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Normalize the discrete valuations so a uniformiser has value one, and write $k_K,k_L$ for the residue fields. The ramification index $e$ is specified by $v_L{|}_K=e{v}_K$, and the residue degree is $f=[k_L:k_K]$. For finite extensions of complete discretely valued fields, $[L:K]=ef$. One way to see the degree equality is that ${\mathcal{O}}_L$ is a finite free ${\mathcal{O}}_K$-module of rank $[L:K]$; reduction modulo a uniformiser of $K$ has $e$ successive quotients isomorphic to $k_L$, hence dimension $ef$ over $k_K$.

An unramified extension has $e=1$ and separable residue extension, equivalently $[L:K]=f$ with separable residue extension. A totally ramified extension has $f=1$, equivalently $[L:K]=e$. The separability condition in the unramified extension definition matters if the residue field is imperfect; it is automatic for finite residue fields.

Suppose $L/K$ is unramified. Choose a primitive element of a field extension $\bar{x}$ for the finite separable extension $k_L/k_K$ and lift it to $x\in {\mathcal{O}}_L$. Since the residue degree of $K(x)/K$ is at least $[k_K(\bar{x}):k_K]=[L:K]$, necessarily $K(x)=L$. Its monic minimal polynomial of an algebraic element $f_x$ has coefficients in ${\mathcal{O}}_K$. Its reduction has degree $[L:K]$ and annihilates $\bar{x}$, whose minimal polynomial over $k_K$ has that same degree. The two coincide, so ${\bar{f}}_x$ is separable.

Conversely, suppose $L=K(x)$, $x\in {\mathcal{O}}_L$, and ${\bar{f}}_x$ is separable. It must be irreducible: otherwise its coprime factors lift by the factorization form of Hensel's lemma, contradicting irreducibility of $f_x$. Hence $\bar{x}$ has degree $[L:K]$ over $k_K$. The degree equality forces $f=[L:K]$ and $e=1$, and the residue extension is separable. We have proved the unramified generator criterion with separable reduction:

Now let $k_L={\mathbb{F}}_q$, $q=p^f$. Every element of $k_L^{\times }$ is a simple root of $X^{q-1}-1$, whose derivative is a unit at each root. Hensel's lemma lifts each of these $q-1$ elements uniquely to a root in ${\mathcal{O}}_L^{\times }$. Thus $L$ contains all $q-1$ roots of $X^{q-1}-1$; these are the nonzero Teichmuller lifts.

More generally, a root of unity of order prime to $p$ reduces injectively into $k_L^{\times }$. Indeed, if such a root reduces to $1$, uniqueness in Hensel's lemma for $X^m-1$ makes it equal to $1$. Consequently every prime-to-$p$ order divides $q-1$.

A root whose order has a nontrivial $p$-part produces a primitive $p$th root of unity $\zeta$. It reduces to $1$ in characteristic $p$. Sinceall the factors have the same positive integral valuation. HenceThis is the ramification bound for a primitive pth root of unity. ThereforeFor $p=2$ the hypothesis $e<1$ cannot occur, so that instance is vacuous rather than a claim excluding the ever-present root $-1$.

An integral domain $A$ is an integrally closed domain if every element of its fraction field which is an integral element over $A$ belongs to $A$. Explicitly, if $z\in \mathrm{Frac}A$ satisfiesthen $z\in A$. Thus $\boxed{A\text{ is integrally closed exactly when its integral closure in its fraction field is }A}$.

Suppose $A$ is a unique factorization domain and $z=a/b$ is integral, with $a,b\in A$, $b\ne 0$, and with no common irreducible element as a factor. Multiplying its monic equation by $b^d$ givesHence $b$ divides $a^d$. Every irreducible element dividing $b$ is a prime element, so it would divide $a$, contrary to the choice of the fraction. Therefore $b$ is a unit, and $z\in A$. We have proved

The going-down theorem is the following. Let $A\subseteq B$ be an integral extension of integral domains with $A$ an integrally closed domain. IfthenWe first prove this for a module-finite ring extension and then remove that extra hypothesis. We use the Lying-over theorem, the Going-up theorem, and the incomparability theorem for integral extensions: primes exist over every base prime, prime chains can be extended upwards, and distinct primes over the same base prime cannot contain one another. The Going-up theorem follows by applying the Lying-over theorem to the quotient by the lower prime. These results concern arbitrary integral extensions; none assumes the going-down theorem.

