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A number field is a finite field extension $K/\mathbb{Q}$ of the rational numbers. Its degree of a field extension $[K:\mathbb{Q}]$ is its dimension as a vector space over $\mathbb{Q}$. Every element is an algebraic number; the primitive element theorem also says that $K=\mathbb{Q}(\theta )$ for some single algebraic number $\theta$.

For example,The basis $1,\sqrt{2}$ makes arithmetic explicit: use $(\sqrt{2}{)}^2=2$ to reduce products. Every nonzero element has its inverse in the same field:The denominator cannot vanish unless $a=b=0$, since $\sqrt{2}$ is not a rational number. For instance, $(3+\sqrt{2}{)}^{-1}=(3-\sqrt{2})/7$ belongs to the number field, although it is not an algebraic integer.

A degree-$n$ number field has exactly $n$ distinct field embeddings into the complex numbers fixing $\mathbb{Q}$, because every finite field extension in characteristic zero is a separable field extension. For $\mathbb{Q}(\sqrt{2})$ these are the two real mapsBoth preserve addition and multiplication. Keeping both values, rather than just the positive choice of $\sqrt{2}$, is what makes the field norm and the Minkowski embedding of a number field useful.

Write $G=\mathrm{Gal}(F/K)$ andEvery member of this decomposition group preserves ${\mathcal{O}}_F$ and $\mathfrak{q}$, so reduction defines a homomorphismIts kernel is the inertia group. We prove surjectivity of decomposition group reduction without presupposing a lift of Frobenius.

If the two residue fields coincide, surjectivity is immediate. Otherwise choose a nonzero $\overline{a}\in k_{\mathfrak{q}}$ generating it over $k_{\mathfrak{p}}$; a generator exists by the primitive element theorem for finite fields. Use the Chinese remainder theorem in ${\mathcal{O}}_F$ to choose $a$ whose residue is $\overline{a}$ at $\mathfrak{q}$ and zero at every other prime above $\mathfrak{p}$. ConsiderIts coefficients are integral and fixed by $G$, hence lie in ${\mathcal{O}}_K$: an element of $K$ integral over ${\mathcal{O}}_K$ belongs to this integrally closed ring of integers.

Reduce at $\mathfrak{q}$. When $\sigma \notin D_{\mathfrak{q}}$, the chosen residue of $a$ at ${\sigma }^{-1}\mathfrak{q}$ is zero, so $\sigma a$ reduces to zero at $\mathfrak{q}$. When $\sigma \in D_{\mathfrak{q}}$, its residue is $\overline{\sigma }(\overline{a})$, which is nonzero. ThereforeAlthough this product is written in $k_{\mathfrak{q}}[T]$, its coefficients belong to $k_{\mathfrak{p}}$. Put $q=|k_{\mathfrak{p}}|$. The finite-field Frobenius automorphism $b\mapsto b^q$ therefore permutes its nonzero roots. In particular, ${\overline{a}}^q$ is $\overline{\sigma }(\overline{a})$ for some $\sigma \in D_{\mathfrak{q}}$. Since $\overline{a}$ generates $k_{\mathfrak{q}}$, that $\overline{\sigma }$ agrees with Frobenius on the whole field. The Galois group of a finite field extension is cyclic and generated by Frobenius, so its generator lies in the image. Thus

A number-field completion is finite. Let $F$ be the given number field and $L=F_{\mathfrak{p}}$ its completion. Normalize its absolute value on a field so that the restriction to $\mathbb{Q}$ is $|\cdot {|}_p$: if $v_{\mathfrak{p}}(p)=e$, use $|x|=p^{-v_{\mathfrak{p}}(x)/e}$. Changing the original absolute value by this positive power does not change its completion. The closure of $\mathbb{Q}$ in $L$ is an isometric copy of ${\mathbb{Q}}_p$.

Take a $\mathbb{Q}$-basis of $F$ and let $V$ be its ${\mathbb{Q}}_p$-span in $L$. Choose a maximal linearly independent subset of that finite spanning set. The compact-sphere proof of finite-dimensional non-Archimedean norm equivalence applies to this basis too, so $V$ is complete and therefore closed in $L$. Since $F\subset V$ is dense in $L$, it follows that $V=L$. Hence

Every finite extension is obtained this way. Let $E/{\mathbb{Q}}_p$ be finite. Use its extending Non-Archimedean absolute value, whose completeness and uniqueness were proved above. In characteristic zero the primitive element theorem gives $E={\mathbb{Q}}_p(\alpha )$; scaling $\alpha$ by a power of $p$, we may arrange that it is integral. Its monic minimal polynomial $f$ belongs to ${\mathbb{Z}}_p[T]$. If the degree is one, $E={\mathbb{Q}}_p$ is already the completion of $\mathbb{Q}$.

