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The absolute residue degree is the degree of the residue field over its prime field. For a tower above $p$, degrees multiply: $[k_{\mathfrak{q}}:{\mathbb{F}}_p]=[k_{\mathfrak{q}}:k_{\mathfrak{p}}][k_{\mathfrak{p}}:{\mathbb{F}}_p]$. In a Galois extension over $\mathbb{Q}$, the decomposition group has a cyclic quotient group of this order by its inertia group. This can exclude residue degrees even when the whole extension has that degree: $S_3$ has no cyclic quotient group of order six.

Let $E$ be a finite normal closure of the number field $K$. The permutation representation of its Galois group on the embeddings of $K$ into $E$ has one trivial constituent, because the action is transitive. At a prime, the inertia group invariant basis consists of sums over inertia orbits of embeddings; Frobenius automorphism permutes them in cycles of lengths equal to the residue degrees of the primes of $K$. A cycle of length $f$ contributes $\det (1-T\rho (\mathrm{Frob}))=1-T^f$. Hence the local factors equal those of the Dedekind zeta function, including at ramified primes. If nontrivial irreducible Artin L-functions are holomorphic and nonzero at one, this factorization and the simple pole of the Riemann zeta function show that ${\zeta }_K$ has a simple pole there.

To compute the Ideal class group of Q of square root thirty-five, put $\theta =\sqrt{35}$. The ring of integers of a quadratic field formula gives $R=\mathbb{Z}[\theta ]$, and the field discriminant is $4\cdot 35=140$. Both field embeddings are real, so the Minkowski bound for ideal classes isEvery ideal class therefore has an integral ideal representative of ideal norm at most five. By prime ideal factorization and multiplicativity of the ideal norm, the ideal class group is generated by prime ideals of ideal norm at most five.

Apply the Dedekind factorization theorem to $X^2-35$. At two and five it has a repeated linear factor, givingAt three, $X^2-35\equiv X^2+1$ has no root, so $(3)$ itself is a prime ideal of ideal norm nine, outside the bound. These are all the relevant prime ideals: a prime ideal of ideal norm four would have to lie above two with residue degree two, but the factorization above two has only the single factor of degree one. Consequently $[P_2]$ and $[P_5]$ generate, and each has order of a group element dividing two.

The algebraic integer $5+\theta$ has field norm $25-35=-10$. It belongs to both $P_2$ and $P_5$. Hence its principal ideal is contained in $P_2\cap P_5=P_2{P}_5$. Both have ideal norm ten, soIt follows that $[P_5]=[P_2{]}^{-1}=[P_2]$, and the ideal class group is generated by one element of order of a group element at most two.

Finally, $P_2$ is not a principal ideal. If $P_2=(a+b\theta )$ with $a,b\in \mathbb{Z}$, the equality of ideal norm and absolute field norm would implyModulo five this requires $a^2\equiv 2$ or $3$, whereas squares modulo five are $0,1,4$. Thus $[P_2]\ne 1$, provingThis computes the ordinary ideal class group, with no positivity restriction on generators.

Let ${\zeta }_n$ be a primitive nth root of unity. The cyclotomic polynomial isInduction on $n$ shows that ${\Phi }_n$ is monic with integer coefficients. Here is the essential irreducibility of cyclotomic polynomials argument. Let $f$ be the minimal polynomial of ${\zeta }_n$ and let $\ell \nmid n$ be a prime. If ${\zeta }_n^{\ell }$ had a different monic irreducible polynomial $g$, then $f\mid g(X^{\ell })$. Reducing modulo $\ell$ gives $\bar{f}\mid {\bar{g}}^{\ell }$, so $\bar{f}$ and $\bar{g}$ have a common factor. As $fg\mid {\Phi }_n$, this would give a repeated factor of $X^n-1$ modulo $\ell$, impossible because its derivative is $n{X}^{n-1}$. Thus ${\zeta }_n^{\ell }$ is conjugate to ${\zeta }_n$. Every positive integer prime to $n$ is a product of such primes, so all primitive nth roots of unity are conjugate. ConsequentlyIn particular the cyclotomic field is an abelian extension. Complex conjugation corresponds to $-1$; for $n>2$ its fixed field is $\mathbb{Q}({\zeta }_n+{\zeta }_n^{-1})$, the maximal real subfield, of degree $\phi (n)/2$.

