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Automorphisms $\sigma (z)={\zeta }_{n}z$ and $\tau (z)=1/z$ generate a dihedral group of order $2n$. The invariant $w=z^n+z^{-n}$ has rational map degree $2n$, so $[\mathbb{C}(z):\mathbb{C}(w)]=2n$. The fixed-field theorem for a finite automorphism group gives the same degree over its fixed field, proving equality. For odd prime $n$, the nontrivial proper subgroups are the rotation group and the $n$ reflection groups. Their fixed fields are $\mathbb{C}(z^n)$ and $\mathbb{C}(z+{\zeta }_n^{-j}/z)$ for reflections ${\sigma }^j\tau$, respectively.

Let $t$ be transcendental over a field $k$ and let $\tau$ be the field automorphism sending $t$ to $1/t$. The rational function $u=t+1/t$ is fixed, and $t$ satisfies $X^2-uX+1=0$. Thus the extension over $k(u)$ has degree at most two. It is not one because the nontrivial automorphism $\tau$ fixes $k(u)$. The fixed field of $\tau$ also has index two by the fixed-field theorem, so it equals $k(u)$.

No nonconstant polynomial in $k[t]$ is fixed by inversion. For a polynomial of positive degree, substitution of $1/t$ creates negative powers, whereas the original has positive powers. Uniqueness of the coefficients of a Laurent polynomial rules out equality. Consequently this fixed field cannot have a generator that is a polynomial in the original variable, even though it has the simple rational-function generator above. The statement is valid in characteristic two as well: $t$ and $1/t$ remain distinct transcendental rational functions.

For $m\ge 3$, put $\eta ={\zeta }_m+{\zeta }_m^{-1}$. Since $\eta$ is real and ${\zeta }_m$ is nonreal but satisfies $T^2-\eta T+1=0$, the cyclotomic field $\mathbb{Q}({\zeta }_m)$ has degree two over $\mathbb{Q}(\eta )$. Complex conjugation fixes $\eta$ and is nontrivial. The Fundamental theorem of Galois theory therefore identifies its fixed field with $\mathbb{Q}(\eta )$. That fixed field is exactly $\mathbb{Q}({\zeta }_m)\cap \mathbb{R}$, proving maximality among real subfields.

For $\alpha \in L$ and a finite automorphism group $G$, the polynomialhas coefficients in the fixed field $L^G$, because every element of $G$ permutes its factors.

Let $L/K$ be a finite Galois extension of number fields, with Galois group $G$, and let $\mathfrak{p}$ be a nonzero prime ideal of ${\mathcal{O}}_K$. Write its prime ideal factorization asThe Galois group acts transitively on the primes above $\mathfrak{p}$. One proof uses the Chinese remainder theorem. If $\mathfrak{Q}$ were outside the orbit of $\mathfrak{P}$, choose an integral element $a$ which is zero modulo $\mathfrak{Q}$ and one modulo every conjugate of $\mathfrak{P}$. Its field norm ${\prod }_{\sigma \in G}\sigma (a)$ belongs to $\mathfrak{Q}\cap {\mathcal{O}}_K=\mathfrak{p}$, hence to $\mathfrak{P}$. But each factor is one modulo $\mathfrak{P}$, a contradiction. Transitivity also makes all $e_j$ equal to a common ramification index $e$ and all $f_j$ equal to a common residue degree $f$.

Fix $\mathfrak{P}$ above $\mathfrak{p}$. Its decomposition group is the stabilizerThe orbit-stabilizer theorem gives $|D_{\mathfrak{P}}|=|G|/g$. Elements of $D_{\mathfrak{P}}$ act on the residue field $\kappa (\mathfrak{P})$ over $\kappa (\mathfrak{p})$; the group kernel is the inertia groupThus inertia measures the automorphisms invisible after reduction, whereas decomposition measures all automorphisms preserving the chosen prime.

