Galois theory

The inverse Galois problem asks which finite groups occur as Galois groups over a given field, with the classical problem taking the base field to be $\mathbb{Q}$. A regular realization over $K(T)$ and Galois-preserving Hilbert specialization solve an instance over a Hilbertian field. Realization over an algebraically closed constant field's rational function field alone does not supply descent to $\mathbb{Q}$.

Fix ordered branch positions and conjugacy classes $C_i$ in a finite group $G$. The generating product-one tuples $(g_1,\dots ,g_r)\in \prod C_i$ form the inner Nielsen set. The type is rigid when this set is nonempty and a single simultaneous-conjugation orbit. Over $\mathbb{C}(T)$ this fixes the connected Galois cover up to isomorphism: tuples prescribe the monodromy epimorphism of the punctured sphere, and conjugate epimorphisms give isomorphic covers; the local completions at punctures are uniquely determined by their cyclic monodromy.

A compact Galois module is a compact profinite abelian group equipped with a continuous Galois group action. An abelian pro-p group is naturally a compact ${\mathbb{Z}}_p$-module. A continuous action of the Galois group of a Zp-extension extends to its Iwasawa algebra of a Zp-extension. This compact topology differs from the discrete topology usually used in Galois cohomology.

A finite separable extension is solvable when its normal closure has a solvable group as Galois group. In characteristic zero it is contained in a radical tower after adjoining suitable roots of unity; containment need not mean it is itself a tower of radical adjunctions.

In the strict tower convention, a radical extension is a field obtained through finitely many adjunctions $a_i^{n_i}\in K_{i-1}$, $K_i=K_{i-1}(a_i)$. Some texts instead use the term for any subfield of such a tower. This distinction matters: solvability by radicals implies containment in a tower, and need not give a radical tower entirely inside the original field.

The absolute Galois group of a field $K$ is $G_K=\mathrm{Gal}(K^{\mathrm{sep}}/K)$. It is a profinite group, the inverse limit of the Galois groups of finite Galois extensions. Its continuous actions on discrete modules are the coefficients for Galois cohomology.

The Galois closure of a finite separable extension $L/K$ is the smallest Galois extension in a fixed separable closure containing $L$. It is generated by all the conjugates of $L$. If $G=\mathrm{Gal}(M/K)$ and $H=\mathrm{Gal}(M/L)$, their subgroup core is trivial. This is distinct from the existing group-theoretic normal closure.

For a finite Galois extension $L/K$, intermediate fields correspond inclusion-reversingly to subgroups of $\mathrm{Gal}(L/K)$. Normal subgroups correspond to Galois intermediate extensions.

When a field $K$ contains the $n$th roots of unity and its characteristic does not divide $n$, Kummer theory identifies cyclic extensions of degree dividing $n$ with suitable classes in $K^{\times }/(K^{\times }{)}^n$. Cohomologically,

For a number field and finite subgroup $\Delta \subset K^{\ast }/K^{\ast n}$, adjoining all the indicated nth roots gives an extension unramified outside the displayed finite set. The valuations modulo $n$ are independent of representatives. At a finite place outside $S$, divide each radicand by a local nth power of a uniformizer to obtain a unit. Its polynomial has separable reduction because $n$ is invertible. A finite unramified extension containing all residue roots contains all lifted roots by Hensel's lemma. This proof does not require ${\mu }_n\subset K$.

For a cyclic Galois extension $M/K$ of degree $m$, choose a generator $g$ and a primitive $m$th root of unity $\zeta \in K$. Reindexing the displayed sum gives $gR(w)=\zeta R(w)$. Linear independence of distinct field embeddings ensures this sum is not identically zero as a function of $w$. Choosing $\alpha =R(w)\ne 0$ therefore gives $m$ distinct conjugates ${\zeta }^j\alpha$, while ${\alpha }^m\in K$. Consequently $M=K(\alpha )$ is the splitting field of $X^m-{\alpha }^m$. Independence follows by taking a shortest supposed linear relation among distinct automorphisms, evaluating at $uv$ and subtracting a suitable multiple of the relation at $v$ to remove one term but not all terms.

