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Every finite extension $L/K$ of non-Archimedean local fields has a unique intermediate field $K_0$ whose residue field is $k_L$ and for which $K_0/K$ is unramified. The extension $L/K_0$ is totally ramified.

If a non-Archimedean local field $K$ has uniformizer $\pi$, valuation ring ${\mathcal{O}}_K$, maximal ideal ${\mathfrak{m}}_K$ and residue field ${\mathbb{F}}_q$, thenThe first factor records the discrete valuation, the Teichmuller representatives split reduction ${\mathcal{O}}_K^{\times }\to {\mathbb{F}}_q^{\times }$, and $1+{\mathfrak{m}}_K$ is the group of principal units.

For an affine plane algebraic curve $F(x,y)=0$, a point over an algebraic closure is nonsingular when the displayed condition holds. Its Zariski tangent space has dimension one, and a local coordinate can be chosen in the direction of a nonzero partial derivative. A smooth algebraic curve has all points nonsingular. Over a complete local field, the Hensel lemma lifts a nonsingular residue point by solving for a coordinate with unit partial derivative.

For a local field $F$ containing the Nth roots of unity, fix the Local Artin map with arithmetic Frobenius convention. The Hilbert norm residue symbol can be defined byIt is independent of the chosen root, continuous and bimultiplicative, factors through $F^{\times }/(F^{\times }{)}^N$ in each variable, and gives a nondegenerate pairing on that finite group. It satisfies $(a,b{)}_{F,N}=(b,a{)}_{F,N}^{-1}$ and $(a,1-a{)}_{F,N}=1$ for $a\ne 1$. Its norm interpretation isFor $N=2$ the quadratic Hilbert symbol is symmetric and sign-valued. It is also characterized by the existence of a nonzero solution of $z^2=a{x}^2+b{y}^2$. Moreover $(a,-a{)}_F=1$, and therefore $(a,a{)}_F=(a,-1{)}_F$. We now compute both the square-class groups and the pairing explicitly.

For odd $p$, write $a=p^{\alpha }u$ with $u\in {\mathbb{Z}}_p^{\times }$. An element is a square exactly when $\alpha$ is even and the residue of $u$ is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency follows from the simple-root form of Hensel lemma applied to $X^2-u$, since $2x$ is then a unit. If $\eta$ is any unit with nonsquare residue, this givesThe nonsquare unit defines an unramified quadratic extension. Its field norms include every unit and have even valuation, as follows from the local norm calculation below. Thus $(\eta ,u{)}_p=1$ for every unit, but $(\eta ,p{)}_p=-1$. Finally $(p,p{)}_p=(p,-1{)}_p=(-1/p)$, by symmetry and the preceding unit calculation. Bimultiplicativity now proves explicit quadratic Hilbert symbols over the p-adic numbers:Here the factors are Legendre symbols of the nonzero residues. The formula works for arbitrary integer valuations $\alpha ,\beta$, since only their parity matters.

At $p=2$, the square of every odd integer is one modulo eight. Conversely, for an odd $u\equiv 1(\mathrm{mod}8)$, apply the strong form of Hensel lemma to $X^2-u$ at $1$: the error has valuation at least three, while the derivative has valuation one. Thus the unit square classes of the p-adic integers are precisely their four residues modulo eight. Combining this with parity of the valuation givesFor an odd unit putInspection of the residues $1,3,5,7$ gives $[u]=[-1{]}^{ϵ(u)}[5{]}^{\omega (u)}$. It remains to compute six basis pairings. Explicit field norms giveindeed $2=1^2+1^2$, $5=1^2+2^2$, $2=2^2-2\cdot 1^2$, and $5=5^2-5\cdot 2^2$.

In contrast $(-1,-1{)}_2=-1$. To justify the obstruction even for rational 2-adic coordinates, write $x=2^t{x}_0$, $y=2^t{y}_0$ with $x_0,y_0$ integral and at least one odd. Then $v_2(x_0^2+y_0^2)$ is zero if exactly one is odd, and one if both are odd. A sum of squares equal to the unit $-1$ consequently forces $t=0$ and exactly one coordinate odd. Its residue modulo four would then be one, whereas $-1$ is three.

Similarly $(2,5{)}_2=-1$. If $x^2-2y^2$ is a unit, the valuations of the two terms have opposite parity, so cannot cancel. It follows that $x$ is an odd integral element and $y$ is integral. Modulo eight its norm is then $1$ or $7$, never $5$. These signs, symmetry, and bimultiplicativity giveThe two boxed formulas determine the quadratic Hilbert symbol on all nonzero elements, including every rational prime $p$.