For the finite case, let $K=\mathrm{Frac}A$ and $L=\mathrm{Frac}B$. Since $B$ is module-finite, $L/K$ is a finite field extension. Take a finite normal field extension $E/K$ containing $L$, obtained as a splitting field of the minimal polynomials of finitely many field generators. The extension $E/K$ need not be a separable field extension. Let $C$ be the integral closure of $A$ in $E$. Then $B\subseteq C$, and $C$ is integral over both $A$ and $B$.

We need one auxiliary fact: the finite group $G={\mathrm{Aut}}_K(E)$ acts transitively on the prime ideals of $C$ over any fixed prime ideal $\mathfrak{p}$ of $A$. First observe that an element $w\in E$ fixed by all of $G$ is a purely inseparable algebraic element over $K$. Indeed, all $K$-embeddings of $K(w)$ into an algebraic closure extend to $K$-automorphisms of the normal field extension $E$. Hence the minimal polynomial of an algebraic element $w$ has only one distinct root. In characteristic zero this says $w\in K$; in characteristic $p>0$ it says $w^{p^e}\in K$ for some $e\ge 0$.

Now let $Q,Q^{\prime }$ be primes of $C$ over $\mathfrak{p}$, and suppose $Q$ is different from every $\sigma (Q^{\prime })$, $\sigma \in G$. By the incomparability theorem for integral extensions, $Q$ is contained in none of these conjugate primes. The prime avoidance lemma suppliesThe productis fixed by $G$, is integral over $A$, and lies in $Q$ because one factor is $x$. Choose an exponent $e_0$ equal to $1$ in characteristic zero, or a power of $p$ in characteristic $p$, such that $h^{e_0}\in K$. Since $A$ is an integrally closed domain, $h^{e_0}\in A$. It also lies in $Q\cap A=\mathfrak{p}$, hence in $Q^{\prime }$. Because $Q^{\prime }$ is a prime ideal, some factor $\sigma (x)$ lies in $Q^{\prime }$. This means $x\in {\sigma }^{-1}(Q^{\prime })$, contradicting our choice. Thus $Q=\sigma (Q^{\prime })$ for some $\sigma$, proving transitivity. The power $e_0$ accounts for the possible purely inseparable field extension and makes the argument valid in every characteristic.

Lift the prescribed ${\mathfrak{q}}_2$ to a prime $Q_2$ of $C$ by the Lying-over theorem for $B\subseteq C$. Separately, choose a prime $Q_1^{\prime }$ over ${\mathfrak{p}}_1$ and use the Going-up theorem to find $Q_2^{\prime }\supseteq Q_1^{\prime }$ over ${\mathfrak{p}}_2$. Transitivity gives $\sigma \in G$ with $\sigma (Q_2^{\prime })=Q_2$. ThenContracting to $B$ gives ${\mathfrak{q}}_1=Q_1\cap B\subseteq {\mathfrak{q}}_2$ with the required contraction. This proves the going-down theorem for the module-finite case.

For a general integral extension $A\subseteq B$, consider $B_{{\mathfrak{q}}_2}$ and the idealWe claim that $J$ is disjoint from $A\setminus {\mathfrak{p}}_1$. Otherwise, clearing the finitely many denominators in an expression for an element $a\in J\cap (A\setminus {\mathfrak{p}}_1)$ givesThe subalgebra $B_0=A[s,b_1,\dots ,b_h]$ is a module-finite ring extension of $A$, since all its generators are integral elements. Apply the finite case to ${\mathfrak{q}}_2\cap B_0$: there is a prime $Q_1\subseteq {\mathfrak{q}}_2\cap B_0$ contracting to ${\mathfrak{p}}_1$. The displayed identity puts $sa$ in $Q_1$, but $a\notin Q_1$. Thus $s\in Q_1\subseteq {\mathfrak{q}}_2$, a contradiction.

