak{p}}$ and admits a $K_{\mathfrak{p}}$-embedding into ${\overline{K}}_{\mathfrak{p}}$. Composing with $L\hookrightarrow L_{\mathfrak{q}}$ gives an embedding inducing $\mathfrak{q}$.

Conversely, suppose two embeddings induce the same prime. Their pulled-back absolute values have the same normalization on $K$, so they extend continuously to embeddings of the same field $L_{\mathfrak{q}}$. To see that their images are still inside the algebraic closure rather than only its completion, the closure of $\tau (L)$ is exactly the finite complete field $K_{\mathfrak{p}}\tau (L)$. The two resulting embeddings of $L_{\mathfrak{q}}$ are conjugate by the stated separable-extension fact. Restricting the conjugacy to $L$ proves that the original embeddings are in the same orbit. Thus the prime ideals and local embedding orbits are naturally identified:

If $L=K(\alpha )$, an embedding is determined by the image of $\alpha$, which may be any root of its minimal polynomial $f$. The orbits on these roots are exactly the root sets of the monic irreducible factors of $f$ over $K_{\mathfrak{p}}$. HenceIf $f_j$ corresponds to ${\mathfrak{q}}_j$, then $L_{{\mathfrak{q}}_j}\cong K_{\mathfrak{p}}[T]/(f_j)$ and $[L_{{\mathfrak{q}}_j}:K_{\mathfrak{p}}]=\mathrm{deg}{f}_j$. This is the local factorization and extended absolute values correspondence and does not require an integral power basis.

For each place of a number field $v$, let $K_v$ be the corresponding completion, and for finite $v$ let ${\mathcal{O}}_v$ be its valuation ring. The adele ring is the restricted productIts restricted product topology has basic open sets ${\prod }_v{U}_v$, where every $U_v\subseteq K_v$ is open and $U_v={\mathcal{O}}_v$ at all but finitely many finite places.

First take $K=\mathbb{Q}$. The neighborhoodof zero meets the diagonal copy of $\mathbb{Q}$ only in zero: a rational number lying in every ${\mathbb{Z}}_p$ is an integer, and the only integer in the indicated real interval is zero. Thus $\mathbb{Q}$ is discrete in ${\mathbb{A}}_{\mathbb{Q}}$.

Every rational adele is congruent modulo $\mathbb{Q}$ to an element ofIndeed, the finitely many negative $p$-adic principal parts can be removed simultaneously by subtracting a rational number, using the Chinese remainder theorem; subtracting an integer then moves the real component into $[0,1]$. This set is compact by the compactness of $[0,1]$, the compactness of every ${\mathbb{Z}}_p$, and the Tychonoff theorem. Its image covers the quotient, so ${\mathbb{A}}_{\mathbb{Q}}/\mathbb{Q}$ is compact.

Now choose a $\mathbb{Q}$-basis of the number field $K$. The given topological isomorphismidentifies the additive pair $({\mathbb{A}}_K,K)$ with $({\mathbb{A}}_{\mathbb{Q}}^n,{\mathbb{Q}}^n)$. A finite product of discrete subgroups is discrete, andis compact.

The idele group iswhere the distinguished subgroup at a finite place is ${\mathcal{O}}_v^{\times }$. It carries the corresponding restricted product topology on the idele group. The inclusion $j:J_K\to {\mathbb{A}}_K$ is continuous: the inverse image of a basic adelic open set is locally a product of open subsets of $K_v^{\times }$, and outside finitely many places every idele component already belongs to ${\mathcal{O}}_v^{\times }\subseteq {\mathcal{O}}_v$.

It is not a homeomorphism onto its image. Let $p_i$ be the $i$th rational prime and define the idele $x^{(i)}$ to equal $p_i$ at the place over $p_i$ and $1$ everywhere else. In the adele topology, $x^{(i)}\to 1$: the difference is zero at every fixed place once $i$ is large, while $p_i-1\in {\mathbb{Z}}_{p_i}$ at the single moving place. In the idele topology the sequence does not converge to $1$, because the open neighborhoodcontains no $x^{(i)}$: its $p_i$-component has positive valuation and is not a unit. Hence the inverse of $j$ on its image is not continuous.

Let $K$ be a number field. Its places consist of its real embeddings, conjugate pairs of complex embeddings, and the finite places associated with nonzero prime ideals $\mathfrak{p}$ of its ring of integers of a number field. At a real or complex place use the usual absolute value. If $\mathfrak{p}$ lies over the prime number $p$ with ramification index $e_{\mathfrak{p}}$, normalize its absolute value byThese normalizations extend the standard absolute values on $\mathbb{Q}$.

Write $d_v=[K_v:{\mathbb{Q}}_v]$ for the local degree of a place. Thus $d_v$ is $1$ at a real place, $2$ at a complex place, and $e_v{f}_v$ at a finite place. The Absolute multiplicative Weil height isThe product formula shows that this is unchanged when $K$ is replaced by a larger number field containing $\alpha$.

For every $k\in \mathbb{Z}$,For $k\ge 0$, this follows directly fromat each place of a number field. The product formula gives $H({\alpha }^{-1})=H(\alpha )$, becauseand this handles negative $k$ as well.

Write $\mathcal{L}(R)$ for the polynomial length. The height bound for a polynomial evaluation isprovided the denominator is nonzero. At non-Archimedean places the integral coefficients and ultrametric inequality give the local estimate without an extra constant; at Archimedean places the triangle inequality gives the polynomial length. Multiplication over every place of a number field and the product formula produce the displayed bound.

Taking $P(X,Y)=X+Y$ and $Q=1$ givesTaking $P(X)=X+1$ gives