Is the extension $\mathbb{F}(t^2) \subset \mathbb{F}(t)$ algebraic?

Question

The question is the same appear in the title. if we have a field $\mathbb{F}$ and we are interested to investigate the possibility of rational function $\mathbb{F}(t^2)$ to be extended over $\mathbb{F}(t)$ to be algebraic. and more generally. what about the case:- $$\mathbb{F}(t^n) \subset \mathbb{F}(t)$$

Answer 1

Yes it's an algebraic extension. A polynomial satisfied by $t$ over $\mathbb{F}(t^n)$ is $X^n - t^n$.

Answer 2

A finitely generated field extension $K(a_1, \ldots, a_n)/K$ is algebraic if and only if the generators $a_i$ are algebraic over $K$.

Taking $K=\mathbb{F}(t^n)$, we see that $\mathbb{F}(t)=\mathbb{F}(t^n)(t)$ is algebraic over $\mathbb{F}(t^n)$ since $t$ is a root of $X^n-t^n \in \mathbb{F}(t^n)[X]$.

$\textit{Sketch of a proof of the above statement}$: Let $L/K$ be a field extension. Note that an element $a \in L$ is algebraic over $K$ if and only if $[K[a]:K]< \infty$, where $[K[a]:K]$ denotes the dimension of $K[a]$ as a $K$-vector space (Try to prove this yourself if you haven't yet). Then do some linear algebra and prove that $[L:k]=[L:K][K:k]$ if $k$ is a subfield of $K$.
Now, let $L:=K(a_1, \ldots, a_n)$ and suppose that the generators are algebraic over $K$. Then, $a_i$ is algebraic over $K(a_1, \ldots, a_{i-1})$ for all $i \geq 2$. It follows that $$ [L:K]=(\prod\limits_{i=2}^{n}[K(a_1, \ldots, a_i):K(a_1, \ldots, a_{i-1})])[K(a_1):K]< \infty.$$