A central simple algebra over $K$ is a finite-dimensional associative $K$-algebra with no nonzero proper two-sided ideals and with center $K$. The Artin–Wedderburn theorem writes it uniquely as $M_r(D)$ with $D$ a central division algebra. After a finite separable splitting field extension it becomes $M_d$, so its dimension is $d^2$, and $d$ is its degree of a central simple algebra. Passing from $D$ to a matrix algebra over $D$ does not change its Brauer class.

The Brauer group consists of these classes, with product induced by tensor product. Extending to a common splitting field shows that a tensor product of central simple algebras is again central simple; its two-sided ideals and center descend from the corresponding full matrix algebra. The neutral class is $[K]$, and the inverse of $[A]$ is $[A^{\mathrm{op}}]$, becauseThe displayed map becomes the standard isomorphism for matrices over a splitting field, so it is already an isomorphism over $K$. Thus Brauer classes measure precisely the obstruction to being a split matrix algebra.

An explicit family is supplied by cyclic algebras. Let $L/K$ be cyclic of degree $n$, with generator $\sigma$, and let $a\in K^{\times }$. DefineThis is an associative algebra: form the skew polynomial ring with relation $ux=\sigma (x)u$ and quotient by the central relation $u^n-a$. Reduction of powers gives the displayed $n^2$-dimensional $K$-basis. An element commuting with every $x\in L$ has no $u^i$ coefficient for $i\ne 0$, since ${\sigma }^i$ is not the identity. Commuting also with $u$ forces the remaining coefficient into $K$. Hence the center is $K$.

Here is an explicit splitting construction. On $L^n$, with basis $e_0,\dots ,e_{n-1}$, putThen $UD(x)=D(\sigma x)U$ and $U^n=aI$. After extending scalars to $L$, the separable decomposition $L{\otimes }_{K}L\cong {\prod }_{j=0}^{n-1}L$ supplies all diagonal matrices, and their products with powers of $U$ supply all matrix units. ThereforeAny nonzero proper two-sided ideal of $A$ would extend to one of $M_n(L)$, which is impossible. Thus this construction really gives a central simple algebra, with splitting field $L$.

Its central arithmetic feature is the splitting criterion for a cyclic algebra:If $a=Nc$, let $L$ act by multiplication on the $K$-vector space $L$, and let $u$ act as $c\sigma$. Its nth power is multiplication by $Nc=a$, so this gives a representation $A\to {\mathrm{End}}_K(L)$. It is injective by simplicity and an isomorphism by equal dimension. Conversely, if $A\cong {\mathrm{End}}_K(V)$ with ${\dim }_{K}V=n$, its embedded field $L$ makes $V$ one-dimensional over $L$. The relation for $u$ says that $u$ is $\sigma$-semilinear. In an $L$-basis it must have the form $c\sigma$, whose nth power is $Nc$. The relation $u^n=a$ forces $a=Nc$.

More generally the crossed-product algebra of a Galois extension attaches an algebra to a multiplicative two-cocycle. Changing the cocycle by a coboundary rescales its basis and preserves the Brauer class. Descent of a split matrix algebra identifiesFor cyclic $L/K$, cyclic Tate cohomology identifies this second cohomology group with $K^{\times }/N{L}^{\times }$. In a crossed product, rescale successive cyclic basis elements to powers of a single $u$; the remaining datum is $u^n=a$, changed by a field norm when $u$ is rescaled. This explains why cyclic algebras give every class split by a cyclic extension, and why multiplication of their parameters is addition of their Brauer classes.

For a finite field $K={\mathbb{F}}_q$, every finite splitting field $L={\mathbb{F}}_{q^n}$ is cyclic over $K$. The field norm is the power map $x\mapsto x^{(q^n-1)/(q-1)}$ on its cyclic multiplicative group, and is onto ${\mathbb{F}}_q^{\times }$. Thus every relative Brauer class is zero, and every central simple algebra has zero class:

For a non-Archimedean local field $K$, the valuation classification of central division algebras shows that every Brauer class splits over a finite unramified extension. For the degree-$n$ unramified extension $L/K$ with arithmetic Frobenius $\sigma$, units are norms and norm valuations are multiples of $n$. Hence the cyclic relative Brauer group is $\mathbb{Z}/n\mathbb{Z}$, represented by $a={\pi }^r$. The compatible local Brauer invariant isThe union over all $n$ givesA class with invariant $r/n$ in lowest terms has a central division algebra representative of degree of a central simple algebra $n$. This is the local valuation classification used above; the explicit cyclic algebras exhibit all of its invariants. At the Archimedean places, $\mathrm{Br}(\mathbb{C})=0$, while $\mathrm{Br}(\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$: its nontrivial class is the quaternion algebra $(\mathbb{C}/\mathbb{R},\text{conjugation},-1)$, since complex norms are positive.