For nontrivial even $\chi$ in the full cyclotomic tower of $\mathbb{Q}$, Kummer theory relates the p-ramified Iwasawa module to the reflected odd unramified Iwasawa module, up to pseudo-isomorphism. Here $\iota (\gamma )={\gamma }^{-1}$ and $(1)$ is the Tate twist. The induced substitution on characteristic series is $T\mapsto u(1+T{)}^{-1}-1$, where $u$ is the cyclotomic value of $\gamma$. Omitting either operation changes the interpolation convention.

When the characteristic differs from $p$, the Kummer sequence gives $0\to H^n(L,K_s^{\times })/p\to H^{n+1}(L,{\mu }_p)\to H^{n+1}(L,K_s^{\times })[p]\to 0$. For $n\ge 1$, the positive-degree multiplicative groups are torsion. Thus the middle group is zero exactly when the first multiplicative group is $p$-divisible and the second has zero $p$-primary torsion.

For a number field $K$, a finite set $S$ and $m\ge 2$, let $K(S,m)$ consist of classes $[a]\in K^{\ast }/K^{\ast m}$ whose valuations outside $S$ are divisible by $m$. It fits into $0\to {\mathcal{O}}_{K,S}^{\times }/({\mathcal{O}}_{K,S}^{\times }{)}^m\to K(S,m)\to \mathrm{Cl}({\mathcal{O}}_{K,S})[m]\to 0$. The S-unit group is finitely generated and the ideal class group is finite, so $K(S,m)$ is finite. When $S$ contains primes dividing $m$, this controls unramified multiplicative Kummer classes outside $S$.

The square-class group identifies two nonzero elements of a field when their quotient is a square. Multiplication makes it an abelian group of exponent two, hence a vector space over ${\mathbb{F}}_2$. It records quadratic extensions and the images of Kummer theory for $n=2$.

The valuation contributes one factor $C_2$. The unit decomposition of the 2-adic Gaussian field contributes ${\mu }_4/{\mu }_4^2\cong C_2$ and the additive quotient ${\mathbb{Z}}_2[i]/2{\mathbb{Z}}_2[i]\cong C_2^2$. Thus there are sixteen square classes and fifteen quadratic extensions over this field.

The square class of $a\in K^{\ast }$ is its image in the square-class group of a field $K^{\ast }/K^{\ast 2}$. Two nonzero elements have the same square class exactly when their quotient is a square. Every element of this quotient has order dividing $2$. Over the rational numbers, a square class has a unique signed squarefree integer representative.

For odd $p$, parity of valuation and the square class of the unit residue give a group $C_2\times C_2$, represented by $1,u,p,pu$ for a unit $u$ with nonsquare residue. For $p=2$, valuation parity and unit residue modulo eight give $C_2^3$, with generators $-1,5,2$. The unit criteria follow from the Hensel lemma, using its strong derivative version for $p=2$.

When a field $K$ contains the relevant roots of unity, adjoining an $n$th root of an element of $K$ gives a Kummer extension. Such extensions describe abelian extensions of exponent dividing $n$.

Galois cohomology is the group cohomology of a Galois group acting on a module. For a discrete $G$-module $M$, the first group is the quotient of crossed homomorphisms $c(\sigma \tau )=c(\sigma )+\sigma c(\tau )$ by maps $c(\sigma )=\sigma m-m$.

For an abelian variety $A$ over a field $K$, its Weil–Châtelet group is the Galois cohomology group $H^1(K,A(K^{\mathrm{sep}}))$, classifying torsors under $A$. Such an algebraic-group torsor becomes $A$ after passing to a separable closure but need not have a $K$-point. A chosen geometric point identifies the torsor with $A$ and its Galois descent with a cocycle; changing that point adds a coboundary. The zero class is precisely the torsor with a rational point. For an elliptic curve, these torsors are genus-one curves with a specified action and Jacobian. Over a number field the kernel of restriction to every completion is the Tate–Shafarevich group.

For a Finite Galois extension $L/K$ and a finite-dimensional $L$-space $V$ with a semilinear action of its Galois group, the map $L{\otimes }_K{V}^G\to V$ is an isomorphism. The vectors ${\sum }_{\sigma }\sigma (a)\sigma (v)$ are invariant and span $V$ by the Artin independence theorem. Choosing an invariant $L$-basis then identifies the invariant space with its $K$-span. The proof does not divide by the extension degree.