Localize $B_{{\mathfrak{q}}_2}$ further at the multiplicative subset $A\setminus {\mathfrak{p}}_1$. The extension of $J$ is a proper ideal, by the disjointness just proved. Choose a maximal ideal containing it and contract back to a prime ideal of $B_{{\mathfrak{q}}_2}$. This prime contains $J$ and avoids $A\setminus {\mathfrak{p}}_1$. Its contraction ${\mathfrak{q}}_1$ to $B$ lies inside ${\mathfrak{q}}_2$ by the prime ideal correspondence for localization, and its contraction to $A$ is exactly ${\mathfrak{p}}_1$. This completes the proof of the going-down theorem without any finite generation assumption on $B$.

A suitable version of the Hensel lemma is this: if $R$ is a complete discrete valuation ring, $h\in R[X]$, and $\overline{a}$ is a root of $\overline{h}$ with ${\overline{h}}^{\prime }(\overline{a})\ne 0$, there is a unique $a\in R$ reducing to $\overline{a}$ with $h(a)=0$. Starting with any lift, Newton iteration over a valued field $a_{j+1}=a_j-h(a_j)/h^{\prime }(a_j)$ keeps the derivative a unit and doubles the error valuation, giving convergence. Uniqueness follows by factoring $h(u)-h(v)$ when two roots have the same residue.

The field in the rest of this question is not assumed complete, so that version cannot simply be applied to it. Instead use the stated uniqueness of the extension of an absolute value. Let ${\alpha }_i,{\alpha }_j$ be any two roots of a polynomial $f$, which is irreducible. The field embedding sending ${\alpha }_i$ to ${\alpha }_j$ identifies $K({\alpha }_i)$ with $K({\alpha }_j)$. Pulling the latter field's absolute value back gives an extension on $K({\alpha }_i)$, which must coincide with the given one. More generally, for every $h\in K[X]$,This proves equal absolute values of algebraic conjugates without assuming separability or a transitive action of a Galois group; repeated roots cause no difficulty.

Every root of the monic $f\in {\mathcal{O}}_K[X]$ has absolute value at most one. Indeed, if $|\alpha |>1$, then each lower term satisfies $|c_j{\alpha }^j|\le |\alpha {|}^j<|\alpha {|}^n$. The ultrametric inequality makes the leading term dominate their sum, contradicting $f(\alpha )=0$. Thus

All roots lie in the valuation ring ${\mathcal{O}}_M$, so their residues are defined in the residue field $k_M$. Let $\phi \in k_K[X]$ be the monic minimal polynomial of an algebraic element of ${\overline{\alpha }}_1$; this residue is algebraic because it satisfies the monic $\overline{f}$. Choose a coefficientwise lift $\Phi \in {\mathcal{O}}_K[X]$. Then $|\Phi ({\alpha }_1)|<1$, and the conjugacy argument gives $|\Phi ({\alpha }_i)|<1$ for every $i$. Hence every ${\overline{\alpha }}_i$ is a root of $\phi$.

Reduction of the splitting-field factorization, with multiplicities, givesEvery monic irreducible factor of $\overline{f}$ over $k_K$ has a root among these residues. That root also satisfies $\phi$, so its minimal polynomial is $\phi$. Unique factorization therefore proves the pure-power reduction of a monic irreducible polynomial:This does not assert that $\phi$ is separable or that the reduction is square-free.

Finally, factor the monic $g$ over $K$ as $g={\prod }_j{h}_j^{e_j}$ with distinct monic irreducible polynomials $h_j$. The same leading-term argument bounds every root of $g$ by one. Coefficients of any $h_j$ are elementary symmetric polynomials in some of these roots, so they lie in ${\mathcal{O}}_K$. By the preceding result each ${\overline{h}}_j$ is a power of a single monic irreducible residue polynomial ${\phi }_j$.