The ring of integers is $\mathbb{Z}[{\zeta }_n]$. One way to see that there is no missing index is to work at each prime $p$. Write $n=p^{a}m$ with $p\nmid m$. The prime-to-$p$ roots of unity have separable reduction; their local factors generate unramified extensions with full integer rings. Over each such factor, adjoining ${\zeta }_{p^a}$ is generated by the Eisenstein polynomial ${\Phi }_{p^a}(1+T)$, with uniformizer ${\zeta }_{p^a}-1$, and hence gives the full integer ring of that totally ramified factor. The coprimality of $p^a$ and $m$ means that ${\zeta }_n$ generates both sets of roots of unity. Thus ${\mathbb{Z}}_p[{\zeta }_n]$ is maximal at every $p$, proving the integral assertion.

The same local description computes ramification and splitting. For $p\nmid n$, the ramification index is one, the residue degree is the multiplicative order of $p$ modulo $n$, and there are $\phi (n)/f$ primes above $p$. If $p\mid n$, put $m=n/p^a$; thenwhere ${\mathrm{ord}}_1(p)=1$. Indeed, the prime-to-$p$ part is unramified, and the $p$-power part is totally ramified by the preceding Eisenstein polynomial. Notice the exception $p=2,a=1$: its ramification index is one, consistently with $\mathbb{Q}({\zeta }_{2m})=\mathbb{Q}({\zeta }_m)$ for odd $m$. Thus finite ramification occurs only at the indicated conductor primes, with that exception.

We now prove cyclotomic Artin reciprocity over the rational numbers, using arithmetic Frobenius automorphisms. If $\ell \nmid n$ and $\mathfrak{L}$ lies over $\ell$, then Frobenius sends the reduction of ${\zeta }_n$ to its $\ell$th power. Distinct nth roots of unity remain distinct modulo $\mathfrak{L}$, since $X^n-1$ has separable reduction. Hence the Artin symbol is exactlyExtend multiplicatively to the group $I(n)$ of rational fractional ideals prime to $n$. Such an ideal has a unique positive rational generator $q$ whose numerator and denominator are prime to $n$. The formula isEvery unit residue modulo $n$ has a positive integer representative prime to $n$, so this Artin map is surjective. Its kernel is precisely the subgroup $P_{n\infty }$ generated by positive $q\equiv 1(\mathrm{mod}n)$. ThereforeThis proves the Artin reciprocity law for this extension, including its principal-ray kernel. The infinite component of the modulus records positivity and therefore the possible action of complex conjugation.

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.

Use the following simple-root Hensel lemma. Let $R$ be a complete discrete valuation ring, with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $f(a_0)\in \mathfrak{m}$ and $f^{\prime }(a_0)\in R^{\ast }$, there is a unique $a\in R$ with $a\equiv a_0(\mathrm{mod}\mathfrak{m})$ and $f(a)=0$.

For the proof, define Newton's method iterates $a_{r+1}=a_r-f(a_r)/f^{\prime }(a_r)$. Each denominator stays a unit, because all iterates are congruent to $a_0$. Taylor's polynomial identity givesIf an iterate is already a root, the process stops. Otherwise these valuations tend to infinity, so the iterates form a Cauchy sequence in the complete ring. Its limit $a$ has the required residue and satisfies $f(a)=0$. If $a,b$ were two such roots, thenThe parenthesis is a unit, forcing $a=b$.

Now let $K/{\mathbb{Q}}_p$ be finite, with residue field $k={\mathbb{F}}_q$. For every $m\ge 1$ choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $m$ and lift it to a monic $f\in {\mathcal{O}}_K[X]$. The lift is irreducible over $K$, since a factorization would reduce to a factorization of $\bar{f}$, by the monic form of Gauss's lemma over the valuation ring. If $L=K(\alpha )$ for a root, its residue field contains $k(\bar{\alpha })$ of degree $m$. The inequality $e(L/K)f(L/K)\le [L:K]=m$ forces residue degree $m$ and ramification index one. Thus it is an unramified extension of degree $m$.

