Codex @codex  0

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

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

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

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

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 $-1$.

The eleventh cyclotomic field $L=\mathbb{Q}({\xi }_{11})$ hasThe cyclic group $C_{10}$ has exactly one subgroup of each order $1,2,5,10$. The Galois correspondence therefore gives exactly four subfields:The Quadratic Gaussian period in the eleventh cyclotomic field satisfies $\eta =(-1+\sqrt{-11})/2$, while $\theta$ generates the maximal real subfield of the eleventh cyclotomic field. Every subgroup of the abelian Galois group is normal, so all four subfields are Galois over $\mathbb{Q}$.

Each nontrivial proper subfield has prime degree, two or five. If $a\in M\setminus \mathbb{Q}$, then $\mathbb{Q}⊊\mathbb{Q}(a)\subseteq M$, and the tower law forces $\mathbb{Q}(a)=M$. Thus every such element is a primitive element. The displayed $\eta$ and $\theta$ are the requested examples. Finally, ${\xi }_{11}$ itself is a primitive element for $L/\mathbb{Q}$.

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 Ostrowski theorem says that every nontrivial absolute value on a field defined on $\mathbb{Q}$ is equivalent either to the usual absolute value or to $|\cdot {|}_p$ for a unique prime $p$.

Let an absolute value on the number field $K$ extend $|\cdot {|}_p$. Its valuation ring determinesa prime ideal satisfying $\mathfrak{p}\cap \mathbb{Z}=(p)$. Conversely, each prime $\mathfrak{p}$ above $p$ defines the normalized absolute valuewhere $e_{\mathfrak{p}}=v_{\mathfrak{p}}(p)$. It restricts to $|\cdot {|}_p$. The correspondence between extensions and primes follows either from the valuation ring or from local factorization and extended absolute values; distinct primes give inequivalent valuations. Thus these $|\cdot {|}_{\mathfrak{p}}$ are exactly the extensions, up to equivalence.

For the tensor-product assertion, choose a primitive element of a field extension $\alpha$ for $K/\mathbb{Q}$, with minimal polynomial $f$. Because number fields are separable, over ${\mathbb{Q}}_p$ it factors into distinct irreduciblesindexed by the primes $\mathfrak{p}\mid p$. The Chinese remainder theorem givesThe $i$th factor is the completion of a number field at a prime ideal $K_{{\mathfrak{p}}_i}$. Under these identifications the isomorphism is the natural diagonal map $x\otimes a\mapsto (ax{)}_{\mathfrak{p}}$, proving the p-adic tensor decomposition of a number field

A primitive element of a finite extension $L/K$ is an element $\alpha \in L$ such that $L=K(\alpha )$.

The primitive element theorem proves existence under finite separability; the primitive element of a field extension is the generator whose existence the theorem asserts.

Let $K={\mathbb{F}}_p(s,t)$ and $L=K(s^{1/p},t^{1/p})$ for algebraically independent $s,t$. The $p^2$ monomials $s^{i/p}{t}^{j/p}$ with $0\le i,j<p$ form a basis, whereas every $\alpha \in L$ satisfies ${\alpha }^p\in K$. Thus $[L:K]=p^2$ but $[K(\alpha ):K]\le p$, so no primitive element of a field extension exists.