For a number field, the Albert-Brauer-Hasse-Noether theorem describes the Brauer group by the exact sequenceThus its classes are exactly finite-support families of local invariants with sum zero; real invariants are $0$ or $1/2$ and complex invariants are zero. Every global Brauer class has a cyclic splitting extension, so cyclic algebras also realize all global classes. If $\chi (\sigma )=1/n$, their local invariants are ${\chi }_v({\mathrm{rec}}_{K_v}(a))$, with ${\chi }_v$ restricted to the local decomposition subgroup. The sum-zero relation for a principal parameter is the invariant form of Artin reciprocity. Injectivity supplies the local-to-global splitting test, which yields the Hasse norm theorem for cyclic extensions as explained in the preceding solution. Here the local valuation classification and the global invariant theorem are the structural theorems used to describe the Brauer groups; the construction and norm-splitting calculation have been proved explicitly.

Fix a prime $\mathfrak{q}$ of $E$ over $\mathfrak{p}$, and a prime $\mathfrak{r}$ of $LE$ over $\mathfrak{q}$. Let $\mathfrak{P}$ be its restriction to $L$. Inside the completed compositum, putThe assumption means $U/F$ is an unramified extension. Its residue degree is some $f$; the completed extension $(LE{)}_{\mathfrak{r}}/J$ is the corresponding compositum $UJ/J$. It suffices to prove that this is unramified, regardless of the ramification of $J/F$.

Here is a proof of unramified extensions remain unramified under base change using the classification already established. Choose an integral generator $\alpha$ of $U/F$ whose monic minimal polynomial $g\in {\mathcal{O}}_F[X]$ has irreducible separable reduction of degree $f$. Write $k_F={\mathbb{F}}_q$ and $k_J={\mathbb{F}}_{q^r}$. Let $J_f/J$ be the degree-$f$ unramified extension. Its residue field ${\mathbb{F}}_{q^{rf}}$ contains ${\mathbb{F}}_{q^f}$, so $\bar{g}$ has a simple root there. Hensel lemma lifts this root to a root of $g$ in $J_f$. This gives an $F$-embedding $U\hookrightarrow J_f$.

An unramified extension of a local field is Galois, so all $F$-embeddings of $U$ into a fixed algebraic closure have the same image. The original $U$ therefore lies in $J_f$, and $UJ$ is an intermediate field of the unramified extension $J_f/J$. Its ramification index is one. Since $\mathfrak{q}$ and $\mathfrak{r}$ were arbitrary,This proof also explains why no linear-disjointness assumption on $L$ and $E$, and no assumption that $E/K$ is unramified, is needed.

For $a,b\in {\mathbb{Q}}_p^{\times }$, the quadratic Hilbert symbol $(a,b{)}_p$ is one if $b$ is a field norm from ${\mathbb{Q}}_p(\sqrt{a})$, and minus one otherwise; a square $a$ gives the trivial extension and symbol one. It depends only on the two square classes. Equivalently, the conic $z^2=a{x}^2+b{y}^2$ has a nonzero rational point precisely when the symbol is one. Interchanging $a$ and $b$ in the conic gives $(a,b{)}_p=(b,a{)}_p$. The Hilbert norm residue symbol is a group homomorphism in each argument; for nonsquare $a$, the cyclic local norm index makes its group kernel a subgroup of index two. The resulting pairing on square classes is nondegenerate. Alsosince $N(\sqrt{a})=-a$. In the language of local class field theory, $(a,b{)}_p={\mathrm{Art}}_p(b)(\sqrt{a})/\sqrt{a}$; in the Brauer group, it records whether the quaternion cyclic algebra $(a,b)$ splits. These are compatible norm-obstruction descriptions.

An unramified extension of local fields of degree two has norm group consisting precisely of elements with even valuation. Indeed,and the norms on units are surjective. For the latter assertion, the norm on the finite residue fields is surjective. On successive higher principal-unit groups, the congruenceallows one to correct a proposed unit norm one digit at a time: the residue field trace is surjective, and completeness makes the corrections converge. An unramified quadratic field is unique up to isomorphism. Therefore, if $d$ is its defining nonsquare class,

For odd $p$, choose the nonsquare unit $u_0$ from part (i). The polynomial $T^2-u_0$ has irreducible separable reduction, so it defines the unramified quadratic extension. ThusThe trivial class is included in ${\Delta }_2$, since a degree-one extension is unramified. The two units pair trivially, while $(u_0,p{)}_p=-1$. Any class outside ${\Delta }_1$ has odd valuation and consequently pairs nontrivially with $u_0$. This proves directly

