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Let $P(x)$ denote the polynomial under consideration. The change $x\mapsto x+1$ is an automorphism of $\mathbb{Q}[x]$ and preserves degrees, so it preserves reducibility. Its transformed polynomial isIt is monic; every other coefficient is divisible by $p$, because $p$ is prime and $1\le j<p$; and its constant coefficient is $p$, not divisible by $p^2$. The Eisenstein criterion proves that $P(x+1)$ is irreducible, and therefore $P$ is irreducible. The argument also includes $p=2$, when the polynomial is linear. This is the prime-index case of an irreducible polynomial arising as a cyclotomic polynomial.

Use the Eisenstein criterion with prime $5$. The leading coefficient is one; every lower coefficient, including the zero coefficients, is divisible by $5$; and the constant coefficient $5$ is not divisible by $25$. The polynomial is monic and hence primitive. Therefore it is irreducible over $\mathbb{Q}$.

A polynomial in $\mathbb{Z}[x]$ is a primitive polynomial when the greatest common divisor of its coefficients is one. The Gauss lemma for polynomials states that a product of primitive integer polynomials is primitive, and that a primitive positive-degree integer polynomial is reducible over $\mathbb{Q}$ exactly when it factors into positive-degree integer polynomials. More generally, polynomial content is multiplicative up to sign.

The Eisenstein criterion states that a primitive polynomial $a_n{x}^n+\cdots +a_0\in \mathbb{Z}[x]$ is irreducible over $\mathbb{Q}$ if there is a prime $p$ such that $p\nmid a_n$, $p\mid a_j$ for every $j<n$, and $p^2\nmid a_0$. These criteria permit integer coefficient tests for rational irreducibility; the following parts apply them with their exact divisibility hypotheses.

The polynomial $p=x^5-4x+2$ satisfies the Eisenstein criterion at the prime two: every nonleading coefficient is even, the leading coefficient is not, and the constant term is not divisible by four. Hence it is irreducible over $\mathbb{Q}$. Its ideal is maximal in $\mathbb{Q}[x]$; explicitly, for any polynomial not divisible by $p$, the Euclidean algorithm gives $ag+bp=1$, furnishing the inverse of its residue class. ThusFor the integer quotient, division by the monic polynomial $p$ gives a unique remainder of degree below five with integer coefficients. If an integer polynomial becomes zero in the rational quotient, its integer remainder is still divisible by $p$ over $\mathbb{Q}$, so must be zero. This gives an injective ring map $\mathbb{Z}[x]/(p)\to \mathbb{Q}[x]/(p)$, proving the monic irreducible integer-polynomial quotient is a domain.

The same unique-remainder argument makes the integer quotient a free abelian group with basis $1,x,x^2,x^3,x^4$. The class of two is nonzero and cannot have an inverse: twice an integer remainder cannot equal the remainder one. Consequently

The Eisenstein criterion says that a primitive integer polynomial $P(x)=a_m{x}^m+\cdots +a_0$ is irreducible over $\mathbb{Q}$ if some prime number $p$ divides every $a_j$ for $j<m$, does not divide $a_m$, and $p^2$ does not divide $a_0$. By Gauss lemma for polynomials, this gives irreducibility in $\mathbb{Z}[x]$ as well.

For prime $n=p$, translate the variable in $P_p(x)=(x^p-1)/(x-1)$:Its leading coefficient is one, all other coefficients are divisible by $p$, and its constant coefficient is exactly $p$. The Eisenstein criterion proves this translated polynomial irreducible. Since $x\mapsto x+1$ is an invertible substitution in $\mathbb{Z}[x]$, $P_p$ is irreducible too.

If $n=ab$ with $a,b>1$, thenBoth factors have positive degree and integer coefficients, proving reducibility. Consequently $\boxed{1+x+\cdots +x^{n-1}\text{ is irreducible exactly when }n\text{ is prime}.}$

Let $L$ be the splitting field of the separable polynomial. Its Galois group is $G={\mathrm{Aut}}_K(L)$, acting faithfully on its $n$ distinct roots, hence as a subgroup of $S_n$. An automorphism preserves each root's minimal polynomial. Conversely an isomorphism $K(\alpha )\to K(\beta )$ for roots of the same irreducible factor extends to an embedding of $L$ in an algebraic closure, and normality of the splitting field makes the image $L$ itself. Thus the root orbits are precisely the irreducible factors' root sets. The action is transitive exactly when the polynomial is irreducible.

