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If $K\subseteq L^G$ and $L=K(\alpha )$, then the coefficients of $f(t,\alpha )$ generate $L^G$ over $K$. Indeed, for the field $E$ generated by those coefficients, $L=E(\alpha )$ and $[L:E]\le |G|$, while the Artin fixed-field theorem gives $[L:L^G]=|G|$.

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.

Put $F=L^G$ and $n=|G|$. The automorphism-count bound for a finite field extension givesFor the reverse inequality, let $x_1,\dots ,x_m$ be linearly independent over $F$. We claim that the $n\times m$ matrixhas linearly independent columns over $L$. If not, choose a nonzero relationwith the fewest nonzero coefficients, and normalize one coefficient to $c_1=1$. Applying $h\in G$ and reindexing the rows gives another relation with coefficients $h(c_j)$. Subtracting eliminates the first term, so minimality forces $h(c_j)=c_j$ for every $h$ and $j$. Thus every $c_j\in F$. The row for the identity automorphism then contradicts the $F$-linear independence of the $x_j$.

Therefore $m\le n$. Taking an $F$-basis of $L$ gives $[L:F]\le n$, and henceThis proves the degree assertion in the Artin fixed-field theorem.

If $\delta \notin \mathbb{Q}(\alpha )$, thenLetThis is the nonidentity automorphism of $K/\mathbb{Q}(\alpha )$. By extending an embedding that sends $\alpha$ to $\beta$, and composing with $s$ if necessary, chooseThenThe eight elementsare distinct, and exhaust the Galois group. Thus the dihedral Galois action on four radical roots gives

Now suppose $c$ is a square in $\mathbb{Q}$. Then $\delta \in \mathbb{Q}$, so every automorphism fixes $\delta$. We are in case (i) with $\tau (\delta )=\delta$, and therefore

Conversely, suppose the group is the Klein four-group. Label the rootsThe three nonidentity elements in its transitive action are the three double transpositions. Each fixes $\delta$: they send the ordered pair $(\alpha ,\beta )$ respectively towhose products all equal $\alpha \beta$. Hence every element of the Galois group fixes $\delta$. By the Artin fixed-field theorem,Therefore $c={\delta }^2$ is a square in $\mathbb{Q}$. This proves the Klein-four criterion for an irreducible even quartic:

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.

Artin fixed-field theorem says that for a finite group $G\le \mathrm{Aut}L$, $L/L^G$ is Galois with group $G$ and degree $|G|$. Let any finite $G$ act faithfully by left translation on variables $X_g$ in $L=\mathbb{C}(X_g:g\in G)$. Artin's theorem then realizes $G$ as $\mathrm{Gal}(L/L^G)$.

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.