A splitting isomorphism of a central simple algebra with $M_n(L)$ transports its Galois action to $T_{\sigma }=c_{\sigma }{\sigma }_0$, giving a nonabelian first cohomology class. Conversely a cocycle gives a semilinear algebra action whose invariants descend to a central simple algebra. Equivalent cocycles give isomorphic fixed algebras. This gives a pointed-set bijection ${\mathrm{CSA}}_n(L/K)\cong H^1(\mathrm{Gal}(L/K),{\mathrm{PGL}}_n(L))$.

For an extension $L/K$ with Galois group $G$, a semilinear action on an $L$-space is an additive group action satisfying $\sigma (av)=\sigma (a)\sigma (v)$. Scalar coefficients are transformed along with vectors. Invariants form a $K$-space, and Galois descent of vector spaces reconstructs the original space from it.

For a finite Galois module $M$ unramified outside a finite set $S$, define $H_S^1(K,M)$ as the classes in Galois cohomology whose restriction to inertia is zero at every finite prime outside $S$. This means zero as a cohomology class, rather than zero for every possible representative. Passing to a finite Galois extension on which $M$ is constant reduces the finiteness problem to characters with bounded ramification and a finite restriction kernel.

A discrete Galois module over $K$ is an abelian group with a continuous action of the absolute Galois group $G_K$. Continuity means that each element has an open stabilizer. Examples are the algebraic points of an elliptic curve, its finite torsion module $E[m]$, and the group of roots of unity ${\mu }_m$. Galois cohomology takes continuous cocycles with values in such a module.

The module ${\mu }_m$ consists of the $m$th roots of unity in a separable closure. If the field characteristic does not divide $m$, it has $m$ elements and fits into the Kummer sequence $1\to {\mu }_m\to K_s^{\times }\stackrel{m}{\to }{K}_s^{\times }\to 1$. Its Galois action is the cyclotomic action; a generator gives a noncanonical identification with $\mathbb{Z}/m\mathbb{Z}$.

For a finite Galois extension $L/K$, multiplicative Hilbert theorem 90 says $H^1(\mathrm{Gal}(L/K),L^{\ast })=0$. A multiplicative cocycle has the form $c_{\sigma }=\sigma (b)/b$. One proof forms a nonzero sum $A={\sum }_{\sigma }{c}_{\sigma }\sigma (a)$, using linear independence of field automorphisms. Its transformation law is $\tau A=c_{\tau }^{-1}A$, so $b=A^{-1}$ gives $c_{\tau }=\tau (b)/b$. Passing to the algebraic closure gives $H^1(K,{\overline{K}}^{\ast })=0$, so the multiplicative Kummer sequence identifies $H^1(K,{\mu }_m)$ with $K^{\ast }/K^{\ast m}$.

For an unramified extension of non-Archimedean local fields, let $\varphi$ generate its cyclic Galois group. The ordinary Hilbert theorem 90 supplies $b\in L^{\times }$ with $x=\varphi (b)/b$. A common uniformizer $\pi \in K$ gives $y={\pi }^{-v_L(b)}b\in {\mathcal{O}}_L^{\times }$, retaining the same quotient because $\varphi$ fixes $\pi$. The reverse implication follows by telescoping the field norm.

In a field of characteristic $p>0$, the Frobenius map $x\mapsto x^p$ is an injective field endomorphism becauseConsequently every element of an extension field has at most one $p$th root.

The formal derivative of $f(T)={\sum }_i{a}_i{T}^i$ is $f^{\prime }(T)={\sum }_{i}i{a}_i{T}^{i-1}$.

In characteristic $p$, the coefficient $i$ in a formal derivative vanishes exactly when $p$ divides $i$.

Over a field of characteristic $p>0$,for some polynomial $g$. Indeed, zero derivative says every exponent with nonzero coefficient is divisible by $p$.

A polynomial over a field is separable when it has no repeated root in a splitting field. Equivalently, it is coprime to its formal derivative.

A nonzero polynomial is factorwise separable if every irreducible factor is a separable polynomial. Unlike the convention that all roots must be distinct, this condition allows repeated irreducible factors. Its splitting field is a separable field extension precisely when the polynomial is factorwise separable: in one direction every root has a separable minimal polynomial; in the other, adjoining finitely many roots with separable minimal polynomials gives a separable extension. Thus the implication from a separable splitting extension to a polynomial with distinct roots needs the extra squarefree hypothesis. The example $(X-1{)}^2$ already distinguishes the conventions.