If $n=p$ is prime, a transitive action has $p\mid |G|$ by orbit-stabilizer, so Cauchy theorem for groups supplies an element of order $p$. Its permutation must be a $p$-cycle. Conversely a $p$-cycle already acts transitively on all $p$ roots. This proves the stated equivalence for prime degree.

The factorization theorem modulo a good prime says that if an integer polynomial retains its degree and has square-free reduction modulo $p$, and that reduction's irreducible factor degrees are $d_1,\dots ,d_r$, its Galois group over $\mathbb{Q}$ contains a permutation with disjoint cycle lengths $d_1,\dots ,d_r$. Equivalently the finite-field Galois group, generated by Frobenius on the roots, occurs as an unramified decomposition subgroup. The degree-retention qualification matters for nonmonic polynomials.

For the given quintic, Eisenstein criterion at three proves irreducibility. Modulo two it factors asNeither factor has a root in ${\mathbb{F}}_2$, so both are irreducible, and they are distinct; the reduction is separable. Thus $G$ contains an element of cycle type $(2)(3)$, whose cube is a transposition. It is also transitive of degree five.

To prove that this forces the full symmetric group here, form the graph on the five roots with an edge for each conjugate in $G$ of that transposition. The graph is $G$-invariant, and transitivity makes its connected components all the same size. That size divides five; there is an edge, so the size cannot be one. Hence the graph is connected. Transpositions along the edges of a connected graph generate all transpositions (move along a path and conjugate successive swaps), and hence generate $S_5$. They all belong to $G$, giving

The first alternative in the supplied analytic-group theorem would give a faithful representation of $\mathcal{N}$ in some ${\mathrm{GL}}_d({\mathbb{Z}}_p)$ or ${\mathrm{GL}}_d({\mathbb{F}}_p[[t]])$. The embedding proved above supplies cyclic subgroups of order $p^a$ for arbitrarily large $a$. Such orders are impossible in either fixed matrix dimension.

In characteristic zero, a matrix $A$ of order $p^a$ is diagonalizable over an algebraic closure, because its minimal polynomial divides the separable polynomial $X^{p^a}-1$. At least one eigenvalue must be a primitive $p^a$th root of unity; otherwise $A^{p^{a-1}}=I$. The cyclotomic polynomial ${\Phi }_{p^a}$ is irreducible over ${\mathbb{Q}}_p$, since ${\Phi }_{p^a}(1+X)$ satisfies the Eisenstein criterion at $p$. Consequently that eigenvalue has degree $\phi (p^a)=p^{a-1}(p-1)$ over ${\mathbb{Q}}_p$, which must be at most the matrix dimension $d$. This bounds $a$.

In characteristic $p$, pass from the power-series ring to its fraction field. A matrix of $p$-power order satisfies $X^{p^a}-1=(X-1{)}^{p^a}$, so it is unipotent. Writing $A=I+B$, nilpotence gives $B^d=0$. For $p^e\ge d$,Thus its order is at most $p^{\lceil {\log }_{p}d\rceil }$, again a fixed bound. Therefore the linear alternative is impossible in both coefficient rings.

Write $q_n(X)=1+X+\cdots +X^{n-1}$, a monic primitive polynomial. If $n=ab$ with $a,b>1$, grouping terms gives a factorization in the polynomial ring $\mathbb{Z}[X]$:Both factors have positive degree, so $q_n$ is not an irreducible polynomial.

Conversely, if $n=p$ is a prime number, change variables by $X=Y+1$. The binomial theorem givesEvery coefficient except the leading one is divisible by $p$, because $p\mid \left(\genfrac{}{}{0ex}{}{p}{k}\right)$ for $1\le k<p$, whereas the constant coefficient $p$ is not divisible by $p^2$. The Eisenstein criterion makes this polynomial irreducible over $\mathbb{Q}$. The substitution $X\mapsto Y+1$ is an invertible ring homomorphism, so $q_p$ is also irreducible over $\mathbb{Q}$; the Gauss lemma for polynomials transfers this to $\mathbb{Z}[X]$. Hence

Put $\alpha =44^{1/11}$ and $L=\mathbb{Q}(\alpha )$. The Eisenstein criterion at $11$ shows that $X^{11}-44$ is an irreducible polynomial, so $[L:\mathbb{Q}]=11$. Its derivative is $11X^{10}$, and the polynomial discriminant isThe discriminant-index formula for an integral lattice implies that the field discriminant has no prime divisor outside $\{2,11\}$. By the discriminant obstruction to ramification, every other prime is unramified.