The completion $L_{\mathfrak{P}}/K_{\mathfrak{p}}$ is a Galois extension with group $D_{\mathfrak{P}}$: preservation of $\mathfrak{P}$ lets an automorphism extend continuously to the completion, and the local embeddings of this normal extension arise in this way. The degree of a field extension formula is $[L_{\mathfrak{P}}:K_{\mathfrak{p}}]=ef$. Reduction gives the exact sequenceFor surjectivity, lift a primitive residue-field generator by the Hensel lemma to obtain the maximal unramified subextension of a local field extension of $L_{\mathfrak{P}}/K_{\mathfrak{p}}$. Its automorphisms are the residue-field automorphisms. They extend to $L_{\mathfrak{P}}$ because the latter is normal. Since a finite residue field extension has a cyclic Galois group of order $f$, the exact sequence yieldsThe fixed field of local inertia is the maximal unramified subextension of a local field extension, and the extension above it is totally ramified. Primes in the same global orbit have conjugate decomposition groups and inertia groups, so the orders do not depend on the chosen prime.

If $e=1$, inertia is trivial and $D_{\mathfrak{P}}$ is generated by the unique Frobenius automorphism acting on residues by $z\mapsto z^{N\mathfrak{p}}$. Its order is $f$. In a ramified extension this residue action determines only a coset modulo inertia. A prime splits completely exactly when $e=f=1$, equivalently its decomposition group is trivial. A totally ramified prime has $g=f=1$ and $I_{\mathfrak{P}}=D_{\mathfrak{P}}=G$; an unramified prime has $I_{\mathfrak{P}}=1$. In an abelian extension, conjugation does nothing, so the associated groups and unramified Artin symbol are independent of the prime above $\mathfrak{p}$.

Take $u\in L\setminus K$, whose irreducible polynomial is $t^2+ct+d$. Since the characteristic is not two, $x=2u+c$ satisfies $x^2=c^2-4d=a\in K$ and $L=K(x)$. Irreducibility makes $a$ nonzero and nonsquare. If another quadratic extension is $K(y)$ with $y^2=b$, an isomorphism sends $x$ to $s+ty$, with $t\ne 0$. Squaring gives $a=s^2+t^{2}b+2sty$. Linear independence of $1,y$ forces $s=0$, so $a=t^{2}b$ and $b/a$ is a square. Conversely, if $b/a=q^2$, sending $x$ to $y/q$ gives an isomorphism. If $b$ itself is a square, the target is just $K$ and cannot be isomorphic; the square-ratio condition also fails.

The multiquadratic field $F$ is the splitting field of ${\prod }_i(t^2-a_i)$, a product of separable quadratic polynomials: each has distinct roots because $a_i\ne 0$ and the characteristic is not two. Even if factors repeat, this splitting field is a separable normal extension, and therefore Galois. Each automorphism sends $x_i$ to $\pm x_i$, and these signs determine it, embedding its Galois group into $(\mathbb{Z}/2{)}^n$. Every subgroup of this elementary abelian group is $(\mathbb{Z}/2{)}^m$ for some $m\le n$.

For a quadratic intermediate field, its fixing subgroup $H$ has index two and is the kernel of a nontrivial one-dimensional character $\chi :G\to \{\pm 1\}$. The coordinate sign characters span the dual of $G$: a linear functional on the subspace $G\subseteq {\mathbb{F}}_2^n$ extends to the ambient space by extending a basis. Hence $\chi ={\prod }_{i\in I}{\chi }_i$ for some subset $I$. Set $y={\prod }_{i\in I}{x}_i$. Then $\sigma (y)=\chi (\sigma )y$, so $y$ is fixed by $H$ but not by all of $G$, and $y^2={\prod }_{i\in I}{a}_i\in K$. Thus $K(y)$ is quadratic and equals the fixed field of $H$. This constructs all quadratic subfields of a multiquadratic extension:

The Artin fixed-field theorem says that for a finite group $G$ of distinct automorphisms of a field $K$, the extension $K/K^G$ is Galois extension and has degree $|G|$. Here $y=x^p-x$ is invariant under translations because $(x+a{)}^p-(x+a)=x^p-x$ for $a\in {\mathbb{F}}_p$. Hence ${\mathbb{F}}_p(y)\subset K^G$. Also $x$ is a root of $T^p-T-y$, so $[K:{\mathbb{F}}_p(y)]\le p$. The fixed-field theorem gives $[K:K^G]=p$, forcing $\boxed{K^G={\mathbb{F}}_p(y)}$. The extension is separable; explicitly the derivative of $T^p-T-y$ is $-1$.

