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For a nonnegative integer $r$, a Cr field is a field over which every homogeneous polynomial of positive degree $d$ in $N>d^r$ variables has a nontrivial zero. Thus a C1 field uses the bound $N>d$, and a C2 field uses $N>d^2$.

The required field theorems are as follows. The Chevalley-Warning theorem implies that every finite field is $C_1$: the number of zeros of a polynomial with degree smaller than its number of variables is divisible by the characteristic, and for a homogeneous polynomial the origin is already a zero. The Lang-Nagata theorem for Ci fields states that a finitely generated extension of transcendence degree $s$ of a $C_r$ field is $C_{r+s}$. The Tsen theorem states that the function field of a curve over an algebraically closed field is $C_1$. In particular, finite fields, algebraically closed fields and fields such as $\mathbb{C}(t)$ are examples of $C_1$ fields. A function field of one variable over a finite field is $C_2$ by the Lang theorem. These are statements of the theorems; no theorem proof is needed here.

For the requested finite extension stability of C1 fields, let $m=[L:K]$ and choose a $K$-basis $u_1,\dots ,u_m$ of $L$. Given a homogeneous $f\in L[X_1,\dots ,X_N]$ of degree $d$ with $N>d$, substitute $X_i={\sum }_j{x}_{ij}{u}_j$ and take the field norm:The field norm is the determinant of multiplication on the $m$-dimensional $K$-space $L$, hence a homogeneous polynomial of degree $m$ in its coordinates. Consequently $F$ is homogeneous of degree $md$ in $mN>md$ variables over $K$. The $C_1$ property gives a nonzero coordinate vector $(x_{ij})$ with $F=0$. Its corresponding vector $(X_i)\in L^N$ is nonzero, since the $u_j$ are a basis. The norm of a field element vanishes only for the zero element, so $f(X_1,\dots ,X_N)=0$. Every finite extension of a $C_1$ field is $C_1$. This proof includes inseparable finite extensions, because the determinant definition of the field norm requires no separability.

For a complete discretely valued field with perfect residue field $k$, the Witt residue sequence is the split exact sequenceThe first map identifies the unramified Brauer classes. The residue map is canonical; a splitting is chosen by a uniformizer $\pi$ and sends an unramified character $\chi$ to the cyclic Brauer class $(\chi ,\pi )$. Changing $\pi$ may change this splitting by an unramified class. This is the version of Witt's theorem used here.

Now suppose $k={\mathbb{F}}_q$. Fix arithmetic Frobenius $F_k:a\mapsto a^q$. Define the local Brauer invariant byHere is an explicit cohomological construction. Set $E=K^{\mathrm{nr}}$ and $G=\mathrm{Gal}(E/K)\cong \hat{\mathbb{Z}}$. The permitted vanishing $\mathrm{Br}(E)=0$, together with Hilbert 90 and inflation, identifies $\mathrm{Br}(K)$ with $H^2(G,E^{\times })$. Normalize the valuation $v(\pi )=1$. Its induced map sends a multiplicative two-cocycle $c$ to the integer-valued two-cocycle $v(c)$, givingThe middle isomorphism is the connecting map for $0\to \mathbb{Z}\to \mathbb{Q}\to \mathbb{Q}/\mathbb{Z}\to 0$, since positive-degree continuous cohomology of the trivial module $\mathbb{Q}$ vanishes.

To see that the valuation map is an isomorphism, its module sequence splits using $\pi$, and the unit part has zero second cohomology. For each finite unramified cyclic extension $E_m/K$, cyclic cohomology identifies that unit contribution with ${\mathcal{O}}_K^{\times }/N_{E_m/K}{\mathcal{O}}_{E_m}^{\times }$. The norm on residue-field units is surjective. On the principal-unit quotient of level $r\ge 1$, the norm is the residue trace, through $N(1+{\pi }^{r}b)\equiv 1+{\pi }^r\mathrm{Tr}(\overline{b})(\mathrm{mod}{\pi }^{r+1})$. The trace of a finite separable field extension is surjective; successive corrections and completeness therefore make the norm surjective on all units. Passing to the direct limit gives the asserted vanishing. Finally, a continuous character of $\hat{\mathbb{Z}}$ is uniquely specified by its value on arithmetic Frobenius, and that value can be any element of $\mathbb{Q}/\mathbb{Z}$. The invariant map is an isomorphism.