Otherwise choosewhere ${\alpha }^{\prime }$ runs over the conjugates in an algebraic closure. Approximate the coefficients of $f$ by integers, obtaining a monic $g\in \mathbb{Z}[T]$ of the same degree. Since $f^{\prime }(\alpha )\ne 0$, sufficiently accurate approximation gives $|g(\alpha )|<|g^{\prime }(\alpha ){|}^2$ and makes $|g(\alpha )/g^{\prime }(\alpha )|<\delta$. The strong form of Hensel lemma in $E$ produces a root $\beta \in E$ of $g$ with $|\beta -\alpha |<\delta$.

For completeness, the required Krasner's lemma follows from the unique extension of an absolute value to a finite extension. In a finite Galois extension containing $\alpha ,\beta$, every automorphism $\sigma$ fixing $\beta$ is an isometry, andIf $\sigma \alpha \ne \alpha$, this contradicts the choice of $\delta$. Thus every automorphism fixing $\beta$ fixes $\alpha$, and Galois correspondence gives ${\mathbb{Q}}_p(\alpha )\subseteq {\mathbb{Q}}_p(\beta )$. Since $\beta \in E$, equality follows. In particular $g$ is irreducible over ${\mathbb{Q}}_p$, and therefore over $\mathbb{Q}$.

Put $F=\mathbb{Q}(\beta )\subset E$. It is a number field, and the closure of $F$ in $E$ contains ${\mathbb{Q}}_p(\beta )=E$, so $F$ is dense. LetEvery element of the ring of integers of a number field has absolute value at most one: otherwise its highest-degree term would strictly dominate all other terms of a monic integer equation. Thus reduction into the residue field of $E$ makes $\mathfrak{p}$ a prime ideal, and $\mathfrak{p}\cap \mathbb{Z}=p\mathbb{Z}$. The restriction of the valuation to $F$ is the valuation of this prime, up to normalization. Indeed, the localized discrete valuation ring $({\mathcal{O}}_F{)}_{\mathfrak{p}}$ maps into ${\mathcal{O}}_E$, its units have absolute value one, and its uniformizer has absolute value less than one. Density and completeness giveThis proves the local realization as a completion of a number field.

A finite separable field extension of degree $d$ has exactly $d$ embeddings into an algebraic closure: adjoin finitely many generators successively, and use part (i) to count extensions of each embedding by the distinct roots of the relevant separable minimal polynomial.

For $L=K(\alpha ,\beta )$, label these embeddings ${\sigma }_1,\dots ,{\sigma }_d$. For each pair $i\ne j$, the equalityexcludes at most one $c\in K$. If the images of $\beta$ agree, the images of $\alpha$ cannot agree too, so it excludes none. Since $K$ is infinite, choose $c$ avoiding the finitely many bad values and put $\eta =\alpha +c\beta$. The $d$ images of $\eta$ are distinct, hence its minimal polynomial has degree at least $d$. But $K(\eta )\subset L$ gives degree at most $d$. Therefore $K(\eta )=L$. Combining generators two at a time proves the primitive element theorem for every finite separable extension over an infinite base field.

A finite field extension $L/K$ is a separable field extension if every element of $L$ has a minimal polynomial over $K$ with distinct roots in an algebraic closure. The primitive element theorem says that a finite separable field extension $L/K$ is $K(\alpha )$ for some $\alpha \in L$.

If $K$ is finite, then so is $L$. The multiplicative group of a finite field is cyclic, so a generator of $L^{\times }$ generates $L$ as a field. To justify the group fact, a finite subgroup of a field's multiplicative group is cyclic: in a finite abelian group its exponent is attained as an element order, while the polynomial root bound forces its size to be at most that exponent.

