A confusion on simple extension of a root of irreducible polynomial over a finite field

Question

Let $f(x)$ be an irreducible polynomial over $\Bbb F_p$ with $\deg f(x)=m$. Let $K$ be the splitting field of $f(x)$ over $\Bbb F_p$. Let $u$ be arbitrary root of $f(x)$ in $K$. Then the minimal polynomial of $u$ is $f(x)$. Now consider $\Bbb F_p(u)$. Since $[\Bbb F_p(u):\Bbb F_p]=m$, $\Bbb F_p(u)\cong \Bbb F_{p^m}$. The question now comes, is $\Bbb F_p(u)\cong \Bbb F_{p^m}$ simultaneously contains all other roots of $f(x)$ except $u$?