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The intended statement uses a nontrivial Non-Archimedean absolute value. Under that convention choose $t$ with $0<|t|<1$. A compact neighborhood $C$ of zero contains an open ball of some radius $r>0$. For sufficiently large $N$, $t^N{\mathcal{O}}_K$ is contained in that ball. The valuation ring is closed, so $t^N{\mathcal{O}}_K$ is a closed subset of $C$ and is compact. Scaling shows that ${\mathcal{O}}_K$ itself is compact.

Its maximal ideal $\mathfrak{m}=\{x:|x|<1\}$ is open. The residue field $k={\mathcal{O}}_K/\mathfrak{m}$ is therefore discrete and compact, hence finite: the open cover by its singleton sets has a finite subcover. Since $\mathfrak{m}$ has finite index, its complement is a finite union of open cosets. Thus $\mathfrak{m}$ is also closed and compact.

The continuous absolute value on a field attains its maximum on $\mathfrak{m}$. Nontriviality supplies a nonzero element there, so this maximum is $\rho =|\pi |$ for some $\pi \ne 0$, with $0<\rho <1$. If $x\in \mathfrak{m}$, then $|x/\pi |\le 1$, so $\mathfrak{m}=\pi {\mathcal{O}}_K$.

For arbitrary $x\ne 0$, choose the integer $n$ for which ${\rho }^{n+1}<|x|\le {\rho }^n$. Then $\rho <|x/{\pi }^n|\le 1$. This element cannot lie in $\mathfrak{m}$, by maximality of $\rho$, so its absolute value on a field is one. Therefore $|x|={\rho }^n$. This proves discrete valuation from nontrivial local compactness:

The nontrivial valuation in the local compactness criterion is essential. If the trivial absolute value is admitted, the displayed source claim is false: take $\mathbb{Q}$ with that absolute value on a field. It is complete and discrete, hence locally compact, but its valuation ring and residue field are both $\mathbb{Q}$, which is infinite. The intended nontrivial case has been proved above.

A local field is nondiscrete and locally compact. In the non-Archimedean case its valuation ring is compact by the scaling argument in part (b). Any Cauchy sequence has a tail in a translate of a compact ball, so has a convergent subsequence; the Cauchy property forces the whole sequence to converge. Thus the field is complete. Part (b) gives a discrete valuation, a uniformizer $\pi$, and a finite residue field $k={\mathbb{F}}_q$, $q=p^f$.

We construct the coefficient field rather than presuming it exists. The polynomial $T^q-T$ has derivative $-1$ in characteristic $p$. Every residue element therefore has a unique polynomial root lift by Hensel's lemma. The set $S$ of these $q$ lifts is a field: qth powers preserve addition and multiplication, so sums and products remain polynomial root lifts; a nonzero lift has inverse $a^{q-2}$. Reduction identifies $S$ with ${\mathbb{F}}_q$. This is the finite coefficient field in positive-characteristic local fields.

For $x\in {\mathcal{O}}_K$, choose $a_0\in S$ with the same residue, write $x=a_0+\pi x_1$, and repeat with $x_1\in {\mathcal{O}}_K$. Completeness givesThe first nonzero coefficient determines the valuation, so the expansion is unique. Multiplying an arbitrary field element by a suitable power of $\pi$ reduces it to this case. Consequently substitution $T\mapsto \pi$ gives a valued-field isomorphismAddition and multiplication agree with those of Laurent series by convergence, and a nonzero series has a nonzero leading coefficient, establishing injectivity as well as surjectivity.

Conversely, ${\mathbb{F}}_q((T))$ is complete for its T-adic absolute value on a field. Its valuation ring ${\mathbb{F}}_q[[T]]$ is compact: fixing finitely many coefficients gives the finite partitions into balls, and the successive coefficient choices realize the inverse limit of the finite rings ${\mathbb{F}}_q[T]/(T^n)$. Equivalently it is the product of countably many finite discrete coefficient sets. Thus it is a non-Archimedean local field. Distinct $q$ give nonisomorphic valued fields because their residue fields have different sizes. Replacing $|T|$ by any number in $(0,1)$ gives an equivalent absolute value on a field.

For $L={\mathbb{Q}}_p({\zeta }_p)$, put $\pi ={\zeta }_p-1$. The polynomialis Eisenstein at $p$. It gives $[L:{\mathbb{Q}}_p]=e=p-1$, total ramification, and $v_L(\pi )=1$. Also ${\mathcal{O}}_L={\mathbb{Z}}_p[\pi ]$: in the power basis, terms $c_i{\pi }^i$ have valuations $(p-1)v_p(c_i)+i$ in different residue classes modulo $p-1$, so cannot cancel at the minimum. Integrality forces all $c_i$ integral.