Since the prescribed ${\overline{g}}_1,{\overline{g}}_2$ are coprime, every ${\phi }_j$ occurs on exactly one side. Assign the entire factor $h_j^{e_j}$ to that side and form the two products. Comparing irreducible-factor multiplicities in the reduction proves the coprime factor lifting from valuation-extension uniqueness:Empty products are one, covering constant prescribed factors. The construction proves the required factor-lifting form of the Hensel lemma directly from extension uniqueness, without adding completeness as an unstated hypothesis.

Because $K$ has characteristic $p$, the prime field ${\mathbb{F}}_p$ lies in $K$, andfor every $c\in {\mathbb{F}}_p$. These are all $p$ roots. Moreover $f^{\prime }(t)=-1$, so $f$ is a separable polynomial.

Let $m(t)$ be the minimal polynomial of an algebraic element of $\alpha$ over $K$. Translation by $c\in {\mathbb{F}}_p$ sends an irreducible factor $g(t)$ of $f(t)$ to the irreducible factor $g(t-c)$. The additive group of ${\mathbb{F}}_p$ therefore acts on the irreducible factors, and every orbit has size $1$ or $p$. An orbit of size $p$ would already contribute $p\mathrm{deg}m$ to the degree-$p$ polynomial $f$, forcing $\mathrm{deg}m=1$. Then $\alpha \in K$, all roots $\alpha +c$ lie in $K$, and $L=K$, contrary to the hypothesis.

Thus $m(t-c)=m(t)$ for every $c\in {\mathbb{F}}_p$. If $1\le d=\mathrm{deg}m<p$, comparison of the coefficient of $t^{d-1}$ in $m(t-c)-m(t)$ for $c\ne 0$ gives $-dc$ times the nonzero leading coefficient, a contradiction. Hence $\mathrm{deg}m=p$, and since $m$ is a monic divisor of the monic degree-$p$ polynomial $f$, we have $m=f$. Thus $f$ is irreducible over $K$.

All roots $\alpha +c$ already belong to $K(\alpha )$, so the splitting field isAs the splitting field of a separable polynomial, $L/K$ is a Galois extension. For each $c\in {\mathbb{F}}_p$, irreducibility gives a distinct $K$-automorphismThese exhaust the $p=[L:K]$ automorphisms. This is an Artin–Schreier extension, and

Write $q=p^r$, where $p=\mathrm{char}K$. Forthe derivative isso $h$ is a separable polynomial and all its roots in its splitting field $L$ are distinct.

Let $R$ be the set of these roots. The Frobenius endomorphism and the identitiesshow that $R$ is closed under addition and multiplication. It is also closed under additive inverses and, for $x\ne 0$, inversion. Thus $R$ is a finite field. Since $h$ splits into $q^n$ distinct linear factors, $|R|=q^n$. The field generated by all roots is therefore $R$ itself, so $L=R$ andHence the polynomial characterization of a finite field gives

Now let $f\in K[X]$ be irreducible of degree $n$, and let $\alpha$ be a root in an algebraic closure. The minimal polynomial of an algebraic element givesso $K(\alpha )$ has $q^n$ elements. Its nonzero elements form a finite group under multiplication, of order $q^n-1$; hence ${\alpha }^{q^n}=\alpha$. Therefore $\alpha$ is a root of $X^{q^n}-X$, and its minimal polynomial $f$ divides every polynomial over $K$ that vanishes at $\alpha$. Thus the irreducible factors of a finite-field Frobenius polynomial satisfy

The seventh cyclotomic polynomial isIt is irreducible over $\mathbb{Q}$, and all its roots ${\zeta }_7^a$, $a=1,\dots ,6$, lie in $L=\mathbb{Q}({\zeta }_7)$. Hence $L/\mathbb{Q}$ is a degree-six Finite Galois extension, withThe group $(\mathbb{Z}/7\mathbb{Z}{)}^{\times }$ is cyclic of order six; for example,

By the Galois correspondence, the subfields are the fixed fields of the four subgroups of $C_6$:Thus these are all the subfields.