For $p=2$, the unit square classes from part (i) give ${\Delta }_1=\{1,-1,5,-5\}$. The element $\omega =(1+\sqrt{5})/2$ satisfies $T^2-T-1$; its reduction $T^2+T+1$ is irreducible and separable over ${\mathbb{F}}_2$. This shows that ${\mathbb{Q}}_2(\sqrt{5})$ is unramified of degree two, givingFor a fully explicit pairing check, use the generators $(-1,5,2)$. We have $(-1,5{)}_2=1$ and $(-1,2{)}_2=1$, since $5=N(1+2i)$ and $2=N(1+i)$ from ${\mathbb{Q}}_2(i)$. But $(-1,-1{)}_2=-1$: a norm equal to $-1$ would be a sum of two squares. If either summand had negative valuation, the sum would have negative valuation (equal negative valuations give valuation $1-2r<0$); if both were integral, their squares could not sum to $-1\equiv 3(\mathrm{mod}4)$. Finally, the unramified norm criterion gives $(5,2{)}_2=-1$. The identity $(a,b{)}_2=(b,a{)}_2$ and $(a,a{)}_2=(a,-1{)}_2$ determine the remaining entries. Writing $(a,b{)}_2=(-1{)}^{B(a,b)}$, the exponent matrix isFor $b=(-1{)}^s{5}^t{2}^r$, annihilating both generators $-1,5$ of ${\Delta }_1$ requires $s=r=0$, so $b\in \{1,5\}$. Annihilating $5$ alone requires $r=0$, so $b$ is a unit class. HenceThese are the Hilbert-symbol annihilators of units and unramified square classes. Annihilators here are taken with respect to this pairing, or equivalently in the character group of a finite abelian group after identifying square classes with their Hilbert-symbol characters.

Interpret the map as the action on ${\mathbb{C}}^n$. A smooth representation requires every vector to have an open stabilizer in the locally profinite group ${\mathrm{GL}}_n(F)$. Take the first standard basis vector $e_1$. Every identity neighbourhood contains matrices $d_a=\mathrm{diag}(1+a,1,\dots ,1)$ with $a\ne 0$ sufficiently small in the local field. We may require $|a|<1$, so $1+a\ne 0$. Injectivity of the field embedding givesThus no identity neighbourhood fixes $e_1$, and its stabilizer is not open. Consequently the representation is not smooth for any $n\ge 1$. This includes $n=1$, when a nonzero vector's stabilizer is just the identity. The field embedding need not be continuous, but no continuity assumption would repair this stabilizer obstruction. The existence of the stated embedding itself forces characteristic zero; in positive characteristic no field embedding into $\mathbb{C}$ exists.

The Weak Mordell-Weil theorem asserts that $E(K)/mE(K)$ is finite for an elliptic curve over a number field $K$ and an integer $m\ge 2$. The role of (i) is to make division by $m$ possible in the local formal group of an elliptic curve whenever the residue characteristic does not divide $m$.

Let $v$ be a finite place with good reduction of an elliptic curve and $v\nmid m$. The kernel of reduction of an elliptic curve is parametrized by $t=-x/y$ in the maximal ideal of the local valuation ring. Multiplication by $m$ is represented by an integral formal power seriesIts linear coefficient is a local unit. Part (i) supplies an integral compositional inverse. Both series converge on the maximal ideal, so multiplication by $m$ is a bijection on the reduction kernel, over the local field and over every finite unramified extension.

This implies unramified division torsors at good primes. For $P\in E(K)$, first divide its reduction by $m$ over the algebraic closure of the residue field. Multiplication by $m$ on the reduced curve is surjective and separable, so the chosen division point is defined over some finite residue extension. Smooth lifting gives a point $Q_0$ over the corresponding unramified local extension with the required reduction. The difference $P-m{Q}_0$ lies in the reduction kernel; the inverse series supplies $Q_1$ with $m{Q}_1=P-m{Q}_0$. Thus $Q=Q_0+Q_1$ divides $P$ over an unramified extension. Applying the same reasoning to all reduced $m$-torsion points lifts all of $E[m]$, so every point of $[m{]}^{-1}P$ lies in the maximal unramified extension. This excludes ramification outside a fixed finite set $S$ consisting of primes of bad reduction and primes over $m$.

For completeness, the remaining global finiteness step in the Kummer-theoretic proof of the weak Mordell-Weil theorem is as follows. Choose $Q\in E(\overline{K})$ with $mQ=P$. The cocycle $\sigma \mapsto \sigma Q-Q$ defines the Kummer map of an elliptic curveChanging $Q$ by an $m$-torsion point changes the cocycle by a coboundary. A zero class makes $Q$ rational after that change, so the map is injective.