In concrete terms, if $E_m/K$ is the unramified extension of degree $m$ and $\sigma$ lifts arithmetic Frobenius, the cyclic algebra with $ub=\sigma (b)u$ and $u^m=a$ hasIndeed its standard two-cocycle has value $a$ when the powers of $\sigma$ wrap past $m$ and value one otherwise. Valuation turns this into the connecting cocycle for the character with Frobenius value $v_K(a)/m$. This also fixes the sign of the normalization.

For restriction and corestriction of local Brauer invariants, let $l$ be the residue field of $L$, $e$ the ramification index and $f=[l:k]$. A finite extension of local fields has $[L:K]=ef$, $v_L{|}_K=e{v}_K$ and $v_K(N_{L/K}b)=f{v}_L(b)$. Every class of $\mathrm{Br}(K)$ is $(\chi ,{\pi }_K)$, with $\chi$ unramified. After restriction to $L$, its character has Frobenius value $f\chi (F_k)$ and its parameter has valuation $e$. ThusFor corestriction, write a class of $\mathrm{Br}(L)$ as $({\chi }_l,{\pi }_L)$. Choose a character ${\chi }_k$ of the finite-field absolute group such that ${\mathrm{Res}}_{l/k}{\chi }_k={\chi }_l$. This is possible because restriction is multiplication by $f$ on $\mathbb{Q}/\mathbb{Z}$, which is divisible. Inflate both characters to the local fields. The norm projection formula for cyclic Brauer pairings givesThis projection identity follows from the cohomological transfer and cup-product identity; on multiplicative degree-zero coefficients transfer is the field norm, and the identity in higher degree follows by the cochain definition or dimension shifting. The norm parameter has $K$-valuation $f$, so its invariant is $f{\chi }_k(F_k)={\chi }_l(F_l)$. ThereforeThis argument uses the character-and-norm description directly and does not cancel multiplication by $[L:K]$ in $\mathbb{Q}/\mathbb{Z}$, where such cancellation would be invalid. For purely inseparable steps in equal characteristic, absolute Galois groups identify, restriction uses inclusion of multiplicative coefficient modules, and Brauer corestriction uses the field norm on them. The same projection calculation applies. Factoring a general finite extension into separable and purely inseparable steps covers all finite extensions in the question.

Lastly, a class with invariant $a/b$ in lowest terms has Brauer period $b$. The unramified extension of degree $b$ splits it, so its division-algebra degree divides $b$. Conversely a maximal subfield of the division algebra splits it and has degree equal to that division-algebra degree; the restriction formula forces $b$ to divide that degree. These standard splitting-field properties are among the permitted inputs. Thus the index of a central simple algebra equals its period over a local field. The central division algebras of degree $n$ are exactly those with invariants $a/n$ and $\gcd (a,n)=1$. The Brauer class determines its division representative uniquely, so there areisomorphism classes, where $\phi$ is the Euler totient function. This includes the single degree-one algebra $K$.

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.

The idele group is the multiplicative restricted productIts restricted product topology has basic open sets ${\prod }_{v\in S}{U}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains every infinite place, and $U_v$ is open in $K_v^{\times }$. In particular, the restriction to almost all local unit groups is part of the topology, not merely the unrestricted product topology.

