Let $\mathbb{F}_{81}$ be the finite field with $81$ elements, $L=Quot(\mathbb{F}_{81}[T])$ and $K=\mathbb{F}_{81}(T^{10})\subseteq L$. Show, that the extension $L/K$ is normal. Hint: How does the factorization of $X^n-a$ look like?
My (old) attempt:
I know that $\mathbb{F}_{p^n}^{\times}$ is cyclic. If we look at $\mathbb{F}_{p^n}$ and a factor $k$ of $n$, I can find subgroups $U$ with order $p^k-1$ so that \begin{align*} X^{p^k}-X=X\cdot\prod_{u\in U} (X-u) \end{align*} In my case, I have $\mathbb{F}_{3^4}$ and $k=1,2,4$
Could that be of any relevance here? I also think that the Frobenius-Automorphism might be useful. My main problem is that I cannot imagine $L$ and $K$.
I think $x\in K$ can be written as $x=\frac{a_nT^{10n}+...+a_1T^{10}+a_0}{b_mT^{10m}+...+b_1T^{10}+b_0}$ and $y\in L$ can be written as $y=\frac{a_rT^{r}+...+a_1T+a_0}{b_sT^{s}+...+b_1T+b_0}$
I'd like to have some ideas/tips/... that push me in the right direction.
My new attempt:
As discussed in the comments, it seems that it suffices that if for any polynomial $Q\in K[X]$ where $K$ doesn't split $Q$, the field extension $L$ is a splitting field of $Q$, then $L/K$ is normal.
I can look at $P(X) = X^{10}-T^{10}\in K[X]$ and this polynomial has obviously a root in $L$ (because $T\in L$). My other roots would be $\zeta_{10}^1 T,...,\zeta_{10}^{9} T$, where $\zeta_{10}$ is the primitive $10-$root of unity.
But all these roots are equal to $T$ when raised to the power of $10$, so $P(X)=X^{10}-T^{10}=(X-T)^{10}$ in $L$ and there is no smaller field extension that splits $P$ because $L=Quot(\mathbb{F}_{81}[T])$ is already the smallest field containing $T$.
Then $L/K$ is normal.
Any flaws in that proof?