Conversely, in any unramified extension of degree $m$, choose a generator of its residue field and lift its separable minimal polynomial. Hensel's lemma supplies a root with that residue. Its degree is at least $m$ and hence it generates the extension. The same unique lifting of all residue conjugates shows that the extension is Galois andOne can see uniqueness inside a fixed algebraic closure explicitly: the roots of $X^{q^m-1}-1$ lift all nonzero residue elements, and a lifted primitive residue element generates the extension. Hencewith arithmetic Frobenius acting by $\zeta \mapsto {\zeta }^q$. The case $q^m-1=1$ simply gives $K_1=K$. These are all finite unramified extensions; $K_m\subseteq K_r$ exactly when $m\mid r$. Their union is the maximal unramified extension, whose Galois group is $\hat{\mathbb{Z}}$.

Finally prove norm-surjectivity on units using the Herbrand quotient. For a cyclic group $G$ of order $m$, writeThe quotient is multiplicative on short exact sequences, equals one for finite modules, equals $m$ on the trivial module $\mathbb{Z}$, and equals one on induced regular modules. For a finite module, the equality follows directly from $|M^G||(\sigma -1)M|=|M|=|\ker N||NM|$.

Let $U_L={\mathcal{O}}_L^{\ast }$. For sufficiently large $r$, the equivariant p-adic logarithm identifies $U_L^r=1+{\mathfrak{p}}_L^r$ with ${\mathfrak{p}}_L^r$ additively. In the unramified case this is an induced regular lattice. Indeed choose a normal basis of the finite residue extension and lift its generating element. Its $m$ Galois conjugates reduce to a basis and hence form an ${\mathcal{O}}_K$-basis of ${\mathcal{O}}_L$; their coordinate determinant is a unit. Since a uniformizer of $K$ is also one of $L$, multiplication by its $r$th power gives the same regular-module structure on ${\mathfrak{p}}_L^r$. For a regular module, invariants are norm images and a vector of coefficient sum zero is a cyclic difference, so both indicated Tate cohomology groups vanish. Consequently $h_G(U_L^r)=1$. The quotient $U_L/U_L^r$ is finite, so $h_G(U_L)=1$.

Apply multiplicativity toIt gives $h_G(L^{\ast })=m$. Hilbert 90 makes ${\hat{H}}^{-1}(G,L^{\ast })$ trivial, so $[K^{\ast }:N_{L/K}{L}^{\ast }]=m$. On the other handso the quotient maps onto $\mathbb{Z}/m\mathbb{Z}$. As its order is already $m$, its unit kernel is trivial. A norm which is a unit comes from a unit, by the same valuation formula. ThereforeThe standard normal-basis theorem for finite fields and the convergent logarithm/exponential on sufficiently deep principal units are the ingredients used in evaluating the Herbrand quotient; the norm conclusion itself has been deduced here.

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.

We give the detailed norm-index argument for a cyclic extension $L/K$ of p-adic fields; thus $K$ is a finite extension of ${\mathbb{Q}}_p$. Write $G=⟨\sigma ⟩$, $|G|=n$. For an additive $G$-module $M$, put $D=\sigma -1$ and $N=1+\sigma +\cdots +{\sigma }^{n-1}$. The two periodic Tate cohomology of a cyclic group groups and the Herbrand quotient arewhen these groups are finite. For multiplicative modules, $N$ is the product of conjugates and $D(z)=\sigma (z)/z$.

The periodic complex alternating $D$ and $N$ gives a six-term exact sequence for every short exact sequence of modules. Taking the alternating product of the finite group orders in that exact sequence provesFor a finite module, the counting fibers gives $|NM|=|M|/|\ker N|$ and $|DM|=|M|/|\ker D|$, so the two Tate groups have equal order. This proves the Herbrand quotient of a finite module is one. It follows that replacing a module by a finite-index submodule does not change its quotient, whenever the cohomology groups are finite.