Use part (b) with ${\Phi }_p(T)$ and differentiate the quotient at ${\zeta }_p$:The root of unity is a unit, while $v_L(p)=p-1$ and $v_L(\pi )=1$. Hence the cyclotomic local different exponent isThis includes $p=2$: that extension is trivial and the exponent is zero.

For $m=p^2-1$, construct an unramified quadratic extension $E/{\mathbb{Q}}_p$ by lifting an irreducible quadratic polynomial over ${\mathbb{F}}_p$. Its residue field is ${\mathbb{F}}_{p^2}$. Each of its nonzero residue elements is a simple polynomial root of $T^m-1$, so Hensel's lemma lifts all $m$ polynomial roots into $E$. A primitive residue element lifts to a primitive mth polynomial root: its reduction already has order $m$, and its mth power is one.

Such a primitive polynomial root cannot generate a proper subfield of $E$, since its residue has order $p^2-1>p-1$ and therefore generates the quadratic residue extension. Thus ${\mathbb{Q}}_p({\zeta }_m)=E$ is unramified of degree two. Part (a) gives ${\mathcal{O}}_E={\mathbb{Z}}_p[{\zeta }_m]$. The reduced minimal polynomial of an algebraic element has degree two and is separable, so its derivative at ${\zeta }_m$ is a unit. Part (b) givesBoth answers use normalized discrete valuations on the top field.

Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with Galois group $G$, normalized valuation $v_L$, integer ring ${\mathcal{O}}_L$, and residue characteristic $p$. The lower ramification numbering isThese higher ramification groups measure successively finer agreement with the identity. The group $G_0$ is the inertia group, the kernel of the action on the residue field. Each $G_i$ is normal in $G$: conjugation permutes integral elements and preserves their valuations. The groups eventually become trivial, since each nonidentity automorphism moves some integral element and its displacement has finite valuation.

Let $K_0=L^{G_0}$, the maximal unramified subextension of a local field extension. Then $G/G_0$ is the Galois group of the residue extension, cyclic and generated by a Frobenius automorphism. For a uniformizer $\pi$ of $L$, the uniformizer criterion for lower ramification groups givesTo see why one uniformizer suffices, the extension $L/K_0$ is totally ramified and its integer ring is ${\mathcal{O}}_{K_0}[\pi ]$. Elements of $G_0$ fix its coefficients. Differences of powers of $\pi$ are divisible by $\sigma \pi -\pi$, so the valuation bound for $\pi$ implies it for every integral element. The converse follows by testing $\pi$ itself.

The tame inertia quotient is seen explicitly fromIt is a homomorphism because inertia acts trivially on the residue field, and its kernel is $G_1$. Thus $G_0/G_1$ is cyclic of order not divisible by $p$. For $i\ge 1$, the homomorphismhas kernel $G_{i+1}$. Indeed composing two automorphisms adds the leading displacements, since $\sigma \pi /\pi \equiv 1$ and inertia fixes residue coefficients. Hence each $G_i/G_{i+1}$ is an elementary abelian p-group. Eventual triviality makes $G_1$ a p-group; the prime-to-p quotient makes it the unique Sylow subgroup of $G_0$. This is the wild inertia group. Tame ramification means $G_1=1$.

The filtration also measures the different exponent. For the totally ramified extension over $K_0$, the minimal polynomial of an algebraic element $g$ of $\pi$ has polynomial roots $\sigma \pi$, $\sigma \in G_0$. The derivative formula givesFor a fixed $\sigma$, its displacement valuation is the number of indices $i\ge 0$ for which it belongs to $G_i$. Interchanging the finite sums therefore gives different exponent from ramification groups:The unramified part contributes zero. More generally different in a tower follows by taking products of bases of a module dual for the trace pairing, and givesThus a tame extension has different exponent $e-1$, while nontrivial wild groups contribute additional positive terms.

The lower numbering restricts to subgroups: if $H=\mathrm{Gal}(L/E)$, its groups are $H_i=H\cap G_i$, directly from the definition using the same top-field valuation. Quotients need a different indexing. For real $s\ge 0$ put $G_s=G_{\lceil s\rceil }$ and define the Herbrand functionThis upper ramification numbering stretches or compresses intervals according to the inertia indices. The quotient theorem says that for a normal subgroup $H$, upper ramification groups commute with quotients:In particular the same upper index has a meaning compatible with passing to a smaller Galois extension, which lower indices generally lack.