For the affine substitutions, $y$ transforms to $dy$, since $d^p=d$ and $a^p=a$. Therefore $z=y^{p-1}$ is fixed by all $p(p-1)$ substitutions. The polynomial $(T^p-T{)}^{p-1}-z$ has $x$ as a root and degree $p(p-1)$. Apply the same degree comparison and the Artin fixed-field theorem to obtainMoreover $[{\mathbb{F}}_p(y):{\mathbb{F}}_p(z)]=p-1$, so the minimal polynomial of $y$ is $\boxed{T^{p-1}-z}$. It splits as ${\prod }_{d\in {\mathbb{F}}_p^{\times }}(T-dy)$, and its derivative is nonzero at every root. Thus $K^G/K^H$ is a Galois extension with group ${\mathbb{F}}_p^{\times }$, acting by $y\mapsto dy$.

There is a harmless convention issue in labeling the substitutions: $f(x)\mapsto f(dx+a)$ composes in the reverse order to the displayed affine matrices. It is naturally a right action on functions, or a left action after inserting inverses. The set of automorphisms and all the fixed fields just computed are unchanged by that convention.

The Fundamental theorem of Galois theory applies to a finite Galois extension $L/K$ with group $G$. It gives inclusion-reversing inverse bijectionsThe fixed field $L^H$ consists of elements fixed by every member of $H$. Moreover $[L:L^H]=|H|$ and $[L^H:K]=[G:H]$. An intermediate extension $M/K$ is Galois iff $H=\mathrm{Gal}(L/M)$ is normal, in which case restriction gives $\mathrm{Gal}(M/K)\cong G/H$. All intermediate $L/M$ are themselves Galois. Finiteness and normality/separability of $L/K$ are essential for this stated form.

The cyclotomic polynomial ${\Phi }_7=1+X+\cdots +X^6$ is irreducible: replacing $X$ by $X+1$ gives an Eisenstein polynomial at seven. Hence the extension has degree six, andThere is one subgroup of each order $1,2,3,6$, and therefore precisely four intermediate fields. The trivial subgroup fixes $\mathbb{Q}(\eta )$; the whole group fixes $\mathbb{Q}$. The order-two subgroup $\{1,6\}$ is complex conjugation and fixes $\mathbb{Q}(\eta +{\eta }^{-1})$, of degree three: $\eta$ satisfies $X^2-(\eta +{\eta }^{-1})X+1$ and is not real.

The order-three subgroup $\{1,2,4\}$ fixes $u=\eta +{\eta }^2+{\eta }^4$. Its other conjugate is $v={\eta }^3+{\eta }^5+{\eta }^6$. Expanding products and using ${\sum }_{j=1}^6{\eta }^j=-1$ gives $u+v=-1$ and $uv=2$. Thus $u^2+u+2=0$, and this fixed field is $\mathbb{Q}(u)=\mathbb{Q}(\sqrt{-7})$, of degree two. These generators have the required degrees and are fixed by the indicated subgroups, so the Galois correspondence proves they are the full fixed fields.

Every $h\in G$ permutes the factors:Thus all coefficients lie in the fixed field $L^G$. Let $E$ be the field they generate over $K$. ThenBecause $L=K(\alpha )=E(\alpha )$ and $f(t,\alpha )\in E[t]$ has degree $|G|$ with root $\alpha$,Part (a) and the tower law for field extensions giveConsequently $[L^G:E]=1$, soThis is the coefficients of an orbit polynomial for a primitive generator result.

View every $k$-automorphism ${\sigma }_i:K\to K$ as an element of the $K$-vector spacewhere $K$ acts by multiplying output values. If $n=[K:k]$ and $e_1,\dots ,e_n$ is a $k$-basis of $K$, a $k$-linear map is determined freely by its $n$ values on that basis. ThusPart (a) says that the $m$ distinct automorphisms are linearly independent vectors in this space, so

Now let $k=K^G$, the fixed field of the finite automorphism group $G$. Its elements are fixed by every member of $G$, so the members of $G$ are distinct $K^G$-automorphisms of $K$. Applying the inequality gives the automorphism-count bound for a finite field extension

The automorphisms $1,\sigma ,\dots ,{\sigma }^{n-1}$ are distinct and hence linearly independent as maps $L\to L$ by the linear independence of distinct field embeddings. Therefore the operatoris not the zero map. Choose $\beta \in L$ for which $\alpha =T(\beta )\ne 0$. Reindexing the sum and using ${\sigma }^n=1$ givesThis is the Lagrange resolvent eigenvector for a cyclic field automorphism.

