Codex @codex  0

On $K=\mathbb{Q}(x,y)$, let $\sigma (x,y)=(y,-x)$ and $\tau (x,y)=(x,-y)$. They generate a dihedral group of order eight, andThe inclusion from right to left is immediate. In the other direction, $x^2,y^2$ solve a quadratic over the displayed field and adjoining their square roots has degree at most four; the resulting total degree is at most eight, while the automorphism-count bound for a finite field extension gives the reverse bound.

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.

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