If $K$ is infinite, first consider $L=K(a,b)$. The minimal polynomials of $a$ and $b$ have distinct roots $a_i,b_j$ in an algebraic closure. Choose $c\in K$, $c\ne 0$, avoiding the finitely many values for which $a_i+c{b}_j=a+cb$ for a pair other than $(a,b)$. Set $\alpha =a+cb$. Over $K(\alpha )$, the polynomials $f(T)$ and $g((\alpha -T)/c)$ have precisely one common root, $a$. Their monic greatest common divisor is therefore $T-a$, so $a\in K(\alpha )$ and $b=(\alpha -a)/c\in K(\alpha )$. Thus $K(a,b)=K(\alpha )$. Induct over finitely many generators of the finite separable field extension. This proves every finite separable extension is simple.

Use the simple-root form of the Hensel lemma: for $F\in {\mathcal{O}}_K[T]$ and $\bar{a}\in k_K$ satisfying $\overline{F}(\bar{a})=0$ and $\overline{F^{\prime }}(\bar{a})\ne 0$, there is a unique $a\in {\mathcal{O}}_K$ with $\bar{a}$ as its reduction and $F(a)=0$.

Let $d=[\ell :k_K]$. The residue field $k_K$ is finite, so $\ell /k_K$ is separable, and the primitive element theorem gives $\ell =k_K(\bar{\alpha })$. Let $\bar{F}$ be the monic irreducible polynomial of $\bar{\alpha }$, of degree $d$, and choose a monic coefficient lift $F\in {\mathcal{O}}_K[T]$. Irreducible reduction implies that $F$ is irreducible over $K$: any monic factorization over $K$ has integral coefficients, since all its roots are integral over the valuation ring, and would reduce to a factorization of $\bar{F}$.

Adjoin a root $\alpha$ and put $L=K(\alpha )$. Then $[L:K]=d$ and $\alpha \in {\mathcal{O}}_L$. Its reduction satisfies $\bar{F}$, so $k_L$ contains a copy of $\ell$ over $k_K$. For finite extensions of non-Archimedean local fields, the fundamental degree relation isThe residue degree is at least $d$, while the left side equals $d$, henceThis constructs the required unramified extension.

For uniqueness, let $L^{\prime }/K$ be another unramified extension with a chosen identification $k_{L^{\prime }}\cong \ell$. The element corresponding to $\bar{\alpha }$ is a simple residue root of $F$, because finite fields are perfect. The Hensel lemma in $L^{\prime }$ gives a root ${\alpha }^{\prime }\in {\mathcal{O}}_{L^{\prime }}$ with this reduction. Since $F$ is irreducible of degree $d$, $[K({\alpha }^{\prime }):K]=d=[L^{\prime }:K]$, so $L^{\prime }=K({\alpha }^{\prime })$. Sending $\alpha$ to ${\alpha }^{\prime }$ yields a $K$-isomorphism $L\to L^{\prime }$, compatible with the prescribed residue identification. ThusThis proves the existence and uniqueness of unramified local extensions using the stated version of Hensel's lemma, without assuming the classification in advance.

Because $\mathrm{char}K=0$, the finite extension $M/K$ is separable. By the primitive element theorem, write $M=K(\theta )$. The minimal polynomial of $\theta$ over $K$ has a root in the algebraically closed field $L$, so sending $\theta$ to that root defines a $K$-field embedding $\iota :M\hookrightarrow L$.

The cyclic group $⟨\sigma ⟩$ acts faithfully on $L$ after replacing $d$ by the order of $\sigma$. The Artin fixed-field theorem givesand says that $L/K$ is a finite Galois extension with cyclic Galois group $⟨\sigma ⟩$. The image $E=\iota (M)$ is an intermediate field. Every subgroup of a cyclic group is normal, so the normal subextension criterion shows that $E/K$ is Galois, and its Galois group is a quotient of $⟨\sigma ⟩$, hence cyclic. Transporting this structure through $\iota$ proves that $M/K$ is Galois with cyclic Galois group. This is the finite extensions of the fixed field of a finite-order automorphism of an algebraically closed field theorem.

Algebraic closedness is necessary. Take $L=\mathbb{Q}$ and $\sigma =\mathrm{id}$, so $K=L^{\sigma }=\mathbb{Q}$, and let $M=\mathbb{Q}(\sqrt[3]{2})$. The extension has degree three but is not normal, since the two nonreal roots of $X^3-2$ do not belong to $M$. It is therefore not Galois.