For a tame totally ramified extension the picture is simple: $G_0=G$, all $G_i$ with $i\ge 1$ are trivial, and the different exponent is $|G|-1$. The extension ${\mathbb{Q}}_p({\zeta }_p)$ for odd $p$ illustrates this, giving exponent $p-2$.

A mixed-characteristic wild family is the lower ramification filtration of a prime-power cyclotomic extension. Take $L={\mathbb{Q}}_p({\zeta }_{p^r})$ with odd $p$ and $\pi ={\zeta }_{p^r}-1$. If ${\sigma }_a(\zeta )={\zeta }^a$ and $a\ne 1$ modulo $p^r$, thenThe last equality follows because ${\zeta }^{a-1}$ has order $p^{r-v_p(a-1)}$ and its uniformizer valuation scales by the ramification index in that cyclotomic tower. The lower breaks are $0,p-1,p^2-1,\dots ,p^{r-1}-1$. Integrating the successive indices $(p-1),p(p-1),\dots$ gives upper breaks $0,1,\dots ,r-1$.

In equal characteristic there is an equally explicit wild example. Start with $K={\mathbb{F}}_q((t))$ and an Artin–Schreier extensionThere is no polynomial root in $K$: a negative-valuation element has $v(x^p-x)=pv(x)$, while the right side has valuation $-m$. Adjoining a polynomial root splits the polynomial, whose polynomial roots are $y+c$ for $c\in {\mathbb{F}}_p$, so the extension is cyclic of degree $p$. The valuation equation forces total ramification, $v_L(t)=p$ and $v_L(y)=-m$. Choose $1\le b\le p-1$ with $-bm\equiv 1(\mathrm{mod}p)$ and set $a=(1+bm)/p$. Then $\pi =t^a{y}^b$ is a uniformizer. For a nonidentity automorphism $y\mapsto y+c$,the first binomial coefficient $b$ is nonzero modulo $p$, and all subsequent terms have strictly higher valuation. ThereforeThis is the ramification break of an Artin–Schreier pole. It shows concretely that wild breaks, unlike tame inertia, can occur arbitrarily far out in the filtration.

Let $K=\mathbb{Q}(\alpha )$ with $\alpha$ an algebraic integer, let $m\in \mathbb{Z}[X]$ be its monic minimal polynomial of an algebraic element, and write $A=\mathbb{Z}[\alpha ]$. The Kummer-Dedekind theorem applies to a rational prime $p$ not dividing the finite index $[{\mathcal{O}}_K:A]$. Ifwith distinct monic irreducible polynomials $g_j$, choose monic integer lifts ${\tilde{g}}_j$. Then the distinct prime ideals above $p$ areIn particular, $p$ is unramified precisely when $\overline{m}$ has no repeated irreducible factor. The exclusion of primes dividing the index is essential.

Here is the quotient-ring proof of the Kummer-Dedekind theorem. The finite group ${\mathcal{O}}_K/A$ has order coprime to $p$. Localizing at the integers coprime to $p$ therefore gives $A\otimes {\mathbb{Z}}_{(p)}={\mathcal{O}}_K\otimes {\mathbb{Z}}_{(p)}$, and reduction yieldsThe second isomorphism follows from $A\simeq \mathbb{Z}[X]/(m)$. The Chinese remainder theorem givesEach factor is a local ring with unique maximal ideal generated by $g_j$ and residue field ${\mathbb{F}}_p[X]/(g_j)$. Pulling its maximal ideal back to ${\mathcal{O}}_K$ gives exactly ${\mathfrak{p}}_j$; these exhaust the primes above $p$. This proves the asserted residue degrees and their independence of the chosen lifts.

It remains to prove the exponents rather than simply identify the primes. Unique prime ideal factorization in the Dedekind domain ${\mathcal{O}}_K$ writes $(p)={\prod }_j{\mathfrak{p}}_j^{b_j}$. At ${\mathfrak{p}}_j$, its discrete valuation ring has $p=u{\pi }^{b_j}$, with $u$ a unit. Thus the maximal ideal of ${\mathcal{O}}_{K,{\mathfrak{p}}_j}/p{\mathcal{O}}_{K,{\mathfrak{p}}_j}$ has nilpotence index of an ideal $b_j$: its $b_j$th power is zero and its $(b_j-1)$st is nonzero. In ${\mathbb{F}}_p[X]/(g_j^{e_j})$ this index is $e_j$, since $g_j^{e_j}=0$ but $g_j^{e_j-1}\ne 0$. The displayed product isomorphism identifies these local factors, hence $b_j=e_j$. This completes the theorem's proof. The same local argument works over a base number field whenever ${\mathcal{O}}_K[\alpha ]$ equals the full relative ring of integers after localizing at the chosen base prime.