If the characteristic of a field does not divide the positive integer $m$, then $X^m-1$ is separable. Its formal derivative is $m{X}^{m-1}$, and a common root would have to be both nonzero and zero.

If $f$ is irreducible over a field of characteristic zero, then $f^{\prime }\ne 0$ and $\mathrm{deg}{f}^{\prime }<\mathrm{deg}f$. Thus $f$ cannot divide $f^{\prime }$, so irreducibility gives $\gcd (f,f^{\prime })=1$. Consequently $f$ has no repeated root in any extension field.

For a finite extension $K/F$, the automorphism group $G={\mathrm{Aut}}_F(K)$ has order dividing $[K:F]$. Indeed, Artin's theorem gives $[K:K^G]=|G|$, while $F\subseteq K^G$ and the tower law gives the divisibility.

Let $L/K$ have degree $d$ with $\gcd (d,n)=1$. If $a=x^n\in K$ for some $x\in L$, thenChoosing $r,s\in \mathbb{Z}$ with $rd+sn=1$ givesso $a$ was already an $n$th power in $K$.

An element $\alpha$ of a field extension $L/K$ is algebraic over $K$ when some nonzero polynomial in $K[X]$ vanishes at $\alpha$.

The minimal polynomial of an algebraic element $\alpha$ over $K$ is the unique monic polynomial of least positive degree in $K[X]$ that vanishes at $\alpha$. It is irreducible, and its degree equals $[K(\alpha ):K]$.

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.

Fix a $K$-basis of a finite extension $L/K$ of degree $n$. The coordinates of $1,\alpha ,\dots ,{\alpha }^{n-1}$ are polynomial functions of the coordinates of $\alpha$, because multiplication in $L$ has fixed structure constants. The determinant $D(\alpha )$ of those coordinate columns is therefore a polynomial, andThus the primitive-element locus is the distinguished Zariski-open set $D(D)\subseteq {\mathbb{A}}_K^n$, possibly empty for a non-simple extension.

Every finite separable extension $L/K$ is generated by one element: $L=K(\alpha )$ for some $\alpha \in L$.

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.

A finite separable extension is simple by the primitive element theorem. A simple separable extension is finite because its generator is separable and hence algebraic. Finite and simple does not imply separable: in characteristic $p$, ${\mathbb{F}}_p(t^{1/p})/{\mathbb{F}}_p(t)$ is finite and simple but purely inseparable.

If $a,b,ab$ are nonsquares in a field $K$ of characteristic other than two, thenis a degree-four Galois extension with group $C_2\times C_2$. Its three quadratic intermediate fields are $K(\sqrt{a})$, $K(\sqrt{b})$, and $K(\sqrt{ab})$.

For distinct primes $p,q\equiv 1(\mathrm{mod}4)$, put $E=\mathbb{Q}(\sqrt{p},\sqrt{q})$ and $K=\mathbb{Q}(\sqrt{pq})$. The algebraic integers $a=(1+\sqrt{p})/2$ and $b=(1+\sqrt{q})/2$ generate an order whose basis has discriminant of elements of a number field $p^2{q}^2$: the trace pairing matrix is, up to basis order, the tensor product of the quadratic trace matrices of determinants $p$ and $q$. If its index in ${\mathcal{O}}_E$ is $j$, the discriminant-index formula for an integral lattice and the relative discriminant tower formula giveBoth $j$ and this ideal norm are positive integers, forcing $j=1$ and ${\mathfrak{d}}_{E/K}={\mathcal{O}}_K$. Thus $E/K$ is unramified at all finite primes. It is also totally real, so no real place ramifies. This quadratic abelian extension is contained in the ordinary Hilbert class field of $K$, and Artin reciprocity gives a quotient of its ideal class group of order two. The extensions $K(\sqrt{p})$ and $K(\sqrt{q})$ are equal to $E$, not two different extensions of $K$.

Up to conjugacy, the transitive subgroups of $S_4$ are $C_4$, $V_4$, $D_8$, $A_4$, and $S_4$.