The diagonal map $K^{\times }\to J_K$ is injective and well-defined by finite support of the principal ideal's valuations; its image consists of principal ideles. To prove discreteness of principal ideles, take the neighbourhood of $1$ which requires $x_v\in {\mathcal{O}}_v^{\times }$ at every finite place and $|x_v-1|<1/2$ in the ordinary real or complex modulus at every infinite place. A diagonal element in it is a global unit, so $x-1\in {\mathcal{O}}_K$. If $x\ne 1$, the nonzero integer $N_{K/\mathbb{Q}}(x-1)$ has absolute value at least one, but the archimedean bounds make its absolute value less than $2^{-[K:\mathbb{Q}]}$. This contradiction proves $K^{\times }$ is a discrete subgroup of $J_K$.

For a modulus of a number field, write $\mathfrak{m}$ for its finite ideal part together with any selected real places. Let $I_{\mathfrak{m}}$ be the group of fractional ideals coprime to the finite part. Let $P_{\mathfrak{m}}$ consist of principal ideals $(a)$ with $a\equiv 1(\mathrm{mod}{\mathfrak{p}}_v^{m_v})$ locally at each finite place in the modulus, and $a>0$ at each selected real place. The ray class group isEquivalently it is the quotient $J_K/(K^{\times }{U}_{\mathfrak{m}})$, with the corresponding principal-unit factors at finite places and positive multiplicative factors at selected real places. The question now assumes that no infinite places occur.

Put $R_{\mathfrak{m}}={\prod }_{v\in S}{\mathcal{O}}_v^{\times }/(1+{\pi }_v^{m_v}{\mathcal{O}}_v)$. Global units map to $R_{\mathfrak{m}}$ by reduction. To map $R_{\mathfrak{m}}$ to the ray class group, choose, by the Chinese remainder theorem, an integral $a$ having the prescribed unit residues at all $v\in S$, and send the tuple to the ray class of $(a)$. A different choice changes the ratio by an element congruent to one at every such place, so the map is well-defined and multiplicative.

The forgetful map to the ordinary ideal class group is surjective: weak approximation for number fields multiplies an ideal by a principal ideal so that all its valuations in $S$ vanish. Its kernel consists exactly of classes represented by principal ideals coprime to $S$, and their generators supply the local unit residues. Finally, a residue tuple maps to the identity ray class precisely when its chosen $a$ differs from a congruence-one generator by a global unit. Thus the ray class exact sequence isThe leftmost arrow need not be injective, explaining the absence of an initial zero.

For $K=\mathbb{Q}(\sqrt{2})$, the ring of integers of Q of square root two is $\mathbb{Z}[\sqrt{2}]$. Its absolute field norm is a Euclidean function. Indeed, for $a+b\sqrt{2}\in K$, choose nearest integers $m,n$; writing $u=a-m$, $w=b-n$, both bounded by $1/2$, gives $|u^2-2w^2|\le 1/2<1$. Applying this to a quotient proves Euclidean division, hence the ring is a principal ideal domain and $\mathrm{Cl}(K)=0$. This is norm-Euclideanity of the integers adjoined square root two.

Write $\mathfrak{p}=(\sqrt{2})$ and $\mathfrak{q}=(1+2\sqrt{2})$. Their ideal norms are $2$ and $7$, respectively; ${\mathfrak{p}}^2=(2)$ and the residue field at $\mathfrak{q}$ is ${\mathbb{F}}_7$, with $\sqrt{2}\equiv 3$ because $1+2\sqrt{2}\equiv 0$. The finite local quotient product for the requested modulus isThe first factor consists of $1$ and $1+t$.

The given fundamental unit $ϵ=1+\sqrt{2}$ and $-1$ generate the global units. Their images areThe first image has order six; its cube is $(1+t,1)$ and its square is $(1,2)$, generating the first factor and the order-three subgroup of the second. The second image supplies the missing order-two element of ${\mathbb{F}}_7^{\times }$. Thus the global unit map is surjective onto all $12$ local unit residues. By the exact sequence, or the unit-surjectivity criterion for a trivial finite ray class group,There is no positivity condition here: the modulus has no infinite part, so using the unit $-1$ is legitimate and essential in the surjectivity calculation.