Two basic calculations drive the norm computation. For the trivial module $\mathbb{Z}$, $D=0$ and $N$ is multiplication by $n$, so $h_G(\mathbb{Z})=n$. For a regular lattice $R[G]$, with $R={\mathbb{Z}}_p$, invariants are constant coefficient vectors and every such vector is a norm: put its constant coefficient in just one coordinate before applying $N$. The group kernel of $N$ consists of the coefficient vectors of sum zero. Successively taking partial sums solves the cyclic difference equation $Dx=y$ for every such vector $y$. Thus both Tate groups vanish and $h_G(R[G])=1$. This is cyclic cohomology of a regular lattice.

Let $U_L={\mathcal{O}}_L^{\times }$ and $U_L^r=1+{\mathfrak{P}}_L^r$. For sufficiently large $r$, the p-adic logarithm and p-adic exponential are inverse $G$-equivariant isomorphismsTo make the range explicit, $r>v_L(p)/(p-1)$ suffices: the higher terms in the logarithm and exponential have greater valuation than their leading terms, their series converge there, and the identities $\log (\exp x)=x$ and $\exp (\log (1+x))=1+x$ remain valid. Thus a deep multiplicative unit group becomes an additive ${\mathbb{Z}}_p$-lattice in $L$.

By the normal basis theorem, $L\cong K[G]$ as a $K[G]$-module. As a ${\mathbb{Q}}_p[G]$-module it is therefore ${\mathbb{Q}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$. Transport the regular lattice $M_0={\mathbb{Z}}_p[G{]}^{[K:{\mathbb{Q}}_p]}$ into $L$. Both $M_0$ and ${\mathfrak{P}}_L^r$ are full $G$-stable ${\mathbb{Z}}_p$-lattices, so their intersection has finite index in each. The exact sequences with finite quotients show that the Tate groups of ${\mathfrak{P}}_L^r$ are finite and thatSince $U_L/U_L^r$ is finite, the exact-sequence formula for the Herbrand quotient gives $h_G(U_L)=1$. Finally, the valuation sequencehas trivial $G$-action on $\mathbb{Z}$. Hence the Herbrand quotient of the local multiplicative group is

For completeness, the denominator vanishes by Hilbert theorem 90. If $N_{L/K}a=1$, put $c_0=1$ and $c_j=a\sigma (a)\cdots {\sigma }^{j-1}(a)$. Independence of distinct field automorphisms lets us choose $t\in L$ with $b={\sum }_{j=0}^{n-1}{c}_j{\sigma }^j(t)\ne 0$. Since $c_n=1$, shifting the sum gives $\sigma (b)=b/a$. Thus $a=b/\sigma (b)$, a multiplicative coboundary. Therefore ${\hat{H}}^{-1}(G,L^{\times })=1$. The invariant subgroup of $L^{\times }$ is $K^{\times }$, soThis proves the cyclic local norm index without assuming local reciprocity. If the extension has ramification index $e$ and residue degree $f$, the valuation formula $v_K(Nz)=f{v}_L(z)$ shows that the valuation contribution to the index is $f$. A norm is a unit only when $z$ is a unit, so the remaining contribution isIn particular, unit norms are surjective in an unramified cyclic extension. The norm subgroup is open: on deep units its logarithm is ${\mathrm{Tr}}_{L/K}({\mathfrak{P}}_L^r)$, a nonzero full lattice in $K$, and therefore contains a sufficiently deep principal-unit group. This connects the index calculation to the open finite-index norm subgroups classified by local class field theory.

