Finite separable extension, algebraically independent elements I was trying to work on this exercise from Lang:

Let $k$ be a field and $k(x_1, \dots, x_n) = k(x)$ 
  a finite separable extension. Let $u_1, \dots, u_n$ be algebraically 
  independent over $k.$ Let 
  $$
w = u_1x_1 + \cdots + u_nx_n.
$$
  Let $k_u = k(u_1, \dots, u_n)$. Show that $k_u(w) = k_u(x)$.

Obviously $k_u(w) \subset k_u(x)$, since $w$ is a linear combination of $x_i$, but it remains to see the reverse inclusion. Fixing an algebraic closure of $k(x)$, I can show that $k_u$ and $K$ (normal closure of $k(x)$) are linearly disjoint with respect to $k$, and free thus: $[k_u(x) : k(x)] = [k(x) : k]$. But I'm not sure if this fact is helpful/where to go from here. Any help would be appreciated!
 A: I will do things in slightly different way from what you said. The proof is inspired by the well-known proof that finite separable extensions are simple.
Fix our discussion in an algebraic closure $K^\mathrm{a}$ of $K=k_u(x)$. Since $k$ is algebraically closed in $k_u$, any irreducible polynomial over $k$ remains irreducible over $k_u$ (lemma 4.10 in the book). In particular, the minimal polynomial of $x$ over $k$ is also its minimal polynomial over $k_u$. We get that $[k_u(x):k_u]=[k(x):k]$. Denote this degree by $d$ and let $\sigma_1, \dots, \sigma_d$ be the distinct $k_u$-embeddings of $k_u(x)$ in $K^\mathrm{a}$, sending $x$ to all of its conjugates, then $\sigma_1|_{k(x)}, \dots, \sigma_d|_{k(x)}$ induce the distinct $k$-embeddings of $k(x)$ in $K^\mathrm{a}$.
Since $k$ is algebraically closed in $k_u$ and $k_u$ is separable over $k$, $k_u$ is linearly disjoint from $k^\mathrm{a}$ over $k$ (REG 1 and REG 2 in section 4 of the book). Now consider
$$\sigma_i(w) = u_1\sigma_i(x_1) + \cdots + u_n\sigma_i(x_n),
\quad i = 1,\dots,d.$$
We claim that they are distinct. In fact if $\sigma_i(w) = \sigma_j(w)$, then
$$u_1(\sigma_i(x_1)-\sigma_j(x_1)) + \cdots + u_n(\sigma_i(x_n)-\sigma_j(x_n)) = 0.$$
Because $k_u$ is linearly disjoint from $k^\mathrm{a}$ over $k$, we get that $\sigma_i(x_1) - \sigma_j(x_1) = \dots = \sigma_i(x_n) - \sigma_j(x_n) = 0$. Therefore $\sigma_i|_{k(x)} = \sigma_j|_{k(x)}$ and we conclude that $i = j$. Now $w$ has at least $d$ conjugates over $k_u$, so the degree $[k_u(w):k_u]$ is at least $d$, implying $k_u(w) = k_u(x)$.