Let $L=K(E[m])$, fixed independently of $P$. Since the torsion is rational over $L$, the cocycle restricted to its Galois group is a homomorphism into $E[m]$. Consequently $L(Q)/L$ is Galois of degree at most $m^2$, contains the entire division fibre, and is unramified outside $S$. Hence $[L(Q):K]\le m^2[L:K]$. There are only finitely many extensions of bounded degree unramified outside a fixed finite set: local extensions of bounded degree have bounded field discriminant exponents, and the Hermite–Minkowski theorem then applies to the bounded global field discriminants. Their finite Galois compositum $M$ contains every such division fibre; each $L(Q)$ is stable over $K$ because it contains $E[m]$ and the whole fibre above the rational point $P$. All the Kummer classes consequently factor through the fixed finite group $\mathrm{Gal}(M/K)$ with values in the fixed finite group $E[m]$. There are only finitely many such cocycles, provingThus (i) provides the local division step that bounds ramification; the global arithmetic finiteness completes the argument.

A rational point of order $2$ can be moved to $(0,0)$, giving an integral modelIts two-isogeny descent partner and the two-isogeny formula areThe maps extend over the identity and $(0,0)$, have these two points as their respective kernels, and direct substitution and the doubling formula give $\hat{\varphi }\varphi =[2]$ and $\varphi \hat{\varphi }=[2]$.

Define the two-torsion square-class homomorphisms byand similarly ${\alpha }^{\prime }$ for $E^{\prime }$, using $b^{\prime }$. To see the homomorphism property, intersect a line $y=mx+c$ with $E$. Its three $x$-coordinates have product $c^2$, since the resulting monic cubic has constant term $-c^2$. The third point is the negative of the sum of the first two, and negation preserves $x$, so the square classes multiply as required. If the line passes through $(0,0)$, the other two coordinates have product $b$, giving exactly the prescribed exceptional value. Vertical lines and repeated intersections give the identity and tangent cases.

We also verify the kernel needed for descent:For a point on $E^{\prime }$, the first coordinate of its image under $\hat{\varphi }$ is $Y^2/(4X^2)$, a square. Conversely, if $P=(s^2,y)$ has square $x$-coordinate, solving $x(\hat{\varphi }(X,Y))=s^2$ givesIts discriminant is $16(s^4+a{s}^2+b)=(4y/s{)}^2$, so it has rational nonzero roots $X=a+2s^2\pm 2y/s$. Taking $Y=\pm 2sX$ gives a point on $E^{\prime }$; choose the signs so its image has second coordinate $y$. The exceptional point $(0,0)$ lies in the image exactly when $b$ is a square, as seen from the points with $Y=0$, $X=a\pm 2\sqrt{b}$. This proves the first kernel formula. Applying the same argument to $E^{\prime }$ and then rescaling the second partner by $(X,Y)\mapsto (X/4,Y/8)$ proves the second.

There are only finitely many candidate square classes. For a rational point write $x=M/e^2$ and $y=N/e^3$, with $\gcd (M,e)=1$. These denominator forms follow because a negative prime valuation of $x$ makes the leading cubic term the unique lowest-valuation term, forcing that valuation to be even. ThenIf a prime divides $M$ but not $b$, the second factor is a unit at that prime, so its valuation in $M$ is even. Thus every class in $\alpha (E)$ has a signed squarefree representative $d\mid b$. This is the prime-support bound in two-isogeny descent; the analogous candidates for ${\alpha }^{\prime }(E^{\prime })$ divide $b^{\prime }$.

The class $d$ is realized by a nonexceptional rational point precisely when its quartic covering in a two-isogeny descenthas a rational point with $UV\ne 0$. Indeed substitution $x=d(U/V{)}^2$, $y=dUN/V^3$ gives the original equation, and conversely the square-class condition gives these parameters. Clearing denominators permits $U,V,N$ integral with $\gcd (U,V)=1$. The zero-coordinate cases account for the already known identity and two-torsion classes.

For clarity, the two-isogeny rank formula contains a factor $4$, even when $E$ has full rational two-torsion. Put $t=\#E(\mathbb{Q})[2]$, which is $2$ or $4$, and let $r$ be the rank. From the two kernel formulas and $2E=\hat{\varphi }\varphi E$,Indeed the kernel of the induced map $E^{\prime }/\varphi E\to \hat{\varphi }{E}^{\prime }/2E$ is $\ker \hat{\varphi }/(\ker \hat{\varphi }\cap \varphi E)$. The intersection is $\varphi (E[2])$, of size $t/2$, so $\delta =4/t$. The Mordell-Weil theorem gives $[E:2E]=2^{r}t$. Therefore

This gives a practical rank procedure. Enumerate the finitely many signed squarefree divisors of $b,b^{\prime }$, and test their quartics over the reals and local fields at $2$ and primes dividing the relevant coefficients or elliptic-curve discriminant. These necessary local conditions give finite Selmer group of an elliptic curve upper bounds. Search the remaining coverings for rational points, or search the curves directly, to realize square classes and obtain lower bounds. One can also exhibit independent points and check independence by the canonical height pairing. When the upper and lower bounds agree, the rank is determined. Local solubility alone does not guarantee a rational point, so a persistent gap can reflect a nontrivial Tate–Shafarevich group rather than a missing congruence test.

