Show that $[k(t): k(t^4 + t) ] = 4$ Let $k$ be the field with $4$ elements, $t$ a transcendental over $k$, $F = k(t^4 + t)$ and $K = k(t).$ Show that $[K : F] = 4.$
I think I have to use the following theorem, but I'm not quite putting it together. 
If $P = P(t), Q = Q(t)$ are nonzero relatively prime polynomials in $F[t]$ which are not both constant, then $[F(t) : F(P/Q)] = $ max(deg $P$, deg $Q$).

Denote $s=t^4 + t$.  Now $t$ is the root of the polynomial $p=x^4 + x -s$. To show $p$ irreducible over $k(s)[x]$, it's enough to show that it's irreducible over $k[s][x]$, since $k[s]$ is a polynomial ring. Equivalently, $p$ irreducible over $k[x][s]$. But this is clear, being a monic polynomial of degree $1$.
The general theorem is proved in the same way.
