Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 18F (Galois Theory)

Suppose otherwise and choose a relationwith the least possible number $m$ of nonzero coefficients. After relabelling and scaling, take every ${\lambda }_i\ne 0$ and ${\lambda }_m=1$. Since the field embeddings ${\sigma }_1$ and ${\sigma }_m$ are distinct, there is a $y\in K$ such that ${\sigma }_1(y)\ne {\sigma }_m(y)$.

Apply the relation to $xy$ and subtract ${\sigma }_m(y)$ times the relation at $x$. Because each field embedding is multiplicative, this givesThe $m$th term has disappeared, while the coefficient of ${\sigma }_1$ is nonzero. This is a shorter nontrivial relation, contradicting minimality. Hence distinct embeddings are linearly independent as functions. This is the Artin independence 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

The actions on the ordered pair $(x,y)$ satisfyThus every element of $G$ has the form ${\sigma }^i$ or $\tau {\sigma }^i$ for $0\le i<4$, so $|G|\le 8$. Their actions areBecause $x,y$ are algebraically independent elements over $\mathbb{Q}$, these eight substitutions are distinct. Henceand the presentation identifies $G$ with the dihedral group $D_8$.

PutBoth generators $\sigma$ and $\tau$ fix $u$ and $v$, so for $F=\mathbb{Q}(u,v)$,The elements $x^2$ and $y^2$ are the two roots ofso $[F(x^2,y^2):F]\le 2$. Adjoining first $x$ and then $y$ takes at most two quadratic extensions, and the tower law givesPart (b) gives $8=|G|\le [K:K^G]$. Since $F\subseteq K^G\subseteq K$, another use of the tower law givesAll inequalities are equalities and $[K^G:F]=1$. Thereforethe dihedral fixed field of a two-variable rational function field.