For the required cyclotomic field, the allowed ring of integers formula makes the index one. Its cyclotomic polynomial isFor $p\notin \{2,5\}$, the polynomial $X^{80}-1$ has derivative $80X^{79}$ coprime to it modulo $p$. Its divisor ${\overline{\Phi }}_{80}$ is therefore squarefree, so the Kummer-Dedekind theorem gives no ramification.

The cyclotomic reduction at a conductor prime is explicit at the remaining two primes:The quartic over ${\mathbb{F}}_2$ has no linear factor. Its only possible irreducible quadratic divisor would be $X^2+X+1$, whose roots have order three, whereas its own roots have order five. Hence it is irreducible. The two quartics over ${\mathbb{F}}_5$ are also irreducible: their roots have multiplicative order sixteen because their fourth powers are $2$ or $3$, both of order four; the finite-field Frobenius automorphism has orbit length four on an element of order sixteen, since $5^4\equiv 1(\mathrm{mod}16)$ and neither $5$ nor $5^2$ is $1$ modulo sixteen.

Consequently the prime ideal factorization above $2$ has one prime with ramification index eight and residue degree four. Above $5$ it has two primes, each with ramification index four and residue degree four. In particular,

Put $a=\sqrt[3]{2}$ and $\zeta ={\zeta }_3$. The polynomial $X^3-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $2$, so $[\mathbb{Q}(a):\mathbb{Q}]=3$. This field is real and cannot contain the nonreal $\zeta$. Adjoining $\zeta$ gives a quadratic extension and hence $[F:\mathbb{Q}]=6$. Since $F$ is the splitting field of $X^3-2$, its Galois group acts faithfully on $a,\zeta a,{\zeta }^{2}a$ and has order six. Consequently

For a prime $\mathfrak{q}$ over a rational prime $p$, its absolute residue degree is the order of the cyclic quotient group $D_{\mathfrak{q}}/I_{\mathfrak{q}}$ from part (i). If that degree were six, then $D_{\mathfrak{q}}$, a subgroup of $S_3$, would have order at least six, so $D_{\mathfrak{q}}=S_3$ and $I_{\mathfrak{q}}=1$. This would make $S_3$ cyclic, a contradiction. The argument applies also at ramified primes. Thus

For the second request printed with the same subpart label, work over $E=\mathbb{Q}(\zeta )$. The Galois group of $F/E$ is cyclic of order three, with generator $\sigma (a)=\zeta a$ and $\sigma (\zeta )=\zeta$. The ideal norm of $(\zeta +3)$ is seven, sinceIn its residue field we have $\zeta \equiv -3\equiv 4(\mathrm{mod}{\mathfrak{p}}_1)$. The polynomial $X^3-2$ has no root in ${\mathbb{F}}_7$, because its cube residues are $0,1,6$. It is therefore irreducible there. Its discriminant $-108$ is a unit at ${\mathfrak{p}}_1$. To check the local index condition, the trace pairing matrix of $1,a,a^2$ has this unit determinant. For any integral element $b$ of $F$, its traces against these three basis elements are integral; solving with this invertible matrix shows that its basis coefficients are integral locally. Thus the relative ring of integers equals ${\mathcal{O}}_E[a]$ at ${\mathfrak{p}}_1$. The relative Kummer-Dedekind theorem gives one unramified prime above ${\mathfrak{p}}_1$, of residue degree three.

Use arithmetic Frobenius, characterized by $b\mapsto b^7$ on this residue field. As $a^3=2$,The three possible images $a,\zeta a,{\zeta }^{2}a$ remain distinct modulo $\mathfrak{q}$, since $a$ is nonzero and $1,4,2$ are distinct modulo seven. This congruence therefore identifies the actual automorphism uniquely:The geometric Frobenius convention would give its inverse. Since $F/E$ is abelian, the arithmetic element is independent of the choice of prime above ${\mathfrak{p}}_1$.

