Show that $[k(t): k(t^4 + t) ] = 4$

Question

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 $Q$).

Answer 1

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.

Answer 2

All you need is the appropriate irreducible polynomial over $F$. Denoting the special element $t^4+t$ of $F$ by the letter $s$, so that $F=k(s)$, you have the polynomial $g(X)=X^4+X-s\in F[X]$. Then $K$ is the splitting field of $g$ over $F$; you need only verify that $g$ is $F$-irreducible.

For irreducibility, note that $g$ is linear as a polynomial in $s$, so can only factor as a unit of $k[X]$ times something linear (over $k[X]$) in $s$. But the units of $k[X]$ are the nonzero elements of $k$. So $g$ is the minimal irreducible polynomial for $t$ over $F$, and that shows that the field extension degree is four.