Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 3 / 18I (Galois Theory)

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.

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.

For either ${\alpha }_j$, the same permutation-of-factors argument shows that every coefficient of $f(t,{\alpha }_j)$ lies in $L^G$.

They need not generate the whole fixed field. Assume $\mathrm{char}K\ne 2$, choose $a_1,a_2\in K$ with independent nonsquare classes, and putLet $G=\{1,\sigma \}$, whereThenso all their coefficients generate only $K$. However $\sqrt{a_1{a}_2}$ is fixed by $\sigma$, and the degree count in part (a) showsThus separate orbit polynomials can miss mixed invariants.