Write $G=\mathrm{Gal}(F/K)$ andEvery member of this decomposition group preserves ${\mathcal{O}}_F$ and $\mathfrak{q}$, so reduction defines a homomorphismIts kernel is the inertia group. We prove surjectivity of decomposition group reduction without presupposing a lift of Frobenius.

If the two residue fields coincide, surjectivity is immediate. Otherwise choose a nonzero $\overline{a}\in k_{\mathfrak{q}}$ generating it over $k_{\mathfrak{p}}$; a generator exists by the primitive element theorem for finite fields. Use the Chinese remainder theorem in ${\mathcal{O}}_F$ to choose $a$ whose residue is $\overline{a}$ at $\mathfrak{q}$ and zero at every other prime above $\mathfrak{p}$. ConsiderIts coefficients are integral and fixed by $G$, hence lie in ${\mathcal{O}}_K$: an element of $K$ integral over ${\mathcal{O}}_K$ belongs to this integrally closed ring of integers.

Reduce at $\mathfrak{q}$. When $\sigma \notin D_{\mathfrak{q}}$, the chosen residue of $a$ at ${\sigma }^{-1}\mathfrak{q}$ is zero, so $\sigma a$ reduces to zero at $\mathfrak{q}$. When $\sigma \in D_{\mathfrak{q}}$, its residue is $\overline{\sigma }(\overline{a})$, which is nonzero. ThereforeAlthough this product is written in $k_{\mathfrak{q}}[T]$, its coefficients belong to $k_{\mathfrak{p}}$. Put $q=|k_{\mathfrak{p}}|$. The finite-field Frobenius automorphism $b\mapsto b^q$ therefore permutes its nonzero roots. In particular, ${\overline{a}}^q$ is $\overline{\sigma }(\overline{a})$ for some $\sigma \in D_{\mathfrak{q}}$. Since $\overline{a}$ generates $k_{\mathfrak{q}}$, that $\overline{\sigma }$ agrees with Frobenius on the whole field. The Galois group of a finite field extension is cyclic and generated by Frobenius, so its generator lies in the image. Thus

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 $K$.

All geometric 2-torsion is rational: its points are $O,(\alpha ,0),(\beta ,0),(\gamma ,0)$. We will construct the required finite quotient directly, without assuming any theorem about finite generation or two-descent on an elliptic curve.

Choose $Q\in E(\overline{K})$ with $2Q=P$ and define the halving cocycle with rational two-torsionSuch a half exists because the nonconstant multiplication morphism $[2]$ on a smooth projective elliptic curve is surjective over an algebraic closure. Since $E[2]$ is rational, $c_P(\sigma \tau )=c_P(\sigma )+c_P(\tau )$. A different half is $Q+T$ with rational $T\in E[2]$ and gives the same cocycle. Replacing $P$ by $P+2R$, with $R\in E(K)$, allows the half $Q+R$ and again changes nothing. Adding chosen halves shows additivity in $P$. If the cocycle is zero, the half is fixed by the absolute Galois group and lies in $E(K)$, so $P\in 2E(K)$. This proves, rather than assumes, an injectionThe fixed field of $\ker c_P$ is exactly $K(Q)$. Its Galois group is the image of $c_P$, so its degree is at most four. This also follows from the permitted biquadratic description.

Let $S$ be the finite set of primes dividing $2(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$. At a place $v\notin S$ the displayed integral cubic has unit elliptic-curve discriminant, hence good reduction of an elliptic curve, and the residue characteristic is odd. Choose a half $\overline{Q}$ of the reduction of $P$ over the algebraic closure of the residue field. Its coordinates belong to a finite residue extension; let $L_v/K_v$ be the corresponding finite unramified extension. Smoothness and the Hensel lemma lift $\overline{Q}$ to some $Q_0\in E(L_v)$. The difference $P-2Q_0$ reduces to zero and belongs to the formal kernel of a minimal Weierstrass equation. In its integral formal group law the doubling series is $2t+O(t^2)$, with unit linear coefficient. The prime-to-residue-characteristic multiplication on a formal group gives a unique $R$ in this kernel such that $2R=P-2Q_0$. Thus $Q_0+R$ is a half of $P$ defined over an unramified local extension. Every other half differs by rational 2-torsion, so it too is unramified. We have proved the unramified halving fields for a split cubic assertion at every $v\notin S$.