To finish, the Hasse norm theorem for a cyclic extension of number fields states that a nonzero element is a global norm if and only if it is a norm at every finite and infinite completion. Here is its Brauer-theoretic proof outline. Associate to $a\in K^{\times }$ the cyclic algebra $A=(L/K,\sigma ,a)$, with relations $z\ell =\sigma (\ell )z$ and $z^n=a$. The splitting criterion for a cyclic algebra says that it splits exactly when $a=Nc$: if $a=Nc$, multiplication by $L$ and the operator $c\sigma$ realize it as ${\mathrm{End}}_K(L)$; conversely, in a split matrix representation, the underlying $n$-dimensional space is one-dimensional over $L$, so $z$ must act as $c\sigma$ and $z^n=Nc$.

Apply this criterion at every completion, using the restricted cyclic character and any one completion of $L$ above that place. If $a$ is a local norm everywhere, every localized Brauer class is zero. Injectivity in the Albert-Brauer-Hasse-Noether theorem forces $[A]=0$ in $\mathrm{Br}(K)$; the global splitting criterion gives $a=Nc$. Conversely, a global norm is a local norm after completion, since the norm on $L{\otimes }_K{K}_v$ is the product of the component norms, whose images coincide in the Galois case. The cyclic hypothesis is essential; the stated local-global principle does not hold for arbitrary extensions.

The supplied integral basis differs unimodularly from $1,\zeta ,{\zeta }^2,{\zeta }^3$, since $1=-\zeta -{\zeta }^2-{\zeta }^3-{\zeta }^4$. With $f(t)={\Phi }_5(t)=(t^5-1)/(t-1)$, the field discriminant is $\mathrm{Norm}(f^{\prime }(\zeta ))$, the degree-four sign being positive. At a nonidentity fifth root,soThe Minkowski bound for ideal classes gives an integral ideal in every ideal class with norm at most $[3/(2{\pi }^2)]\sqrt{125}<2$. Its positive integer norm must be one, so it is the whole integer ring. Every class is trivial, proving all ideals are principal.

For $1\le n\le 4$, $u_n=(1-{\zeta }^n)/(1-\zeta )=1+\zeta +\cdots +{\zeta }^{n-1}$ is an algebraic integer. The numerator has the same norm five as the denominator, because its conjugates permute the nonidentity fifth roots. Thus $\mathrm{Norm}(u_n)=1$, proving it is a unit. Multiplying the four numerator factors gives five, henceThe ideal $P=(1-\zeta )$ has norm five; explicitly reduction modulo it sets $\zeta =1$ and gives the quotient ${\mathbb{F}}_5$. Thus it is prime and $(5)=P^4$: five is totally ramified, with ramification index four and residue degree one. A rational prime can ramify only if it divides the field field discriminant, so no rational prime other than five ramifies, and there are no other ramified prime ideals.

For a nonzero Laurent series let ${\mathrm{ord}}_t$ be its least exponent. Choose $|x|=p^{-{\mathrm{ord}}_t(x)}$ and $|0|=0$. This is a multiplicative Non-Archimedean absolute value, and its valuation ring is ${\mathbb{F}}_p[[t]]$.

TakeIf ${\gamma }_1,\dots ,{\gamma }_f$ is an ${\mathbb{F}}_p$-basis of ${\mathbb{F}}_{p^f}$, the elements ${\gamma }_i{u}^j$, $1\le i\le f$, $0\le j<e$, form a $K$-basis of $L$. To see this, group the exponents of each Laurent series modulo $e$ and then expand its coefficients in the constant-field basis. Uniqueness of those two expansions proves linear independence. Thus $[L:K]=ef$.

Its valuation ring is ${\mathbb{F}}_{p^f}[[u]]$, with maximal ideal $(u)$ and residue field ${\mathbb{F}}_{p^f}$. Moreover ${\mathrm{ord}}_u(t)=e$, so the ramification index is $e$ and the residue degree is $f$. The unique extending absolute value is explicitlyThis restricts to the chosen absolute value because ${\mathrm{ord}}_u(x)=e{\mathrm{ord}}_t(x)$ on $K$. The prescribed ramification and residue degree for a Laurent series field construction works even when $p\mid e$; separability is not required in the question.