Now take $a=0$, $b=-p^2$ for an odd prime $p$. The curve has four rational two-torsion points $O,(0,0),(p,0),(-p,0)$. Their images under $\alpha$ are $1,-1,p,-p$. The prime-support bound allows no other classes, soIts partner is $E_p^{\prime }:Y^2=X^3+4p^{2}X$. Any nonidentity rational point with $X\ne 0$ has $X>0$, because $X^2+4p^2>0$. Also the exceptional point has square class $4p^2=1$. Thus its image is contained in the four positive candidate classes $\{1,2,p,2p\}$. The rank bound for a prime congruent-number elliptic curve follows:

For an exact rank-one example, take $p=5$. The point $(X,Y)=(5,25)$ on $E_5^{\prime }:Y^2=X^3+100X$ realizes the class $5$. The five-adic obstructions for the congruent-number isogeny covers exclude the other two possible nontrivial classes. For $d=2$, the quartic isModulo $5$, if $5\nmid U$ the right side is $2$, a nonsquare. Hence $U=5U_1$ and $N=5N_1$; primitivity gives $5\nmid V$. Dividing by $25$ then gives $N_1^2=50U_1^4+2V^4\equiv 2(\mathrm{mod}5)$, again impossible. For $d=10$, the quartic is $N^2=10(U^4+V^4)$. At least one of $U,V$ is nonzero modulo $5$, and their fourth powers are $0$ or $1$, so their sum is $1$ or $2$ modulo $5$. The right side has five-adic valuation $1$, impossible for a square. Thus ${\alpha }^{\prime }(E_5^{\prime })=\{1,5\}$ exactly, andThe dual isogeny maps $(5,25)$ to $(25/4,-75/8)$ on $E_5$, giving an explicit rational point associated with this rank-one computation.

The structural facts used about a non-Archimedean local field are that it is complete for a normalized discrete valuation $v:K^{\times }\to \mathbb{Z}$, its valuation ring ${\mathcal{O}}_K=\{x:v(x)\ge 0\}$ is a discrete valuation ring, its maximal ideal is ${\mathfrak{m}}_K=\pi {\mathcal{O}}_K$ for a uniformizer $\pi$, and its residue field is a finite field $k={\mathbb{F}}_q$. Polynomial operations are continuous for this valuation topology.

A simple-root version of Hensel lemma is: if $f\in {\mathcal{O}}_K[X]$, $a_1\in {\mathcal{O}}_K$, $f(a_1)\in {\mathfrak{m}}_K$, and $f^{\prime }(a_1)\notin {\mathfrak{m}}_K$, then there is a unique $a\in {\mathcal{O}}_K$ with $f(a)=0$ and $a\equiv a_1(\mathrm{mod}{\mathfrak{m}}_K)$.

Construct $a_n$ with $f(a_n)\in {\pi }^n{\mathcal{O}}_K$. Given $a_n$, put $a_{n+1}=a_n+{\pi }^n{b}_n$. Taylor expansion over the valuation ring givesThe omitted terms contain ${\pi }^{2n}$, and $2n\ge n+1$ for $n\ge 1$. The residue of $f^{\prime }(a_n)$ remains the nonzero residue of $f^{\prime }(a_1)$, so choose $b_n$ withThis makes $f(a_{n+1})\in {\pi }^{n+1}{\mathcal{O}}_K$. The congruences $a_{n+1}\equiv a_n(\mathrm{mod}{\pi }^n)$ make $(a_n)$ a Cauchy sequence; completeness gives its limit $a\in {\mathcal{O}}_K$. Continuity and $v(f(a_n))\ge n$ give $f(a)=0$, and the first congruence gives its specified residue class.

For uniqueness, if $a,b$ are two roots in that class, polynomial division of their difference givesThe second factor is a unit, since $f^{\prime }(b)$ is a unit and $a-b\in {\mathfrak{m}}_K$. Thus $a=b$. This proves both existence and uniqueness in the stated Hensel lemma.