There are only finitely many bounded-degree extensions with restricted ramification of $K$ of degree at most four. This is the permitted number-field finiteness consequence of the Hermite–Minkowski theorem: bounded local degrees bound the discriminant exponents at the finitely many allowed ramified primes, so absolute discriminants are bounded. Each possible halving field $L$ is Galois, and there are only finitely many homomorphisms $\mathrm{Gal}(L/K)\to E[2]$. The injection constructed above therefore has finite image. Consequently $E(K)/2E(K)$ is finite. No elliptic-curve finiteness or descent theorem was used in this argument.

For $K=\mathbb{Q}$, write $s=\#S$. Each coordinate of $c_P$ in a basis of $E[2]\cong (\mathbb{Z}/2\mathbb{Z}{)}^2$ is a quadratic Galois character unramified outside $S$. Its quadratic extension has a signed square-free integer representative $d=\pm {\prod }_{p\in S}{p}^{e_p}$, $e_p\in \{0,1\}$: an odd valuation outside $S$ would ramify there. There are at most $2^{s+1}$ choices, including the trivial character. Hence the pair of characters gives at most $2^{2s+2}$ cocycles. Each prime in $S$ is at most $2m$, because the three root differences have absolute values at most $2m$ and $2\le 2m$. Among the integers up to $2m$ there are at most $m$ primes: two, together with at most $m-1$ odd candidates. The rational halving quotient bound from root size is therefore

Take the formal Laurent series field $K=\mathbb{C}((T))$, with $v_T$ the least exponent of a nonzero term and $|x|=2^{-v_T(x)}$. Its value group is $\mathbb{Z}$, so this is a discrete valuation. A Cauchy sequence has an eventually stable coefficient at every exponent, and its tail has a common lower bound on its exponents. The stabilized coefficients therefore give a formal Laurent series field element to which the sequence converges. Hence $K$ is complete.

Its valuation ring is $\mathbb{C}[[T]]$, and its residue field is the infinite field $\mathbb{C}$. If $K$ were locally compact, the previous compact-ball argument would make $\mathbb{C}[[T]]$ compact. Its quotient by the open ideal $T\mathbb{C}[[T]]$ would then be compact and discrete, hence finite, contradicting its identification with $\mathbb{C}$. Thus $\boxed{\mathbb{C}((T))\text{ is complete and discretely valued but not locally compact}}$.

For each finite place $v\in S$, choose $d_v\in K_v^{\times }$ such that $K_v(\sqrt{d_v})/K_v$ is the unramified quadratic extension. Such a choice exists even at residue characteristic two. At odd residue characteristic use a unit with nonsquare residue. At residue characteristic two, choose $\overline{c}$ for which $X^2-X-\overline{c}$ is irreducible over the finite residue field, lift $c$ to ${\mathcal{O}}_v$, and set $d_v=1+4c$. The lifted polynomial has unit derivative at every residue point and gives the unramified degree-two extension; its root is $(1+\sqrt{d_v})/2$.

The square classes are open in $K_v^{\times }$. Indeed the strong form of Hensel lemma applied to $X^2-u$ at $1$ shows that $u$ is a square whenever $v(u-1)>2v(2)$. Thus any sufficiently close approximation $a$ to $d_v$ has $a/d_v$ square. By weak approximation for number fields, choose $a\in K$ satisfying this at every finite $v\in S$. Then the local extensions $K_v(\sqrt{a})$ are all unramified of degree two. At least one such condition ensures $a$ is not a square in $K$. If $S$ contains no finite place, impose the same condition at one auxiliary finite place.

Set $L=K(\sqrt{a})$. At every specified finite prime there is one prime of $L$ above it, with ramification index one and residue degree two. Hence $\boxed{\mathfrak{p}{\mathcal{O}}_L\text{ is prime for every finite }\mathfrak{p}\in S}$. The dyadic case is covered by the unramified construction, rather than by a nonexistent nonsquare residue unit in characteristic two.

We count extensions up to $K$-isomorphism. An extension of degree $n$ has a maximal unramified extension $E/K$ inside it, with $[E:K]=f\mid n$, and $L/E$ is totally ramified of degree $e=n/f$. The unramified extension of each degree is unique, so there are only finitely many possibilities for $E$.