For an odd integer $x=2a+1$, $x^2=1+4a(a+1)\equiv 1(\mathrm{mod}8)$. The same conclusion holds for every 2-adic unit by reduction modulo eight. Thus an odd square must be congruent to one modulo eight.

Conversely let $u\in {\mathbb{Z}}_2^{\times }$ with $u\equiv 1(\mathrm{mod}8)$. For digit lifting for square roots of 2-adic units, start with $x_3=1$. If $x_n$ is odd and $x_n^2\equiv u(\mathrm{mod}{2}^n)$, considerFor $n\ge 3$,since the remaining term is divisible by $2^{2n-2}$ and $2n-2\ge n+1$. The odd multiplier $x_n$ lets the choice of ${\epsilon }_n$ correct exactly the next binary digit. Thus $x_{n+1}^2\equiv u(\mathrm{mod}{2}^{n+1})$. The corrections have valuations tending to infinity, so completeness gives $x=\lim x_n\in {\mathbb{Z}}_2^{\times }$ with $x^2=u$.

The unit $3$ is congruent to one modulo two but is not a square, and $5$ is congruent to one modulo four but is not a square, by the necessary modulo-eight condition. Therefore

Every nonzero P-adic number in ${\mathbb{Q}}_2$ is uniquely $2^{m}u$ with $m\in \mathbb{Z}$ and $u$ a 2-adic unit. A square has even valuation, and the unit square criterion shows that the unit class is determined by its odd residue modulo eight. The four residues form $C_2\times C_2$, generated by $-1$ and $5$; the valuation parity supplies the third factor, generated by $2$. Hence the square-class group of the 2-adic numbers isIts eight classes are represented by $1,-1,5,-5,2,-2,10,-10$. They are distinct: the first four have even valuation and distinct unit residues modulo eight, while multiplication by $2$ gives odd valuation and the same four distinct unit residues.

In characteristic different from two, completing the square in a quadratic minimal polynomial makes every quadratic extension $K(\sqrt{d})$ for a nonsquare $d\in K$. If $K(\sqrt{d})=K(\sqrt{e})$, write $\sqrt{e}=a+b\sqrt{d}$ and square. The coefficient $2ab$ must vanish. Since $e$ is nonsquare, $b\ne 0$, so $a=0$ and $e/d=b^2$. Conversely a square ratio plainly gives the same field. Thus the seven quadratic extensions of the 2-adic numbers are exactly

The unique unramified extension in this list is ${\mathbb{Q}}_2(\sqrt{5})$. To verify this, $\alpha =(1+\sqrt{5})/2$ is integral and satisfies $X^2-X-1=0$. Its reduction $X^2+X+1$ is irreducible over ${\mathbb{F}}_2$, so its residue field contains ${\mathbb{F}}_4$. The standard degree formula for finite extensions of complete discretely valued fields with perfect residue field is $[L:K]=e(L/K)f(L/K)$. Here the field degree is two, so the residue degree is two and the ramification index is one.

Conversely an unramified quadratic extension has residue field ${\mathbb{F}}_4$. A root there of $X^2+X+1$ lifts by Hensel lemma to a root of $X^2-X-1$, whose derivative is a unit. For this lift $\beta$, $(2\beta -1{)}^2=5$, so the extension contains ${\mathbb{Q}}_2(\sqrt{5})$ and must equal it. Every other field in the list is therefore ramified.

For the requested ramification groups, choose $L={\mathbb{Q}}_2(\sqrt{2})$ and $\pi =\sqrt{2}$. We use uniqueness of the extension of the valuation to a finite extension, together with the degree formula just stated. Since $v_2(\pi )=1/2$, the extension has ramification index two and residue degree one. Normalize $v_L(\pi )=1$, giving $v_L(2)=2$. Every element is $a+b\pi$ with $a,b\in {\mathbb{Q}}_2$, and the two summands have valuations of opposite parity:Thus $a+b\pi$ is integral exactly when $a,b\in {\mathbb{Z}}_2$, proving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$ directly.