Krasner's lemma says that if $K$ is complete, $\alpha$ is separable over $K$, andthen $K(\alpha )\subseteq K(\beta )$.

To prove it, let $\tau$ be an automorphism of $\overline{K}$ fixing $K(\beta )$. The absolute value has a unique extension to every finite extension of the complete field $K$, so it is invariant under $\tau$. If $\tau (\alpha )\ne \alpha$, the ultrametric inequality and the displayed strict inequality giveBut $\tau (\beta )=\beta$ and invariance gives $|\tau (\alpha )-\beta |=|\alpha -\beta |$, a contradiction. Every automorphism fixing $K(\beta )$ therefore fixes $\alpha$, which proves the field inclusion by Galois correspondence.

Now let $L/{\mathbb{Q}}_p$ be finite. By the primitive element theorem, write $L={\mathbb{Q}}_p(\alpha )$ with separable minimal polynomial $f\in {\mathbb{Q}}_p[X]$. Approximate the coefficients of $f$ closely by those of a polynomial $g\in \mathbb{Q}[X]$ of the same degree. Continuity of roots over a non-Archimedean field gives a root $\beta$ of $g$ arbitrarily close to $\alpha$. Choose it close enough for Krasner's lemma; thenThe degree bound from $\mathrm{deg}g=\mathrm{deg}f$ forces equality. If $F=\mathbb{Q}(\beta )$ and $\mathfrak{p}$ is the prime selected by the embedding $F\hookrightarrow {\overline{\mathbb{Q}}}_p$, its completion isThus every finite extension of ${\mathbb{Q}}_p$ is the completion of a number field at a prime ideal.

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

For a ring homomorphism $A\to B$, the module of Kähler differentials ${\Omega }_{B/A}$ is the $B$-module generated by symbols $db$, subject toEquivalently, it represents $A$-linear derivations:For $A\to B\to C$, the Transitivity exact sequence for Kähler differentials isIf $B\to C=B/I$ is surjective, the Conormal exact sequence for Kähler differentials is

Let $L/K$ be a finite field extension. By the primitive element theorem, its maximal separable field extension $L_s/K$ is simple, and transitivity reduces the calculation to a simple algebraic extension. If $L=K(\alpha )$ with minimal polynomial $m_{\alpha }$, thenThus a separable simple extension has zero differentials. Conversely, if $L/K$ is not separable, the purely inseparable part has a generator whose minimal polynomial has zero formal derivative in positive characteristic, producing a nonzero differential. Hence

Now let $\mathrm{char}K=p>0$. If $L=K(a)$, $a^p\notin K$, and $a^{p^2}\in K$, then the minimal polynomial is $T^{p^2}-a^{p^2}$ and has zero derivative, soIf $L=K(a,b)$, where $a\notin K$, $b\notin K(a)$, and $a^p,b^p\in K$, then $[K(a):K]=[L:K(a)]=p$. Both defining equations have zero derivative, and

For the three morphisms, compute the Sheaf of relative Kähler differentials on coordinate rings.

If $\mathrm{char}k\ne 2$, this is $k[y]/(y)dy$ and its support is the origin $V(y)=V(x,y)$. If $\mathrm{char}k=2$, it is free of rank one and its support is all of $X$.

Its support is the origin $V(x)$ in every characteristic; the module is $k[x]/(x)$ unless $\mathrm{char}k=2$, when it is $k[x]/(x^2)$.

Its support is the entire component $V(x)\subset X$.

If $L/K$ is finite separable, the primitive element theorem writes it as $K(b)$ with $f^{\prime }(b)\ne 0$. The relation in part a has a nonzero $dt$ component, so projection along that relation gives an isomorphism

For an arbitrary simple finite extension, part a presents ${\Omega }_{L/k}$ as a quotient of a vector space of dimension ${\dim }_K{\Omega }_{K/k}+1$ by the image of a space of dimension at most one. Its dimension is therefore at least ${\dim }_K{\Omega }_{K/k}$. Applying this one generator at a time through a finite tower proves the same inequality for every finite extension.

Strict inequality occurs in characteristic $p$. Take $k=K={\mathbb{F}}_p(u)$ and $L=K(u^{1/p})$. Then ${\Omega }_{K/k}=0$, while the relation $b^p-u=0$ has zero differential relative to $k$, sohas dimension one.

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.