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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.

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