Let $k$ be a field and $k(x_1,x_2,...,x_n)= k(x)$ a finite separable extension. Let $u_1,u_2,..., u_n$ be algebraically independent over $k$. Let $w= u_1x_1 + u_2 x_2 +\cdots +u_n x_n .$ Let $k_u = k(u_1,u_2,..., u_n).$ Show that $k_u (w)= k_u (x)$.
Please give some suggestion on how to do this problem. Thanks in advance.
It should be pretty clear that $k_u(x) = k_u(x_1, \cdots, x_n)$, using the fact that each set can be generated from the other using polynomials with coefficients in $k$.
The assumption that $k(x_1, \cdots, x_n) = k(x)$ is a finite separable extension implies the same for $k_u(x_1, \cdots, x_n) = k_u(x)$, which allows you to pass to a normal closure $L$ of the extension $k_u(x)/k_u$, so $L/k_u$ is Galois. Now $k_u(x)$ and $k_u(w)$ are two intermediate extensions of $L/k_u$; can you see a way of solving the problem from this approach?