Put $K=\mathbb{C}((t))$. This is a complete discretely valued field, with valuation ring $\mathbb{C}[[t]]$ and residue field $\mathbb{C}$; its infinite residue field means it is not a locally compact local field. The finite-extension facts needed here are uniqueness of the extended valuation, completeness of the extension, and $[L:K]=ef$ for complete discretely valued fields with perfect residue field. They apply in this characteristic-zero setting.

Let $[L:K]=n$. The finite residue extension of $\mathbb{C}$ must be $\mathbb{C}$ itself, so $f=1$ and $e=n$. Normalize $v_L$ integrally and choose a uniformizer $\pi$. Since $v_L(t)=n$, write $t=u{\pi }^n$ with $u\in {\mathcal{O}}_L^{\times }$.

The residue of $u$ has an $n$th root $b_0\in {\mathbb{C}}^{\times }$. Apply Hensel lemma to $X^n-u$: its derivative at $b_0$ is the unit $n{b}_0^{n-1}$ because the residue characteristic is zero. Hence there is $v\in {\mathcal{O}}_L^{\times }$ with $v^n=u$. Setting $s=v\pi$ givesThe elements $1,s,\dots ,s^{n-1}$ are linearly independent over $K$. In a nontrivial relation ${\sum }_{i=0}^{n-1}{a}_i{s}^i=0$, the nonzero summands have valuations $n{v}_K(a_i)+i$, distinct modulo $n$, so exactly one has least valuation and cannot cancel. Since $L$ has dimension $n$, this proves $L=K(s)$.

The constants already embed $\mathbb{C}$ into ${\mathcal{O}}_L$ with residue field $\mathbb{C}$. Subtracting the residue coefficient and dividing by $s$ repeatedly, exactly as in Question 1(c), expresses every integral element uniquely as ${\sum }_{j\ge 0}{c}_j{s}^j$ with $c_j\in \mathbb{C}$. Completeness and the first-nonzero-coefficient valuation show that allowing finite negative powers givesThis proof of finite extensions of the complex Laurent series field did not need a Galois hypothesis. All $n$ distinct $n$th roots of unity lie in the constant field, and the substitutions $s\mapsto \zeta s$ preserve $s^n=t$. They give all $n$ automorphisms, so every such finite extension is cyclic Galois:

Choose compatible roots $s_n=t^{1/n}$ in an algebraic closure, first choosing them successively along the factorial-degree tower and then taking powers to define the other roots, with $s_n=s_m^{m/n}$ whenever $n\mid m$. The fields $K_n=K(s_n)$ are nested under divisibility. Every finite extension inside the closure equals one of them: its constructed root $s$ of $X^n-t$ differs from $s_n$ by a constant root of unity. Therefore their union is the entire algebraic closure. Explicitly, every algebraic element generates a finite extension, so lies in the union; any polynomial over the union has its coefficients in one finite extension, and its roots are still algebraic over $K$, hence also lie there.

For $n\mid m$, restriction of automorphisms sends ${\zeta }_m\in {\mu }_m$ to ${\zeta }_m^{m/n}\in {\mu }_n$. Using the compatible primitive roots $\exp (2\pi i/n)$, these are the reduction maps on exponent groups. Thus the absolute Galois group of the complex Laurent series field isThe inverse limit carries its profinite topology; choosing compatible roots gives the displayed identification with the profinite completion of the integers.

All the local fields here carry normalized discrete valuations. Write ${\mathcal{O}}_F$ for the valuation ring, ${\mathfrak{m}}_F$ for its maximal ideal, and $k_F$ for its residue field. We use the basic degree identity $[B:A]=e(B/A)f(B/A)$ for finite extensions of non-Archimedean local fields, with $e$ the ramification index and $f$ the residue degree.

First construct the maximal unramified subextension of a local field extension inside an arbitrary finite extension $L/K$. Choose a primitive element of a field extension $\bar{z}$ for the finite residue field extension $k_L/k_K$. Let $\bar{h}$ be its monic irreducible minimal polynomial of an algebraic element, of degree $f$, and lift its coefficients to obtain a monic $h\in {\mathcal{O}}_K[T]$ of the same degree. Finite finite fields are perfect fields, so ${\bar{h}}^{\prime }(\bar{z})\ne 0$. By Hensel lemma applied in $L$, there is $z\in {\mathcal{O}}_L$ reducing to $\bar{z}$ with $h(z)=0$. For $U=K(z)$ we haveEvery inequality is therefore an equality. In particular $U/K$ is unramified, $k_U=k_L$, and $h$ is the minimal polynomial of an algebraic element $z$.

