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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 $[L:L^G]=|G|$.

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 $\mathrm{Gal}(L/L^G)$.