Exercise 6.3.11 (Galois group property)

Proof. The map $𝜃$ associate to a permutation $\sigma \in S_k$ is (well) defined by

• We show that $𝜃$ is surjective. Let $\beta$ be any element of $\mathrm{SF}(K)$. By Corollary 5.1.14, there is some polynomial $q\in K[t]$ such that

  $$\beta =q({\alpha }_1,\dots ,{\alpha }_k).$$

  Define

  $$p(t_1,\dots ,t_k)=q(t_{{\sigma }^{-1}(1)},\dots ,t_{{\sigma }^{-1}(k)})\in K[t_1,\dots ,t_k],$$

  and $\alpha =p({\alpha }_1,\dots ,{\alpha }_k)=q({\alpha }_{{\sigma }^{-1}(1)},\dots ,{\alpha }_{{\sigma }^{-1}(k)})\in \mathrm{SF}(K)$. Then

  $$\begin{array}{llll}\hfill 𝜃(\alpha )& =p({\alpha }_{\sigma (1)},\dots ,{\alpha }_{\sigma (k)})& \hfill & \\ \hfill & =q({\alpha }_{\sigma ({\sigma }^{-1})(1)},\dots ,{\alpha }_{\sigma ({\sigma }^{-1})(k)})& \hfill & \\ \hfill & =q({\alpha }_1,\dots ,{\alpha }_k)& \hfill & \\ \hfill & =\beta .& \hfill & \end{array}$$

  This shows that $𝜃$ is surjective.

• We verify that $𝜃$ is a homomorphism of fields.

  Let $\alpha =p({\alpha }_1,\dots ,{\alpha }_k)\in \mathrm{SF}(K)$, and $\beta =q({\alpha }_1,\dots ,{\alpha }_k)$.

  $$\begin{array}{llll}\hfill 𝜃(\alpha +\beta )& =𝜃(p({\alpha }_1,\dots ,{\alpha }_k)+q({\alpha }_1,\dots ,{\alpha }_k))& \hfill & \\ \hfill & =𝜃((p+q)({\alpha }_1,\dots ,{\alpha }_k))& \hfill & \\ \hfill & =(p+q)({\alpha }_{\sigma (1)},\dots ,{\alpha }_{\sigma (k)})& \hfill & \\ \hfill & =p({\alpha }_{\sigma (1)},\dots ,{\alpha }_{\sigma (k)})+q({\alpha }_{\sigma (1)},\dots ,{\alpha }_{\sigma (k)})& \hfill & \\ \hfill & =𝜃(p)+𝜃(q).& \hfill & \end{array}$$

  Similarly, replacing $\text{+}$ by $\text{×}$, we obtain

  $$𝜃(\mathit{\alpha \beta })=𝜃(\alpha )𝜃(\beta ).$$

  Moreover, $1=p({\alpha }_1,\dots ,{\alpha }_k)$, where $p$ is the constant polynomial equal to $1$. Then $𝜃(1)=p({\alpha }_{\sigma (1)},\dots ,{\alpha }_{\sigma (k)})=1$.

  So $𝜃:\mathrm{SF}(K)\to \mathrm{SF}(K)$ is a field homomorphism.

We can conclude as in the text that $𝜃{\text{∈}}_K(f)$. □