For the Quadratic Gaussian period in the seventh cyclotomic field, putThe cyclotomic relation gives $\eta +\bar{\eta }=-1$, while direct multiplication gives $\eta \bar{\eta }=2$. HenceIts discriminant is $-7$, so

For the real cubic subfield of the seventh cyclotomic field, divideby ${\zeta }_7^3$. With $\theta ={\zeta }_7+{\zeta }_7^{-1}$,soThis cubic has no rational root, and therefore

The requested primitive elements, minimal polynomials, and automorphism groups are consequentlyMore explicitly, the nontrivial automorphism of $\mathbb{Q}(\eta )$ sendsand the three automorphisms of $\mathbb{Q}(\theta )$ cyclically permute

Finally, $G$ is abelian, so every subgroup is normal. The subextensions of an abelian Galois extension theorem shows that

The element $\alpha$ is a root ofBy the rational root theorem, any rational root of this monic polynomial must be one ofDirect substitution shows that none is a root. A reducible cubic over a field has a linear factor, so $f$ is an irreducible polynomial over $\mathbb{Q}$. It is therefore the minimal polynomial of $\alpha$, and

The minimal polynomial of $\beta$ isand part (b) showed that ${\mathcal{O}}_K=\mathbb{Z}[\beta ]$. Modulo $3$,The quadratic has no root in ${\mathbb{F}}_3$, so it is irreducible. The Dedekind factorization theorem therefore givesThe two factors are distinct prime ideal, with norms $3$ and $3^2=9$, respectively.

Modulo $2$,and the quadratic is irreducible over ${\mathbb{F}}_2$. HenceThis is a product of two distinct proper prime ideals, so $(2)$ is not prime. These decompositions are collected in prime ideals above two and three in the cubic field of discriminant minus 307.

Put $K=\mathrm{Frac}R$, letbe the minimal polynomial of an algebraic element $y$ over $K$, and let $B$ be the integral closure of $R$ in a finite normal extension containing all roots of $f$. Since $y$ is integral over $R$ and $R$ is integrally closed domain, every $a_i$ belongs to $R$.

Write $y={\sum }_j{p}_j{z}_j$ with $p_j\in \mathfrak{p}$ and $z_j\in A$. Every $K$-embedding into the normal extension fixes the $p_j$ and sends each $z_j$ to an element integral over $R$. Thus every conjugate of $y$ lies in the extended ideal $\mathfrak{p}B$. Each nonleading coefficient of $f$ is, up to sign, an elementary symmetric polynomial in those conjugates, so it lies in $\mathfrak{p}B\cap R$.

For an integral extension, extension followed by contraction preserves a prime ideal:Indeed, the determinant trick gives $r^m\in \mathfrak{p}$ for $r\in \mathfrak{p}B\cap R$, and primality then gives $r\in \mathfrak{p}$. Hence $a_i\in \mathfrak{p}$ for every $i<d$.

Suppose a Non-Archimedean absolute value extends uniquely to every finite extension. If $f\in {\mathcal{O}}_K[X]$ is monic and irreducible, every root has absolute value at most one: otherwise its leading power dominates all lower terms, contradicting $f(\alpha )=0$ by the ultrametric inequality. In a splitting field, let $\phi$ be the minimal polynomial of an algebraic element of the residue of one root. Lift its coefficients to ${\mathcal{O}}_K$. The lift has absolute value less than one at that root and hence, by equal absolute values of algebraic conjugates, at every root. Consequently every root residue is a zero of $\phi$. Since reduction preserves the product of the linear factors with multiplicities, every irreducible factor of $\overline{f}$ is $\phi$, proving the displayed identity for some $m\ge 1$.

A finite extension of complete discretely valued fields is unramified exactly when it has an integral generator whose reduced minimal polynomial of an algebraic element is separable. In one direction, lift a residue-field primitive element. In the other, the factorization form of Hensel's lemma makes the separable reduction irreducible, forcing residue degree to equal the entire extension degree.