If $r$ cyclically sends $\alpha \mapsto \beta \mapsto -\alpha \mapsto -\beta$ and $s$ fixes $\alpha$ while negating $\beta$, then $r^4=s^2=1$ and $srs=r^{-1}$. These automorphisms realize $D_8$ as the Galois group of the corresponding quartic splitting field.

Let $g(X)=X^4+b{X}^2+c\in \mathbb{Q}[X]$ be irreducible, let $\alpha$ be a root, and put $\delta =\sqrt{c}$. Its roots areso its splitting field is $K=\mathbb{Q}(\alpha ,\delta )$. If $\delta \notin \mathbb{Q}(\alpha )$, then $[K:\mathbb{Q}]=8$ and the Galois group is $D_8$. If $\delta \in \mathbb{Q}(\alpha )$, then $[K:\mathbb{Q}]=4$ and the group is $V_4$ when $\delta \in \mathbb{Q}$, and $C_4$ otherwise.

For irreducible $X^4+b{X}^2+c\in \mathbb{Q}[X]$, the Galois group is $C_2\times C_2$ exactly when $c$ is a square in $\mathbb{Q}$. In the transitive Klein-four action on the roots $\alpha ,-\alpha ,\beta ,-\beta$, all three nonidentity double transpositions fix $\alpha \beta =\sqrt{c}$, forcing it into the fixed field $\mathbb{Q}$.

A field homomorphism preserves addition, multiplication, and the multiplicative identity.

A field isomorphism is a bijective field homomorphism; its inverse is also a field homomorphism. For fields extending a specified field $K$, an isomorphism may be required to fix $K$ pointwise, a stronger requirement than an abstract field isomorphism.

A K-isomorphism between extensions of $K$ is a field isomorphism fixing $K$ pointwise. Simple algebraic extensions $K(\alpha )$ and $K(\beta )$ admit one sending $\alpha$ to $\beta$ exactly when their minimal polynomials agree: both then identify with the same quotient $K[X]/(m)$.

A field embedding $L\to \Omega$ is an injective field homomorphism. A $K$-embedding fixes every element of the base field $K$.

An embedding of a subfield into an algebraically closed field extends over an algebraic extension: for a simple extension, map its generator to a root of the transported minimal polynomial, then iterate or use Zorn lemma. If the larger extension is normal over the fixed base, its image under the extended embedding is itself.

Distinct field embeddings ${\sigma }_1,\dots ,{\sigma }_n:L\to \Omega$ are linearly independent over $\Omega$ as functions on $L$. Thus a nontrivial linear combination ${\sum }_i{a}_i{\sigma }_i$ cannot vanish on every element of $L$.

For the proof, choose a vanishing relation with the fewest nonzero coefficients and normalize one coefficient to one. If ${\sigma }_1\ne {\sigma }_m$, choose $y$ with ${\sigma }_1(y)\ne {\sigma }_m(y)$. Evaluating at $xy$ and subtracting ${\sigma }_m(y)$ times the relation evaluated at $x$ eliminates its $m$th term but leaves a nonzero first coefficient, contradicting minimality.

If $K/k$ is finite, then any distinct $k$-automorphisms ${\sigma }_1,\dots ,{\sigma }_m$ of $K$ are linearly independent in the $K$-vector space ${\mathrm{Hom}}_k(K,K)$, whose dimension over $K$ is $[K:k]$. Hence $m\le [K:k]$. In particular, a finite automorphism group $G$ satisfies $|G|\le [K:K^G]$.

If $\sigma$ has order $n$ and the fixed field contains a primitive $n$th root $\zeta$, thensatisfies $\sigma (T_{\zeta }(\beta ))=\zeta T_{\zeta }(\beta )$. Independence of the powers of $\sigma$ ensures that $T_{\zeta }(\beta )$ is nonzero for some $\beta$.

An embedding of $K({\alpha }_1,\dots ,{\alpha }_i)$ extends across ${\alpha }_{i+1}$ by sending it to a root of its transformed minimal polynomial. There are between one and the relative degree many choices, with equality for a separable extension.

For finite extensions $K\subseteq E\subseteq L$, the degrees multiply:If $(u_i)$ is a $K$-basis of $E$ and $(v_j)$ is an $E$-basis of $L$, then the products $(u_i{v}_j)$ form a $K$-basis of $L$: spanning follows by expanding first over $E$ and then over $K$, while $K$-linear independence follows by grouping coefficients of each $v_j$ and using both basis properties.