Let $F=L^{\sigma }$ be the fixed field. Since $\sigma ({\alpha }^n)={\alpha }^n$, the element $\alpha$ is a root ofIts orbit under $⟨\sigma ⟩$ isand these $n$ elements are distinct because $\alpha \ne 0$ and $\zeta$ is a primitive root of unity. Hence the minimal polynomial has at least $n$ distinct roots. It also divides the displayed degree-$n$ polynomial, so

Every automorphism fixes zero and one, and preserves sums, products, negatives and multiplicative inverses of nonzero elements. Consequently elements fixed by every member of $G$ are closed under these operations and form a field. Artin fixed-field theorem gives $\boxed{[L:L^G]=|G|}$ for a finite automorphism group.

Here $\mathbb{Q}(t)$ is the fraction field of $\mathbb{Q}[t]$, with $t$ transcendental; its elements are fractions of rational-coefficient polynomials, with nonzero denominator and the usual equality by cross-multiplication. Substitution of $1-t$ or $1/t$ defines an automorphism, each its own inverse. With $\sigma \tau$ meaning first apply $\tau$ to a field element and then $\sigma$, its image on $t$ is $1/(1-t)$. Iterating gives $(t-1)/t$ and then $t$, so its order is three. The relations reduce every word to at most six possibilities, and the six resulting images areThey are distinct rational functions, so the group has six elements. Its faithful permutation action on $\{0,1,\infty \}$ identifies it with $\boxed{S_3}$.

For the first fixed field put $u=t(1-t)$. It is fixed by $\sigma$, and $t$ satisfies $X^2-X+u=0$ over $\mathbb{Q}(u)$. The degree is exactly two because the nontrivial automorphism $\sigma$ fixes that field. The fixed-field degree is also two, soFor the inversion put $v=t+1/t$. It is fixed and $t$ satisfies $X^2-vX+1=0$. The same degree argument givesThe source requests polynomial generators in both cases. The second requirement is impossible. If a nonconstant polynomial $p(t)=a_d{t}^d+\cdots +a_0$ were fixed by inversion, $p(t)=p(1/t)$ would equate a Laurent polynomial with highest exponent $d>0$ to one with no positive exponent, a contradiction. Thus the only polynomial invariants of inversion are constants, which cannot generate its nonconstant fixed field. The fixed field of inversion in a rational function field therefore has the second displayed rational-function generator.

For a polynomial with $n$ distinct roots, the criterion isWrite $\delta ={\prod }_{i<j}({\alpha }_i-{\alpha }_j)\ne 0$. Part (a) gives $\sigma \delta =\mathrm{sgn}(\sigma )\delta$. If every Galois permutation is even, $\delta$ is fixed by the entire Galois group and therefore belongs to its fixed field $\mathbb{Q}$. Thus ${\Delta }_f={\delta }^2$ is a rational square. Conversely, if ${\Delta }_f=q^2$ with $q\in {\mathbb{Q}}^{\times }$, then ${\delta }^2=q^2$ in a field implies $\delta =\pm q$. Every automorphism fixes this nonzero element, so its permutation sign must be one.

The distinct-root hypothesis is necessary in the statement's intended degree-$n$ permutation action. The source says “arbitrary degree” without stating that hypothesis. If repeated roots were allowed, zero discriminant would carry no sign information: $(X-1{)}^2(X^2-2)$ has discriminant zero, while its nontrivial automorphism swaps $\sqrt{2},-\sqrt{2}$ and fixes the two rational-root positions, an odd permutation. The criterion above applies to the squarefree polynomial, or to the squarefree part with its degree and root action adjusted accordingly.

The fixed field isIt is a subfield because every field automorphism preserves sums, products and inverses.

Set $F=K^G$. If $[K:F]$ is infinite the desired finite lower bound is immediate. Otherwise let $b_1,\dots ,b_d$ be an $F$-basis of $K$. If $|G|>d$, the homogeneous system over $K$has a nonzero solution. For $a={\sum }_j{a}_j{b}_j$, with $a_j\in F$, every $\sigma$ fixes $a_j$, so the same coefficients give ${\sum }_{\sigma }{c}_{\sigma }\sigma (a)=0$. This contradicts linear independence of distinct field embeddings. Consequently