To prove that neither candidate disappears through an index correction, let $\mathfrak{p}$ lie above $p\in \{2,11\}$ and normalize its valuation to have value group $\mathbb{Z}$. If $e=e_{\mathfrak{p}/p}$, the relation ${\alpha }^{11}=44$ givesIn both cases $11\mid e$. The fundamental identity for prime decomposition, ${\sum }_{\mathfrak{p}\mid p}{e}_{\mathfrak{p}/p}{f}_{\mathfrak{p}/p}=11$, now forces exactly one prime ideal above each candidate, with $e=11$ and $f=1$. This is ramification forced by a coprime root valuation; at $2$ it avoids an invalid direct application of the Eisenstein criterion, since $4\mid 44$. Hence

Write $\alpha ={\lambda }^{1/5}$, $M=\mathbb{Q}(\alpha )$ and $E=\mathbb{Q}({\zeta }_5)$. The Eisenstein criterion at $\lambda$ gives $[M:\mathbb{Q}]=5$, whereas the prime cyclotomic field degree gives $[E:\mathbb{Q}]=4$. The splitting field $F=E(\alpha )$ is a Galois extension of $E$, and an automorphism fixing $E$ sends $\alpha$ to ${\zeta }_5^b\alpha$. Thus its Galois group embeds in $C_5$. It is nontrivial, since $M$ cannot be contained in the degree-four field $E$. Therefore $[F:E]=5$ and $[F:\mathbb{Q}]=20$.

Index the five roots by $j\in {\mathbb{F}}_5$, with root ${\zeta }_5^j\alpha$. Every automorphism has the form ${\zeta }_5\mapsto {\zeta }_5^u$, $\alpha \mapsto {\zeta }_5^b\alpha$ and acts on indices as $j\mapsto uj+b$, with $u\in {\mathbb{F}}_5^{\times }$. This is a faithful inclusion of $G=\mathrm{Gal}(F/\mathbb{Q})$ into the affine group over a finite field, and both groups have order $20$, so it is an isomorphism. This is the affine Galois group of a prime-radical splitting field.

Let $D$ be the specified decomposition group. Its nonidentity element is an involution. If its linear coefficient were $1$, it would be a nontrivial translation, which has order $5$. Hence it acts as $j\mapsto -j+b$. There is exactly one fixed root, at $j=b/2$, and the other four roots form two pairs. By involution orbits in an affine group of odd prime degree, $D$ has three orbits on the five roots. Those roots identify with $G/\mathrm{Gal}(F/M)$, so the double-coset formula gives three primes in $M$.

For $E$, the embeddings identify with $G/\mathrm{Gal}(F/E)\simeq {\mathbb{F}}_5^{\times }$. The image of $D$ is $\{1,-1\}$ and acts by multiplication, giving the two orbits $\{1,-1\}$ and $\{2,-2\}$. ThereforeThe orbit argument counts primes whether or not $p$ is ramified; cyclicity of the given order-two decomposition group does not require that it be generated by an unramified Frobenius element.

A primitive sixth root of unity satisfieswhich is irreducible over $\mathbb{Q}$ because its discriminant is $-3$. For the other polynomial choose $\alpha =i{3}^{1/6}$, so ${\alpha }^6=-3$ and ${\alpha }^3=-i\sqrt{3}$. Then ${\zeta }_6=(1-{\alpha }^3)/2\in \mathbb{Q}(\alpha )$, so this field already contains all six roots. The Eisenstein criterion at $3$ proves $x^6+3$ irreducible, giving a splitting field of degree six.

Put $r(\alpha )={\zeta }_6^2\alpha$ and $s(\alpha )={\zeta }_6\alpha$. The first automorphism fixes ${\zeta }_6$ and has order three. The second sends ${\alpha }^3$ to its negative, so sends ${\zeta }_6$ to ${\zeta }_6^{-1}$; hence $s^2=1$ and $srs=r^{-1}$. These six distinct transformations exhaust the Galois group. ThereforeThe presence of sixth roots of unity inside the splitting field is why this particular binomial has degree six rather than a generic twelve-degree splitting field.