The splitting field of a polynomial over $K$ is the smallest field extension of $K$ in which the polynomial is a product of linear factors.

For a polynomial $f\in K[x]$ of degree $n$, its splitting field $L$ has $[L:K]\mid n!$, including in positive characteristic when the polynomial is inseparable. Prove this by induction on the degree. If $f$ is irreducible, adjoining a root gives degree $n$ and leaves a polynomial of degree $n-1$ to split, so the tower law for field extensions gives the divisibility. If $f=gh$ with positive degrees $r,s<n$, split $g$ first and then $h$ over the resulting field. The degree divides $r!s!$, which divides $(r+s)!$. This separate reducible case avoids the invalid step of assuming that the degree of an arbitrarily chosen irreducible factor must divide $n$.

A root of an irreducible polynomial in an extension field defines a base-field embedding of the polynomial quotient by evaluation. Conversely the image of the residue class of the variable determines the embedding and is a root. This counts embeddings through separable minimal polynomials.

Every nonconstant polynomial $f\in K[X]$ has a splitting field of finite degree over $K$. It is constructed by adjoining a root of an irreducible factor and repeating over the enlarged field. Any two splitting fields of $f$ over $K$ are isomorphic by an isomorphism that fixes $K$. Inside one fixed algebraic closure, the field generated by all roots is unique as a subfield.

If $\alpha =\sqrt[3]{2}$ is the positive real root and ${\zeta }_3=e^{2\pi i/3}$, then the roots of $X^3-2$ areTheir splitting field over $\mathbb{Q}$ is $\mathbb{Q}(\alpha ,{\zeta }_3)$.

A finite extension $L/K$ is normal exactly when it is the splitting field over $K$ of some polynomial. For the forward direction, multiply the minimal polynomials of finitely many generators of $L$; normality puts all their roots in $L$. The reverse direction follows because a $K$-embedding permutes the roots of a polynomial and hence preserves its splitting field.

Let $g\in K[T]$ in characteristic $p$, let $f(T)=g(T^p)$, let $L$ split $g$ over $K$, and let $M$ split $f$ over $L$. If $y_i$ are the roots of $g$ and $x_i^p=y_i$, then the Frobenius endomorphism makes each $x_i$ unique andbecause $y_i=x_i^p$. Thus $M$ is already the splitting field of $f$ over $K$, and every root of $f$ is purely inseparable over $L$.

Let $f\in {\mathbb{F}}_q[X]$ have irreducible factors of degrees $d_1,\dots ,d_r$. A root of a degree-$d_i$ factor lies in ${\mathbb{F}}_{q^n}$ exactly when $d_i$ divides $n$. Consequently the splitting field of $f$ is

An algebraic extension $L/K$ is normal when every irreducible polynomial over $K$ with a root in $L$ splits in $L$. For finite extensions, equivalently every $K$-embedding into an algebraic closure has image $L$.

An algebraic extension is separable when every element has a minimal polynomial with distinct roots.

An algebraic element $x$ over $K$ is separable when its minimal polynomial over $K$ has no repeated root in a splitting field. For an irreducible polynomial $m_x$, this is equivalent to $m_x^{\prime }\ne 0$, because then $\gcd (m_x,m_x^{\prime })=1$.

If $\alpha$ and $\beta$ are separable algebraic elements, put $K=F(\alpha )$ and $L=K(\beta )$. The minimal polynomial of $\beta$ over $K$ divides its separable polynomial over $F$, and is therefore separable. The extension count for field embeddings gives $[K:F]$ embeddings of $K$ into an algebraic closure, each with $[L:K]$ extensions to $L$. Hence $L$ has $[L:F]$ embeddings. For $E=F(\alpha +\beta )$, each embedding of $E$ has at most $[L:E]$ extensions, so there are at least $[E:F]$ embeddings of $E$ and hence exactly that number. Since $E$ is a simple field extension, its minimal polynomial has $[E:F]$ distinct roots. Thus $\alpha +\beta$ is a separable algebraic element.