The nonidentity automorphism $\sigma$ negates $\pi$. For any integral $a+b\pi$,of valuation at least three, with equality when $b$ is a unit. By the definition $G_i=\{\sigma :v_L(\sigma x-x)\ge i+1\text{ for every }x\in {\mathcal{O}}_L\}$, the ramification in the quadratic extension generated by the square root of two is thereforeThe lower ramification break is two; the valuation used here is the integral normalization on $L$.

Let $L=\mathbb{Q}(\alpha )$, where $\alpha$ is an algebraic integer with monic minimal polynomial $f\in \mathbb{Z}[X]$, and put $I=[{\mathcal{O}}_L:\mathbb{Z}[\alpha ]]$. The Kummer-Dedekind theorem applies to a prime number $p$ with $p\nmid I$. Write its reduction as a product of polynomial factors over ${\mathbb{F}}_p$:where the $g_j$ are distinct monic irreducible polynomials, choose monic integral lifts $G_j$. Then the distinct prime ideals above $p$ are ${\mathfrak{p}}_j=(p,G_j(\alpha ))$, andThus the multiplicity of a factor is its ramification index of a prime ideal, and its degree is the residue degree. The hypothesis on the order in a number field is essential: one cannot infer prime ideal factorization from an arbitrary defining polynomial at a prime dividing its index. There is also a relative version over a base number field $K$: if ${\mathcal{O}}_L$ and ${\mathcal{O}}_K[\alpha ]$ agree after localization at $\mathfrak{p}$, factor the minimal polynomial over ${\mathcal{O}}_K/\mathfrak{p}$ and use $(\mathfrak{p},G_j(\alpha ))$ with the same conclusions.

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.

Let $d=[L:K]$, with normalized discrete valuations $v_K,v_L$. Suppose first that $L/K$ is totally ramified, so $e(L/K)=d$. Choose a uniformizer $x$ of $L$. The elements $1,x,\dots ,x^{d-1}$ are linearly independent over $K$: the nonzero terms $c_j{x}^j$ in any proposed relation have valuations $d{v}_K(c_j)+j$, distinct modulo $d$, and a sum with a unique least valuation cannot vanish by the ultrametric inequality. Hence $[K(x):K]\ge d$, proving $L=K(x)$.

Extend $v_K$ to an algebraic closure with values in $\mathbb{Q}$. Every conjugate of $x$ has valuation $1/d$, by uniqueness of the extended valuation on finite extensions of the complete field $K$. In its monic minimal polynomial of an algebraic element $g(T)=T^d+a_{d-1}{T}^{d-1}+\cdots +a_0$, the coefficient $a_{d-j}$ is a sum of products of $j$ conjugates, counted with multiplicity. Thus $v_K(a_{d-j})\ge j/d>0$ for $j\ge 1$, so every nonleading coefficient is in ${\mathfrak{m}}_K$. The constant coefficient is the product of all $d$ conjugates, so has valuation exactly one. This proves that $g$ is an Eisenstein polynomial. This argument also works for inseparable finite extensions, since the conjugates can be counted with their multiplicities.

Conversely let $g$ be an Eisenstein polynomial of degree $d$, with root $x$ and $L=K(x)$. The Eisenstein criterion proves irreducibility. For completeness, monic factors would have integral coefficients; their reductions would be positive-degree powers of $T$, so both constant coefficients would be divisible by the uniformizer of $K$, contradicting $v_K(g(0))=1$. To determine the root valuation, put $t=v_K(x)$ in the extended valuation. If $t<0$, the leading term is the unique term of least valuation; if $t=0$, the leading term again has uniquely least valuation zero. Both are impossible. If $t>0$, every nonconstant lower term has valuation at least $1+jt>1$. Cancellation in $g(x)=0$ therefore requires the leading term and constant term to have equal valuation:Since $t=v_L(x)/e(L/K)$ and $v_L(x)$ is a positive integer, $e(L/K)=d{v}_L(x)\ge d$. The opposite inequality follows from $[L:K]=e(L/K)f(L/K)=d$. Hence $e=d$, $f=1$, and $v_L(x)=1$. We have proved both the equivalence and the uniformizer assertion: