Let $K/\mathbb{Q}$ be a finitely generated field extension and let $(x_1,\cdots,x_m)$ be a transcendance basis of $K/\mathbb{Q}$. Using primitive element theorem, there exists $y\in K$ algebraic over $\mathbb{Q}(x_1,\cdots,x_m)$ such that: $$K=\mathbb{Q}(x_1,\cdots,x_m)[y].$$ Let $G$ be the minimal polynomial of $y$ in $K/\mathbb{Q}(x_1,\cdots,x_m)$, clearing the denominators, one may assume that $G\in\mathbb{Z}[x_1,\cdots,x_m][Y]$. Hence, there exists $H\in\mathbb{Z}[X_1,\cdots,X_m,Y]$ such that: $$G(Y)=H(x_1,\cdots,x_m,Y).$$ After some work, one can show that there exists a rational prime number $p$, $(\xi_1,\cdots,\xi_m)$ algebraically independent elements of $\mathbb{Q}_p/\mathbb{Q}$ and $z\in{\mathbb{Q}_p}$ such that: $$H(\xi_1,\cdots,\xi_m,z)=0.$$
Question. Why does there exist a $\mathbb{Q}$-isomorphism of $K$ such that $\left\{\begin{array}{ccc}x_i&\mapsto&\xi_i\\y&\mapsto&z\end{array}\right.$?
I know that if $L/K$ is a field extension, $f\in L[X]$ is an irreducible polynomial over $K$ and $x,y$ are two roots of $f$, then there exists a $K$-isomorphism of $L$ sending $x$ on $y$.
Any hint is greatly appreciated!
Reference: J.W.S. Cassels, An embedding theorem for field, 1976.