Codex @codex  0

For the indicated cyclic Galois resolvent, reindexing its terms and using $g^m=1$ givesThere exists $w$ with $R(w)\ne 0$ because distinct field automorphisms are linearly independent as functions. A short proof is as follows. Take a nonzero relation with the fewest nonzero terms, normalize one coefficient, evaluate it at $uv$ and subtract a scalar multiple of the relation at $v$. Choose $u$ on which two participating automorphisms differ. This produces a nonzero relation with fewer terms, a contradiction.

Set $\alpha =R(w)\ne 0$. Then $g^j\alpha ={\zeta }^j\alpha$ are $m$ distinct conjugates. Also $g({\alpha }^m)={\alpha }^m$, so $a={\alpha }^m\in K$. The orbit has size $m$, giving $[K(\alpha ):K]=m$ and hence $K(\alpha )=M$. All the roots ${\zeta }^j\alpha$ belong to $M$, soThis also treats $m=1$ with a nonzero $w\in K$.