The idele group is the multiplicative restricted productwith respect to ${\mathcal{O}}_v^{\times }$ at the finite places of a number field; $K_v$ is the completion of a valued field at the place $v$. Thus each tuple has nonzero components, and all but finitely many finite components are units. Its restricted product topology on the idele group has basic open sets ${\prod }_{v\in S}{W}_v\times {\prod }_{v\notin S}{\mathcal{O}}_v^{\times }$, where $S$ is finite and contains the infinite places, and each $W_v$ is open in $K_v^{\times }$. In particular $U_K$ is an open subgroup.

Embed $K^{\times }$ diagonally. Take a neighbourhood of one whose finite components all lie in ${\mathcal{O}}_v^{\times }$ and whose infinite components satisfy $|x_v-1|<1/2$ in the usual real or complex modulus. A diagonal element there is an algebraic unit $a$. If $a\ne 1$, then $a-1$ is a nonzero algebraic integer, so its field norm is a nonzero integer. Buta contradiction. Thus $K^{\times }$ is discrete. It is also closed: in a topological group, a subgroup with an isolated identity cannot have an external accumulation point, since quotients of two nearby subgroup elements would approach the identity.

Send an idele to its associated fractional ideal byOnly finitely many exponents are nonzero. This homomorphism is onto, by choosing powers of local uniformizers, and its kernel is $U_K$. Diagonal elements map to principal fractional ideals. The resulting quotient givesIt is a topological isomorphism when the ideal class group is given the discrete topology, since $U_K$ is open.

Use normalized local moduli: real modulus, squared complex modulus, and $|{\pi }_v{|}_v=(N{\mathfrak{p}}_v{)}^{-1}$ at a finite place. The idelic modulus $\Vert x\Vert ={\prod }_v|x_v{|}_v$ defines $J_K^1=\ker \Vert \cdot \Vert$, the norm-one idele group. The product formula puts $K^{\times }$ inside this kernel. Every ideal class has a representative in $J_K^1$, because an infinite component can be rescaled to correct the modulus without altering its fractional ideal. The compact space $J_K^1/K^{\times }$ therefore maps continuously onto the discrete ideal class group. Its image must be finite, proving $\mathrm{Cl}(K)$ is finite.

For the Dirichlet unit theorem, put $U^1=U_K\cap J_K^1$ and $E=K^{\times }\cap U^1={\mathcal{O}}_K^{\times }$. Let $r_1$ count real embeddings and $r_2$ count conjugate complex pairs. Infinite logarithms define a continuous surjectionusing $\log |x_v|$ at real places and $2\log |x_v|$ at complex places. Its kernel is compact: it consists of real signs, complex unit circles, and the product of compact finite-place unit groups. More generally the inverse image of a bounded closed subset of $H$ is compact. Since $E$ is closed and discrete, its intersection with each such inverse image is finite. Hence $\Lambda =\ell (E)$ is discrete in $H$, and the kernel $E\cap \ker \ell$ is a finite group. It is exactly the roots of unity $\mu (K)$, since every element of a finite multiplicative group has finite order and every root of unity has all local moduli one.

The image of $U^1$ in $J_K^1/K^{\times }$ is an open subgroup, hence also closed, and is homeomorphic to $U^1/E$. The assumed compactness therefore makes $U^1/E$ compact, and its continuous quotient $H/\Lambda$ is compact. A discrete cocompact subgroup of a real vector space is a full Euclidean lattice, of rank $\dim H=r_1+r_2-1$. Thus $E/\mu (K)\cong {\mathbb{Z}}^{r_1+r_2-1}$, and lifting a lattice basis splits off the free factor:This derives both finiteness and the unit rank from the stated compactness assumption, rather than assuming either conclusion to prove compactness.