The polynomial $T^4-2$ is irreducible by Eisenstein criterion at two, so $1,\alpha ,{\alpha }^2,{\alpha }^3$ is a rational basis and each of these elements is integral. The nontrivial automorphism of $K/k$ sends $\alpha$ to $-\alpha$. For an algebraic integer $\theta$, its relative trace and the trace of the algebraic integer $\alpha \theta$ areBoth are in ${\mathcal{O}}_k$, proving $2a,2b,2c,4d\in \mathbb{Z}$. Multiplying $\theta$ by its conjugate gives the field normAgain this is in ${\mathcal{O}}_k$, so both displayed coefficients are integers.

To finish the denominator argument, put $A=2a$, $B=2b$, $C=2c$, $D=4d$, all integers. The two norm coefficients implyThe second congruence modulo two first gives $D$ even. Modulo four it then gives $B$ even. Therefore $2BD$ vanishes modulo four, and the first congruence reads $A^2+2C^2\equiv 0(\mathrm{mod}4)$. The possible squares modulo four force $A$ even and then $C$ even. Returning to the second congruence now gives $D^2\equiv 0(\mathrm{mod}8)$, so $D$ is divisible by four. Thus $a,b,c,d$ are all integers.

Conversely, every integer linear combination of these powers is integral. We have proved the integral basis of the fourth-root-of-two field:

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:

Now take $q=p^a$ with $a\ge 1$ and $\zeta ={\zeta }_{p^a}$. The translated cyclotomic polynomial isIn characteristic $p$, putting $U=T^{p^{a-1}}$ reduces this to ${\sum }_{j=0}^{p-1}(1+U{)}^j=U^{p-1}=T^{p^{a-1}(p-1)}$. Its constant coefficient over the integers is $p$. It is therefore Eisenstein at $p$, of degree $d=p^{a-1}(p-1)$. Consequently ${\mathbb{Q}}_p({\zeta }_{p^a})/{\mathbb{Q}}_p$ is totally ramified of degree $d$, with uniformizer ${\zeta }_{p^a}-1$. If $q=1$, the field is simply ${\mathbb{Q}}_p$, a trivial totally ramified extension.

For a general positive integer $n=p^{a}m$ with $p\nmid m$, the field generated by the prime-to-$p$ root of unity ${\zeta }_m$ is unramified. Indeed $T^m-1$ has distinct roots modulo $p$. In a finite unramified field whose residue field contains those roots, Hensel lemma lifts all of them uniquely. Reduction preserves their orders: a prime-to-$p$ root reducing to one must itself be one, by Hensel uniqueness at the root one of $T^m-1$. The arithmetic Frobenius therefore acts as ${\zeta }_m\mapsto {\zeta }_m^p$, and the extension degree is the multiplicative order of $p$ modulo $m$, with degree one when $m=1$.

The identityfollows by taking powers of a primitive nth root of unity and using the coprimality of $p^a,m$ in the opposite direction. The prime-to-$p$ field is unramified and the $p$-power field is totally ramified; unramified base change preserves total ramification proves that their compositum has the same residue degree as the prime-to-$p$ field. Thus the total ramification criterion for local cyclotomic extensions gives the full classification:This includes trivial extensions when $a=0$. For $p=2$ only $m=1$ is possible.

For a finite totally ramified Galois extension $L/K$ with Galois group $G$, define its lower ramification numbering byHere $G_0$ is the inertia group, which equals $G$, and $G_1$ is the wild inertia group. For a uniformizer $\pi$ the condition is equivalent to $v_L(\sigma (\pi )-\pi )\ge i+1$. To justify this uniformizer criterion for lower ramification groups, expand every $b\in {\mathcal{O}}_L$ as a convergent series ${\sum }_{j\ge 0}{c}_j{\pi }^j$ with the coefficients $c_j$ chosen from representatives of $k_K$ in ${\mathcal{O}}_K$. Such an expansion is obtained by successive subtraction of a residue representative and division by $\pi$, since total ramification gives $k_L=k_K$. The automorphism fixes each $c_j$, and $\sigma (\pi {)}^j-{\pi }^j$ is divisible by $\sigma (\pi )-\pi$. Thus the required valuation bound for $\pi$ implies it for every $b$, including the limit by completeness.

