I copy the lemma and then describe my confusion.
Lemma 5.28 Let $B/k$ be a splitting field of some polynomial $g(x)\in k[x]$. If $p(x) \in k[x]$ is irreducible, and if $$p(x)=q_1(x)\cdots q_t(x)$$ is the factorization of $p(x)$ into irreducibles in $F[x]$, then all the $q_i(x)$ have the same degree.
Proof. Regard $p(x)$ as a polynomial in B[x], and let $E=B(z_1,\cdots,z_n)$ be a splitting field of $p(x)$, where $z_1,\cdots,z_n$ are the roots of $p(x)$. If $p(x)$ does not factor in $B[x]$, we are done. Otherwise, choose $z_1$ to be a root of $q_1(x)$ and, for each $j\neq1$, choose $z_j$ to be a root of $q_j(x)$. Since both $z_1$ and $z_j$ are roots of the irreducible $p(x)$, Proposition 3.116 gives an isomorphism $\varphi_j:k(z_1)\rightarrow k(z_j)$ with $\varphi_j(z_1)=z_j$ which fixes $k$ pointwise. Now Proposition 5.22 says that $\varphi_j$ extends to an automorphism $\Phi_j$ of $E$, and Corollary 5.19 gives $\Phi_j(B)=B$.Hence, $\Phi_j$ induces an isomorphism $\Phi_j^{\ast}:B[x]\rightarrow B[x]$(by letting $\Phi_j$ act on the coefficients of a polynomial). It follows that $$p^{\ast}(x)=q_1^{\ast}(x)\cdots q_t^{\ast}(x)$$, where $p^\ast(x)=\Phi^\ast_j(p(x))$ and $q^\ast_i(x)=\Phi_j^\ast(q_i(x))$ for all i...
My main confusion occurs here, so I don't copy the rest part. If you want check the whole proof, please click this picture Lemma 5.28 in the textbook.
My confusion is that since $\Phi_j^\ast$ is from $B[x]$ to $B[x]$, how it can operate $q_i(x)$. $q_i(x)$ may not in $B[x]$. Actually $q_i(x)$ is in $F[x]$.