Fix $E$ and $e$. The coefficient space of monic degree-$e$ Eisenstein polynomials is compact: its nonconstant nonleading coefficients lie in the compact maximal ideal, and its constant coefficient lies in the compact set ${\mathfrak{m}}_E\setminus {\mathfrak{m}}_E^2$. Each such polynomial $F$ has a root $\alpha$ generating a degree-$e$ extension. For sufficiently close coefficients of a polynomial $G$, the numbers $G(\alpha )$ and $G^{\prime }(\alpha )-F^{\prime }(\alpha )$ are arbitrarily small, while $F^{\prime }(\alpha )\ne 0$ because the characteristic is zero. The strong form of Hensel lemma, applied in $E(\alpha )$, supplies a root $\beta$ of $G$ as close to $\alpha$ as desired. Choose the neighborhood so that the Krasner's lemma inequality holds. Then $E(\alpha )\subseteq E(\beta )$, and both degrees are $e$, so the extensions are equal. Compactness gives a finite cover by such neighborhoods, proving finiteness of degree-$n$ extensions. If extensions are realized as subfields of a fixed algebraic closure rather than counted up to isomorphism, each has only finitely many embeddings, so that count is finite too.

For odd $p$, write $K^{\times }={\pi }^{\mathbb{Z}}{\mathcal{O}}_K^{\times }$. The square map is bijective on $1+{\mathfrak{m}}_K$ by Hensel lemma, and the finite residue field has cyclic multiplicative group of even order. ThusThe quadratic extensions from square classes correspond to its three nontrivial classes. Representatives are $u,\pi ,u\pi$, where $u$ is a unit with nonsquare residue. There are $\boxed{3}$ quadratic extensions: $K(\sqrt{u})$ is unramified, while $K(\sqrt{\pi })$ and $K(\sqrt{u\pi })$ are totally ramified. For completeness, two nonsquare parameters give the same extension only if their ratio is a square: writing $\sqrt{b}=x+y\sqrt{a}$ and squaring forces $x=0$, hence $b/a=y^2$.

Use $y$ on the target and $x$ on the source, so the map of coordinate rings sends $y$ to $x^4$. A closed point of the real affine line corresponds to a maximal ideal generated either by $y-a$, $a\in \mathbb{R}$, or by a monic irreducible quadratic $q(y)$. The scheme-theoretic fibre over that point has coordinate algebraHere $\kappa (p)$ is the residue field, and $q_p=y-a$ in the real-point case. The fibre is a scheme, so its nilpotents must be retained.

For $a>0$, put $r=a^{1/4}$. The factors $x-r$, $x+r$, $x^2+r^2$ are pairwise coprime, with the last one irreducible over $\mathbb{R}$. The Chinese remainder theorem identifies $B_p$ with $\mathbb{R}\times \mathbb{R}\times \mathbb{C}$, giving three irreducible components.

For $a<0$, put $r=(-a{)}^{1/4}$. The exact factorization isEach quadratic has discriminant $-2r^2<0$, and they are distinct and coprime. Thus $B_p\cong \mathbb{C}\times \mathbb{C}$, giving two irreducible components.

For $a=0$, $B_p=\mathbb{R}[x]/(x^4)$. It has one prime ideal, generated by the class of $x$, so one irreducible component. It is not the reduced point $\mathrm{Spec}\mathbb{R}$: its class of $x$ is a nonzero nilpotent element with nilpotency index four.

Finally write the irreducible quadratic as $q_p(y)=(y-z)(y-\overline{z})$, with $z\notin \mathbb{R}$. Over $\mathbb{C}$, $q_p(x^4)$ has four roots above $z$ and four above $\overline{z}$. They are all simple, because $z\ne 0$, and the two sets are disjoint. None is real. Conjugation pairs them into four distinct irreducible real quadratics. Hence $B_p\cong {\mathbb{C}}^4$ and this fibre has four irreducible components.

The real fourth-power fibre classification is:

All fibres in each row are isomorphic as affine schemes over $\mathbb{R}$. The rows are distinct: their numbers of irreducible components already distinguish them, and the zero fibre is additionally nonreduced. ThereforeThese are ordinary scheme components, not geometric components counted after complexifying the base. Counting only real roots would miss both the complex-residue factors and the nonreal closed points.

A nilpotent element belongs to every prime ideal and therefore to every maximal ideal, soFor the reverse containment, let $r$ be nonnilpotent. The localization of a ring $S=R[1/r]$ is nonzero: if $1=0$ there, some power of $r$ would annihilate $1$ in $R$. This is still a finite-type integer algebra, because it can be presented as $R[T]/(rT-1)$.

Choose a maximal ideal $\mathfrak{n}$ of $S$. Its residue field $L=S/\mathfrak{n}$ is a finite-type integer algebra, hence finite by part (a). Consider the image $F$ of $R$ in $L$. It is a finite integral domain containing $1$, so it is a field: multiplication by a nonzero element is injective on the finite set $F$, hence surjective, and therefore has an inverse in $F$.