If $L/K$ is unramified, then $[L:K]=f$, so this construction gives $U=L$. Taking $x=z$ supplies the required generator, and $h^{\prime }(x)$ is a unit. Conversely suppose $L=K(x)$ with $x$ integral and its monic minimal polynomial of an algebraic element $g$ satisfying $g^{\prime }(x)\notin {\mathfrak{m}}_L$. Its coefficients lie in ${\mathcal{O}}_K$, since all conjugates of an integral element are integral and $K$ is complete. In the constructed unramified extension $U$ the element $\bar{x}\in k_U=k_L$ is a simple root of $\bar{g}$. Hensel lemma in $U$ produces a root $x^{\prime }\in U$ with residue $\bar{x}$. In $L$ both $x$ and $x^{\prime }$ lift this same simple residue root, so Hensel uniqueness gives $x=x^{\prime }$. Thus $L\subseteq U\subseteq L$, proving

For lifting residue-field embeddings of unramified extensions, use the generator $z$ and polynomial $h$ just constructed when $L/K$ is unramified. Given a $k_K$-field homomorphism $\iota :k_L\to k_M$, its value $\iota (\bar{z})$ is a simple root of $\bar{h}$. Hensel lemma in $M$ produces the unique root $z^{\prime }\in {\mathcal{O}}_M$ with that residue. The assignment $z\mapsto z^{\prime }$ defines a $K$-field homomorphism $K[T]/(h)\to M$; it induces $\iota$ because $\bar{z}$ generates $k_L$. Conversely any such field homomorphism preserves the unique extended valuation, hence induces a residue field homomorphism, and its image of $z$ is uniquely forced by Hensel uniqueness. This proves the asserted bijection. The same proof only requires that $M$ be a non-Archimedean local field containing $K$ with its compatible valuation; its finiteness over $K$ is not essential.

Now set $M=L$. The finite residue field extension has exactly $f=[L:K]$ automorphisms, cyclically generated by $a\mapsto a^{|k_K|}$. Their unique lifts give $[L:K]$ automorphisms of $L$. Consequently $L/K$ is a Galois extension, and reduction identifies its Galois group with that cyclic residue Galois group. Its generator is the arithmetic Frobenius:

For the compositum, choose a uniformizer ${\pi }_M$ of the totally ramified extension $M/K$. The Eisenstein generator of a totally ramified extension characterization, proved in the next solution, says that $M=K({\pi }_M)$ and its minimal polynomial of an algebraic element is Eisenstein over ${\mathcal{O}}_K$. A uniformizer of $K$ remains a uniformizer in the unramified extension $L$. All its Eisenstein coefficient conditions therefore remain true over ${\mathcal{O}}_L$. Thus the same Eisenstein polynomial is irreducible over $L$ and generates $LM=L({\pi }_M)$ as a totally ramified extension. This proves $LM/L$ is totally ramified, of degree $[M:K]$.

Here is an explicit example showing that the converse decomposition of an arbitrary extension can fail. Take $K={\mathbb{Q}}_3$ and $L=K(i)$ with $i^2=-1$. Since $T^2+1$ is irreducible and separable modulo three, the degree and residue-degree argument above shows that $L/K$ is unramified quadratic. PutThe element $1+i$ is a unit in ${\mathcal{O}}_L$, so $T^2-3(1+i)$ is Eisenstein over $L$. Hence $[N:L]=2$, $e(N/K)=2$, and $f(N/K)=2$. Moreover $i={\alpha }^2/3-1$ gives $N=K(\alpha )$, withWe verify its Galois group, since that is what rules out a totally ramified complement. By Hensel lemma there is $r\in K$ with $r^2=-2$: the residue root $1$ of $T^2+2$ is simple modulo three. Set $\beta =(1+i)/r$. ThenThe assignments $\sigma (i)=-i$ and $\sigma (\alpha )=\beta \alpha$ respect the defining relation, because $(\beta \alpha {)}^2=-3i(1+i)=3(1-i)$. They therefore define a $K$-automorphism of $N$. Its square satisfies ${\sigma }^2(i)=i$ and ${\sigma }^2(\alpha )=\bar{\beta }\beta \alpha =-\alpha$, so $\sigma$ has order four. There are thus four automorphisms of the degree-four extension, andThis is the cyclic quartic local extension without a totally ramified complement. Its unique quadratic subfield is $L$, fixed by ${\sigma }^2$, and is unramified. If $N=L^{\prime }{M}^{\prime }$ with $L^{\prime }/K$ unramified and $M^{\prime }/K$ totally ramified, then $[L^{\prime }:K]\le f(N/K)=2$ and $[M^{\prime }:K]\le e(N/K)=2$. To obtain the full degree four both would have degree two, forcing $M^{\prime }$ to be a totally ramified quadratic subfield. There is no such subfield. Thus $N/{\mathbb{Q}}_3$ is the required counterexample.