Exercise 12.2.8

Proof. Write $\chi :\mathrm{Gal}(\mathit{KL}∕K)\to \mathrm{Gal}(L∕F)$ the injective homomorphism (12.26) defined by $\sigma \mapsto \sigma {\text{|}}_L$.

Let ${\alpha }_1,\dots ,{\alpha }_n$ be a numbering of the roots of $f$, and $\phi :\mathrm{Gal}(L∕F)\to S_n,$ the isomorphism defined for every $\tau \in \mathrm{Gal}(L∕F)$ by

Similarly, since $\mathit{KL}$ is the splitting field over $K$ of the same polynomial $f$, the isomorphism $\psi :\mathrm{Gal}(\mathit{KL}∕K)\to S_n$ is defined for every $\sigma \in \mathrm{Gal}(\mathit{KL}∕K)$ by

If $\tau =\sigma {\text{|}}_L$, and $\tilde{\tau }=\phi (\tau ),\tilde{\sigma }=\psi (\sigma )$, then for all $i$, $\tau ({\alpha }_i)=(\sigma {\text{|}}_L)({\alpha }_i)=\sigma ({\alpha }_i)$, therefore

Since the roots ${\alpha }_1,\dots ,{\alpha }_n$ are distinct and ${\alpha }_{\tilde{\tau }(i)}={\alpha }_{\tilde{\sigma }(i)}$ for every $i$, then $\tilde{\tau }=\tilde{\sigma }$, so

for every $\sigma \in \mathrm{Gal}(\mathit{KL}∕K)$. Hence $\psi =\phi \circ \chi$.

As a conclusion, $\tau \in \mathrm{Gal}(\mathit{KL}∕K)$ and $\tau {\text{|}}_L\in \mathrm{Gal}(L∕F)$ map to the same element of $S_n$ under (12.28) and (12.29). □