Exercise 13.1.1

Proof.

(a)

By definition of the isomorphism ${\varphi }_1:\mathrm{Gal}(L∕F)\to S_n$ in Section 6.3, if ${\tau }_1={\varphi }_1(\sigma )$, then $$\begin{array}{lll}\hfill \sigma ({\alpha }_i)={\alpha }_{{\tau }_1(i)},i=1,\dots ,n.& & \hfill \text{(1)}\end{array}$$

As ${\beta }_1,\dots ,{\beta }_n$ are the same roots in a different order, there exists a permutation $\gamma \in S_n$ such that

$$\begin{array}{lll}\hfill {\beta }_i={\alpha }_{\gamma (i)},i=1,\dots ,n.& & \hfill \text{(2)}\end{array}$$

This numbering of the roots is associate to the isomorphism ${\varphi }_2$. If ${\tau }_2={\varphi }_2(\sigma )$, then

$$\begin{array}{lll}\hfill \sigma ({\beta }_i)={\beta }_{{\tau }_2(i)},i=1,\dots ,n.& & \hfill \text{(3)}\end{array}$$

Therefore, for all $i=1,\dots ,n$, using (2), (3), and (2) again,

$$\begin{array}{lll}\hfill \sigma ({\alpha }_{\gamma (i)})& =\sigma ({\beta }_i)={\beta }_{{\tau }_2(i)}={\alpha }_{\gamma ({\tau }_2(i))}.& \hfill \text{(4)}\end{array}$$

Now, with the substitution $i\to \gamma (i)$ in (1), we get

$$\begin{array}{lll}\hfill \sigma ({\alpha }_{\gamma (i)})& ={\alpha }_{{\tau }_1(\gamma (i))}.& \hfill \text{(5)}\end{array}$$

Thus , by (4),(5), ${\alpha }_{\gamma ({\tau }_2(i))}={\alpha }_{{\tau }_1(\gamma (i))}$ for all $i$. Since $i\mapsto {\alpha }_i$ is one-to-one,

$$\gamma ({\tau }_2(i))={\tau }_1(\gamma (i)),i=1,\dots ,n,$$

so

$$\gamma {\tau }_2={\tau }_1\gamma .$$

Therefore ${\tau }_2={\gamma }^{-1}{\tau }_1\gamma$, so ${\varphi }_2(\sigma )=\hat{\gamma }({\varphi }_1(\sigma ))$, for all $\sigma \in \mathrm{Gal}(L∕F)$:

$${\varphi }_2=\hat{\gamma }\circ {\varphi }_1.$$

(b)

Let $G$ the image of ${\varphi }_1$ in $S_n$: $G=\{{\varphi }_1(\sigma )|\sigma \in \mathrm{Gal}(L∕F)\}\subset S_n$.

Similarly the image of ${\varphi }_2$ is $G^{\prime }=\{{\varphi }_2(\sigma )|\sigma \in \mathrm{Gal}(L∕F)\}\subset S_n$.

Since ${\varphi }_2(\sigma )={\gamma }^{-1}{\varphi }_1(\sigma )\gamma$ for all $\sigma \in \mathrm{Gal}(L∕F)$ by part (a),

$$G^{\prime }={\gamma }^{-1}\mathit{G\gamma }.$$

So, if we change the labels, then $G$ gets replaced with a conjugate subgroup.