The Primitive Element Theorem states that if $E/F$ is a finite separable field extension, then there exists an element $a$ such that $E=F(a)$.
There's a similar result I found, that I don't quite fully understand. For instance, let $F$ be a field and $F[a_1,\dots,a_n]$ be a finite separable extension. Suppose also that $F_u=F(u_1,\dots,u_n)$ is a purely transcendental extension of $F$, with $u_1,\dots,u_n$ algebraically independent over $F$. Why is it true that $F_u[a_1,\dots,a_n]=F_u[u_1a_1+\cdots+u_na_n]$?
I get that $u_1a_1+\cdots+u_na_n\in F_u[a_1,\dots,a_n]$, and so $F_u[u_1a_1+\cdots+u_na_n]\subseteq F_u[a_1,\dots,a_n]$. However, why is the converse true? I tried reproducing the argument for the primitive element theorem without success. Thank you.
Since the $u_i$ are algebraically independent over $F$, and $F_u[a_1,\dots,a_n]$ is finitely generated over $F_u$, I was trying to use the corollary Bill Cook pointed me to, to conclude that by setting $y_1=\sum_{j=1}^n u_ja_j$, then $F_u[a_1,\dots,a_n]$ is integral over $F_u[y_1]$. From this can I conclude the equality? I wary of how to proceed, as I do not know where separability comes into use.