Why is $\mathbb F_5(\root{15}\of t)$ not normal over $\mathbb F_5(t)$?
(I'm asking this question in order to understand this answer).
My idea for a proof so far:
$f:=X^{15}-t\in F_5(t)$ is the minimal polynomial of $\root{15}\of t$ in $F_5(t)$, so it suffices to show that not all roots of $f$ are elements of $F_5(\root{15}\of t)$.
Let $L_1$ and $L_2$ be roots of $f$ in $F_5(\root{15}\of t)$. Now $(\frac{L_1}{L_2})^{15}=1$, so $\frac{L_1}{L_2}$ is a 15th root of unity and therefore, because we work in characteristic $5$, a 3rd root of unity.
Frome here on, I'd like to argue that $1$ is the only 3rd root of unity in $F_5(\root{15}\of t)$, therefore $L_1=L_2$, so $f$ has only one root in $F_5(\root{15}\of t)$ but $3$ roots in an algebraic closure, so $F_5(\root{15}\of t)$ is not normal over $F_5(t)$.
Now my questions are: Does this proof work as outlined above? If yes, how can I justify that $1$ is the only 3rd root of unity in $F_5(\root{15}\of t)$? If not, how else can I prove that $F_5(\root{15}\of t)$ is not normal over $F_5(t)$? Thanks for your help.
Yes, the method of proof works. It is easy to see that the only 3rd root of unity in $\Bbb{F}_5$ is $1$. To complete the proof, all you need is the following fact:
Claim: Let $F$ be a field. Then, the only elements of $F(t)$ that are algebraic over $F$ are those in $F$.
I will sketch the proof. Each element $\alpha\in F(t)$ is of the form $\alpha=f(t)+\frac{g(t)}{h(t)}$ for $f(X),g(X),h(X)\in F[X]$, with $\deg g<\deg h$ or $g=0$. Let $p(X)\in F[X]$, $\deg p>0$. If $g\neq 0$, expand $p(\alpha)$ and clear denominators to show that $p(\alpha)\neq 0$. If $g=0$ and $p(\alpha)=0$, then $f$ is constant by considering the coefficient of the highest power of $t$.