For $L={\mathbb{Q}}_p({\zeta }_{p^a})$, all primitive conjugates lie in $L$ and the Eisenstein degree is $\phi (p^a)$. HenceIf ${\sigma }_b\ne 1$, put $s=v_p(b-1)$, so $0\le s<a$. The element ${\zeta }^{b-1}$ is a primitive $p^{a-s}$th root of unity. Applying the already proved uniformizer result to its field, and multiplying its valuation by the ramification index of $L$ over that field, givesThe identity automorphism has infinite valuation. Therefore the lower ramification filtration of a prime-power cyclotomic extension is explicitlyThe order of the middle group is $p^{a-j}$. At odd $p$, $G_0/G_1$ has order $p-1$; at $p=2$, the condition $b\equiv 1(\mathrm{mod}2)$ is automatic, so $G_0=G_1$ when $a\ge 2$. The case $p=2,a=1$ has trivial Galois group, and the displayed formula remains valid. For $a=0$ all groups are trivial. The valuation calculation also provides the upper ramification numbering, if desired, by integrating the reciprocal indices with the Herbrand function; no change of numbering is implicit in the displayed answer.

A K-isomorphism $L\to L^{\prime }$ is a field isomorphism fixing every element of $K$. The field automorphism group ${\mathrm{Aut}}_K(L)$ consists of such isomorphisms $L\to L$, with composition.

If an isomorphism sends $\alpha$ to $\beta$, it takes every polynomial relation over $K$ for $\alpha$ to the same relation for $\beta$; the monic minimal polynomials coincide. Conversely, if their common minimal polynomial is $m$, evaluation induces field isomorphisms $K[X]/(m)\cong K(\alpha )$ and $K[X]/(m)\cong K(\beta )$. Their composite sends $\alpha$ to $\beta$ and fixes $K$.

An automorphism of $K(\alpha )$ sends $\alpha$ to a root of $m$ inside $K(\alpha )$ and is determined by this image. More precisely the field automorphism group permutes the set of roots of $m$ lying in this field, a set of size $s\le d$. This action is faithful because fixing $\alpha$ fixes the generated field. Thus ${\mathrm{Aut}}_K(K(\alpha ))$ embeds in $S_s$, and then in $S_d$ by fixing $d-s$ extra labels. In particular it is finite. This argument also covers inseparable minimal polynomials.

For an example not isomorphic even as abstract fields, take $K={\mathbb{F}}_p(t)$, let $\alpha$ generate ${\mathbb{F}}_{p^p}$ over ${\mathbb{F}}_p$, and put $\beta =t^{1/p}$. The constant-field minimal polynomial of $\alpha$ remains irreducible over the rational function field: if it factored over ${\mathbb{F}}_p(t)$, each monic factor's coefficients would be algebraic over ${\mathbb{F}}_p$, being symmetric functions of constant algebraic roots. A rational function algebraic over ${\mathbb{F}}_p$ is constant, so those coefficients would lie in ${\mathbb{F}}_p$, contradicting the original irreducibility. The polynomial $X^p-t$ is irreducible by Eisenstein criterion at $t$ in ${\mathbb{F}}_p[t]$. Both degrees are $p$, butThe first field has $p^p$ roots of $X^{p^p}-X$; the second has only $p$. Indeed a rational function algebraic over a finite constant field must be constant: a nonconstant rational function makes the transcendental variable algebraic over its own rational-function field, and therefore cannot itself be algebraic over the constants. The root counts are isomorphism invariants. These extensions are not isomorphic.

If $K={\mathbb{F}}_q$ is finite, every degree-$d$ extension is a field with $q^d$ elements. Such fields are splitting fields of $X^{q^d}-X$ over $K$, so uniqueness of the splitting field gives a K-isomorphism. There is no finite-base-field example of this kind.

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 ${\mathfrak{p}}_1$.

For the first field, set $K=\mathbb{Q}(\sqrt{5},i)$. Its field extension degree is four, since $\mathbb{Q}(\sqrt{5})$ is real and adjoining $i$ doubles its degree. The polynomial $X^3-5$ is irreducible over $K$: a reducible cubic would have a root in $K$, but every root has degree three over $\mathbb{Q}$ by Eisenstein criterion, impossible by the tower law because $3\nmid 4$. Thus

For the second field, put $\alpha =\sqrt[4]{5}>0$. The splitting field is $F=\mathbb{Q}(\alpha ,i)$ and has degree eight, since $X^4-5$ is Eisenstein and $\mathbb{Q}(\alpha )$ is real. Define automorphisms $r(\alpha )=i\alpha ,r(i)=i$ and $s(\alpha )=\alpha ,s(i)=-i$. They satisfy $r^4=s^2=1$ and $srs=r^{-1}$, and give eight distinct automorphisms. Hence $\mathrm{Gal}(F/\mathbb{Q})\cong D_8$.