The kernel $\mathfrak{m}$ of $R\to F$ is consequently maximal. The image of $r$ is nonzero, since $r$ became a unit in $S$ and remains a unit in its nonzero residue field. Thus $r\notin \mathfrak{m}$. Every nonnilpotent element can therefore be avoided by a maximal ideal, givingThis radical equality for finitely generated integer algebras uses part (a) to ensure that the contraction of the localized maximal ideal is maximal. Such contraction is not generally maximal for arbitrary localization of a ring; the finite-field image is the decisive additional step. The zero ring is immediate, with the intersection of its empty set of maximal ideals understood as the whole ring. No general theorem that integer algebras are Jacobson rings is being quoted.

Let $L/K$ be an extension of number fields, let $\alpha \in {\mathcal{O}}_L$ be a primitive element of a field extension, and let $f\in {\mathcal{O}}_K[X]$ be its monic minimal polynomial of an algebraic element. Put $R={\mathcal{O}}_K[\alpha ]$ and $S={\mathcal{O}}_L$. The Kummer-Dedekind theorem applies at a nonzero prime ideal $\mathfrak{p}$ of ${\mathcal{O}}_K$ when $R_{\mathfrak{p}}=S_{\mathfrak{p}}$: equivalently, $\mathfrak{p}$ is outside the support of the finite-index quotient $S/R$. Write the factorization over the residue field $k={\mathcal{O}}_K/\mathfrak{p}$ aswith distinct monic irreducible polynomials ${\overline{g}}_j$, and choose monic lifts $g_j\in {\mathcal{O}}_K[X]$. ThenHere $f({\mathfrak{P}}_j/\mathfrak{p})$ denotes the residue-field degree, and the ramification index of a prime ideal is $m_j$.

To prove this, the local index condition givesThe final isomorphism is the Chinese remainder theorem. Each factor is a local ring with unique maximal ideal generated by ${\overline{g}}_j$, so its inverse image in $S$ is exactly ${\mathfrak{P}}_j$, and its residue field is $k[X]/({\overline{g}}_j)$. This proves the list of primes and the residue-field degrees. For the exponents, localize at ${\mathfrak{P}}_j$. The ring of integers of a number field is a Dedekind domain, so $S_{{\mathfrak{P}}_j}$ is a discrete valuation ring. If $\mathfrak{p}{S}_{{\mathfrak{P}}_j}={\mathfrak{P}}_j^{e_j}{S}_{{\mathfrak{P}}_j}$, the quotient has $e_j$ successive layers isomorphic to $S/{\mathfrak{P}}_j$, and hence $k$-dimension $e_j\mathrm{deg}{\overline{g}}_j$. The corresponding polynomial factor has $k$-dimension $m_j\mathrm{deg}{\overline{g}}_j$. Thus $e_j=m_j$, completing the proof rather than merely identifying the radicals.

For the calculation, set $K=\mathbb{Q}(\sqrt{5})$ and $t=(\sqrt{5}-1)/2$. The ring of integers of a quadratic field gives ${\mathcal{O}}_K=\mathbb{Z}[t]$, with $t^2+t-1=0$. The algebraic norm of $4+\sqrt{5}$ is $11$, so its principal ideal has ideal norm $11$. Its quotient ring has prime order and is therefore the finite field ${\mathbb{F}}_{11}$; consequently $(4+\sqrt{5})$ is a prime ideal. In that residue field, $\sqrt{5}=-4=7$ and $t=3$.

Let $L=\mathbb{Q}({\zeta }_5)$. Since $t={\zeta }_5+{\zeta }_5^{-1}$ lies in $\mathbb{Z}[{\zeta }_5]$, the permitted identity ${\mathcal{O}}_L=\mathbb{Z}[{\zeta }_5]$ gives ${\mathcal{O}}_L={\mathcal{O}}_K[{\zeta }_5]$. Thus the local index condition holds at every prime. The relative minimal polynomial of an algebraic element is $X^2-tX+1$, whose reduction isThe roots are distinct. Applying the Kummer-Dedekind theorem givesFor the last equality, evaluation at $5$ or $9$ in ${\mathbb{F}}_{11}$ gives ${\zeta }_5+{\zeta }_5^{-1}=3$ and therefore $4+\sqrt{5}=5+2t=0$. Both descriptions are the same kernels of these evaluation maps. Each prime has ideal norm $11$, residue-field degree one and ramification index of a prime ideal one.