Artin's fixed-field upper bound $[K:K^G]\le \mathrm{\mid }G\mathrm{\mid }$

Define the $K$-linear map $T:K^{m+1}\to K^m$ by $T(c_1,\dots ,c_{m+1})=({\sum }_j{c}_j{\sigma }_i(x_j){)}_{i=1}^m$. Since the domain has dimension $m+1$ and the codomain dimension $m$, [L1] gives a nonzero vector in $\ker T$.

Among nonzero vectors in $\ker T$, choose $c=(c_j)$ with least support and scale it so its first supported coordinate is $1$. For $\tau \in G$, applying $\tau$ to all equations $T(c)=0$ and reindexing the rows by ${\sigma }_i\mapsto \tau {\sigma }_i$ shows that $\tau (c)=(\tau (c_j))$ also lies in $\ker T$. The vector $\tau (c)-c$ has a zero in the normalized coordinate and support strictly smaller than that of $c$ unless it vanishes; minimality therefore gives $\tau (c_j)=c_j$ for every $j$ and every $\tau \in G$, so all $c_j$ lie in $K^G$.

The identity row of $T(c)=0$ is ${\sum }_j{c}_j{x}_j=0$, a nontrivial $K^G$-linear dependence among the arbitrary $m+1$ elements. Thus no $m+1$ elements of $K$ are linearly independent over $K^G$, and $[K:K^G]\le m=\mathrm{\mid }G\mathrm{\mid }$. This includes $m=1$ and also covers repeated or zero $x_j$.