The Galois correspondence gives all the subfields in the following table. The middle column records generators, not merely their degrees.Each displayed generator is fixed by the indicated subgroup. The two last quartic generators have fourth power $-20$, so their degree is four by Eisenstein criterion; the other quartic degrees are immediate. These are all the subgroups of $D_8$ and hence all the subextensions.

An isomorphism fixing $L$ fixes its subfield $K$, so an $L$-isomorphism is a $K$-isomorphism. The converse is false because an isomorphism over $K$ may act nontrivially on $L$.

Take $K=\mathbb{Q}$, $L=\mathbb{Q}(\sqrt{2})$, $\alpha =2^{1/4}>0$, $F=L(\alpha )$ and $F^{\prime }=L(i\alpha )$. Both are degree-two extensions of $L$. The polynomial $X^4-2$ is irreducible over $\mathbb{Q}$ by the Eisenstein criterion, and both $\alpha$ and $i\alpha$ are roots. Sending $\alpha$ to $i\alpha$ therefore gives a $\mathbb{Q}$-isomorphism $F\to F^{\prime }$; on $L$ it sends $\sqrt{2}$ to $-\sqrt{2}$.

There is no $L$-isomorphism. The two quadratic extensions adjoin square roots of $\sqrt{2}$ and $-\sqrt{2}$. Their square classes would agree only if $-1$ were a square in $L$. To check this criterion directly, a square root of $\sqrt{2}$ in $L(i\alpha )$ would have form $a+b(i\alpha )$; its square forces $ab=0$. The case $b=0$ would put $\alpha$ in $L$, while $a=0$ would require $b^2=-1$. Both are impossible, the latter since $L\subseteq \mathbb{R}$. Thus the extensions are isomorphic over $K$ but not over $L$.

The filtration and its meaning. Let $L/K$ be a finite Galois extension of non-Archimedean local fields, with group $G$, normalized valuation $v_L$, and residue fields $k_L/k_K$ of characteristic $\ell$. The lower ramification numbering isThe higher ramification groups are normal, decrease with $j$, and eventually become trivial: for each nonidentity automorphism some integral element is moved by a nonzero amount. The first group $G_0$ is the inertia group, withIts fixed field is the maximal unramified extension inside $L$. The wild inertia group is $G_1$. For a uniformizer ${\pi }_L$, there are injective homomorphismsThe kernels are the indicated next groups; the additive homomorphism assertion follows by expanding a product of automorphisms modulo ${\pi }_L^{j+2}$. These maps show that $G_0/G_1$ is cyclic of order prime to $\ell$, while $G_1$ is an $\ell$-group. Thus tame ramification is exactly the case $G_1=1$.

To calculate the groups, the uniformizer criterion for lower ramification groups saysIndeed, ${\mathcal{O}}_L$ is generated by ${\pi }_L$ over the integers of the maximal unramified subextension, which $G_0$ fixes. Differences of powers of ${\pi }_L$ are divisible by $\sigma {\pi }_L-{\pi }_L$, so the one test implies all the defining inequalities.

Eisenstein facts used in the calculations. For a complete discrete valuation ring $A$ with uniformizer $\varpi$, a monic polynomialis an Eisenstein polynomial if $\varpi \mid a_j$ for every $j<d$ and ${\varpi }^2\nmid a_0$. The Eisenstein criterion makes it irreducible. If $\theta$ is a root, its valuation relative to the base normalization is $1/d$, the extension has degree and ramification index $d$, its residue field is unchanged, andThe root valuation follows by comparing the terms of its equation. The ramification index must then be at least $d$, hence exactly $d$. In the basis $1,\theta ,\dots ,{\theta }^{d-1}$, the valuations of $b_j{\theta }^j$ are distinct modulo $d$, so an integral linear combination has every $b_j\in A$. This proves the integer-ring assertion. Conversely, a uniformizer in a totally ramified extension generates the field, and its minimal polynomial is Eisenstein; this follows from its value $1/d$, equal valuations of its conjugates, and the valuation one of its norm.