For an algebraic element $\alpha$ over a field $F$ of characteristic $p>0$, suppose first that $\alpha$ is a separable algebraic element. Its minimal polynomial over $E=F({\alpha }^p)$ divides its original separable polynomial and also $T^p-{\alpha }^p=(T-\alpha {)}^p$. It is therefore separable with only one distinct root, so has degree one and $\alpha \in E$. Conversely, if the irreducible minimal polynomial $m$ is inseparable, its formal derivative vanishes and $m(T)=g(T^p)$ for some $g$ of degree $\mathrm{deg}(m)/p$. Thus $[F({\alpha }^p):F]\le \mathrm{deg}(m)/p<[F(\alpha ):F]$, proving the converse.

An algebraic extension in characteristic $p>0$ is purely inseparable when every one of its elements is purely inseparable over the base field.

An algebraic element $x$ is purely inseparable over $K$ when $x^{p^n}\in K$ for some $n\ge 0$.

An element is purely inseparable over $K$ exactly when its minimal polynomial has the formTo prove the nontrivial direction, repeatedly factor a zero formal derivative through $T\mapsto T^p$ until the remaining irreducible polynomial has nonzero derivative. Since $T^{p^r}-x^{p^r}$ has only the root $x$, that remaining separable polynomial must be linear.

Over $K={\mathbb{F}}_p(t)$, the polynomial $X^p-t$ is irreducible by the Eisenstein criterion in ${\mathbb{F}}_p[t][X]$. In any extension containing a root $\alpha$, characteristic $p$ givesIt therefore has exactly one distinct root, of multiplicity $p$.

If $M/L$ is purely inseparable, an embedding of $L$ into an algebraic closure has at most one extension to $M$: if $x^{p^n}\in L$, the image of $x$ must be the unique $p^n$th root of the image of $x^{p^n}$. Existence in a normal overfield therefore implies uniqueness.

A field automorphism is a bijective ring homomorphism from a field to itself. The automorphisms fixing a specified subfield form a Galois group when the extension is Galois.

An automorphism of a field extension $K/F$ is a field automorphism $\sigma :K\to K$ fixing every element of $F$. Such maps form the field automorphism group ${\mathrm{Aut}}_F(K)$ under composition. For a Finite Galois extension this is the Galois group $\mathrm{Gal}(K/F)$. Specifying the fixed base field distinguishes these maps from arbitrary field automorphisms of $K$.

The field automorphism group fixing $K$ consists of all field automorphisms of $L$ fixing $K$ pointwise, with composition. It is defined without assuming that $L/K$ is Galois. For $L=K(\alpha )$ with algebraic degree $d$, the action on the roots of the minimal polynomial inside $L$ is faithful, so this group embeds in $S_d$ and is finite.

A finite extension $L/K$ is Galois when it has $[L:K]$ automorphisms fixing $K$.

For a prime $p=n^2+1$, this real number field is a Galois extension of degree four with cyclic Galois group. If $\theta =\sqrt{p+\sqrt{p}}$, its minimal polynomial of an algebraic element is $X^4-2p{X}^2+p^2-p$, irreducible by Eisenstein's criterion at $p$. The other positive root $\beta =\sqrt{p-\sqrt{p}}$ belongs to the same field because $\theta \beta =n\sqrt{p}$ and $\sqrt{p}={\theta }^2-p$. Thus all four roots lie in the field, proving normality. The Galois automorphism $\theta \mapsto \beta$ sends $\sqrt{p}$ to its negative and consequently sends $\beta$ to $-\theta$. It has order four, so the group is $C_4$, rather than a product of two order-two groups.

The prime condition is used for irreducibility, while the square condition on $p-1$ supplies the missing conjugate inside the field. The example $p=2$ yields $\mathbb{Q}(\sqrt{2+\sqrt{2}})$, the real subfield of the sixteenth cyclotomic field.

For a finite Galois extension $L/K$ with group $G$, there is $\theta \in L$ such that the elements $g(\theta )$ form a $K$-basis. Equivalently the additive $K[G]$-module $L$ is a regular representation. Over p-adic fields, its p-adic lattices are thus commensurable with regular lattices; this is useful in Herbrand quotient calculations after applying a p-adic logarithm to deep principal units.

The Galois group $\mathrm{Gal}(L/K)$ consists of the field automorphisms of $L$ that fix $K$ pointwise. For a finite Galois extension, its order is $[L:K]$.