For a monogenic separable integer ring $A[\theta ]$, the different ideal is generated by $f^{\prime }(\theta )$. For a Galois extension, the different exponent from ramification groups isIn the totally ramified case these agree directly: $f^{\prime }({\pi }_L)={\prod }_{\sigma \ne 1}({\pi }_L-\sigma {\pi }_L)$, and an automorphism with displacement valuation $h$ contributes once to each of $G_0,\dots ,G_{h-1}$.

Upper numbering. Extend the lower indexing to real $s\ge 0$ by $G_s=G_{\lceil s\rceil }$, and setOn $[-1,0]$ take $\phi (s)=s$. This Herbrand function slows the indexing as the groups shrink. Lower numbering is compatible with subgroups, while upper ramification numbering is compatible with quotients: for normal $H$, $(G/H{)}^u=G^{u}H/H$. This is the fact that upper ramification groups commute with quotients. Thus upper breaks are particularly useful when comparing intermediate Galois extensions.

The eighth-root cyclotomic extension. Put $\zeta ={\zeta }_8$ and $\pi =\zeta -1$. The shifted cyclotomic polynomialis Eisenstein at two. Hence $L={\mathbb{Q}}_2(\zeta )$ is totally ramified of degree four, $v_L(2)=4$, and $\pi$ is a uniformizer. Its Galois group consists of ${\sigma }_a(\zeta )={\zeta }^a$ for $a=1,3,5,7$, and is $C_2\times C_2$.

Since $\zeta$ is a unit,For $a=3,7$, this is the valuation of $i-1$ or $-i-1$, namely two, because their squares are units times two. For $a=5$, it is $v_L(-2)=4$. The ramification groups of the eighth-root cyclotomic extension of the 2-adic field are thereforeThe lower breaks are one and three, andThus the upper breaks are one and two:The different exponent from ramification groups is $3+3+1+1=8$, agreeing with $v_L(4{\zeta }^3)=8$ from the derivative of $T^4+1$.

The cubic splitting field at three. Let ${\alpha }^3=2$, $\zeta ={\zeta }_3$, and putThe polynomialsare Eisenstein at three. They give totally ramified subextensions of degrees three and two. Their intersection is the base field, so $L={\mathbb{Q}}_3(\alpha ,\zeta )$ has degree six. Its ramification index is divisible by both three and two, hence equals six; therefore $L/{\mathbb{Q}}_3$ is totally ramified. It is the splitting field of $T^3-2$, with Galois group $S_3$.

Normalize $v_L$ so that $v_L(3)=6$. Then $v_L(\beta )=2$ and $v_L(\lambda )=3$, makinga uniformizer. Let $\sigma (\alpha )=\zeta \alpha ,\sigma (\zeta )=\zeta$ and $\tau (\alpha )=\alpha ,\tau (\zeta )={\zeta }^2$. They generate $S_3$. Since $\tau (\lambda )=(\zeta +1)\lambda$,For the order-three automorphism,Here $\alpha$ is a unit and $v_L(\zeta \alpha +1)=v_L(\sigma \beta )=2$. The same calculation gives valuation two for ${\sigma }^2$. The three transpositions are conjugate and the defining filtration is normal, so they all have displacement valuation one. The ramification groups of the splitting field of T3 minus 2 over Q3 areThe lower breaks are zero and one. Since $[G_0:G_s]=2$ for $0<s\le 1$,and the upper groups areThe different exponent from ramification groups is $5+2=7$. As an independent check, the cubic subfield has different exponent $v_{{\mathbb{Q}}_3(\alpha )}(3{\alpha }^2)=3$, and the quadratic extension above it is tame with different exponent one. Transitivity of the different ideal gives $1+2\cdot 3=7$, as required.

The Fundamental theorem of Galois theory gives an inclusion-reversing bijection between subgroups $H\le \mathrm{Gal}(L/K)$ and intermediate fields $M$, by $H\mapsto L^H$ and $M\mapsto \mathrm{Gal}(L/M)$, for a finite Galois extension. One has $[L:L^H]=|H|$ and $[L^H:K]=[\mathrm{Gal}(L/K):H]$. The field $L^H$ is Galois over $K$ exactly when $H$ is normal; then its Galois group is the quotient by $H$.