I am stuck on the following exercise:
Let $K$ be any field and $s$ an indeterminate. Then $K(s)$ is a field extension of $K(s^n)$. Prove that $[K(s):K(s^n)]=n$. Hence show that the minimum polynomial of $s$ over $K(s^n)$ is $t^n-s^n$.
[Hint: first show that $s$ satisfies a polynomial of degree $n$ over $K(s^n)$; this gives $\leq$. Then show that $\{1,s,\ldots,s^{n-1}\}$ is linearly independent over $K(s^n)$; this gives you $\geq$.]
I am trying to use the hint:
I let $f(t) = t^n-s^n$ hence $f(s) = 0$ which means that $[K(s):K(s^n)] \leq n$
Now I try to prove the linear independence:
Suppose $1+\sum_{i=1}^{n-1} \alpha_{i} s^i = 0$ I am trying to show all $\alpha_i = 0$. My plan was to somehow apply $f$ to that but I am having no luck.
Suppose toward a contradiction that $\{1,s,\ldots,s^{n-1}\}$ is linearly dependent over $K(s^n)$. Then there exist $\alpha_0(s^n),\ldots,\alpha_{n-1}(s^n)\in K(s^n)$, not all zero, such that $\sum_{i=0}^{n-1}\alpha_i(s^n)\cdot s^i=0$. We may write $\alpha_i(s^n)=\frac{p_i(s^n)}{q_i(s^n)}$ with $p_i(s^n),q_i(s^n)\in K[s^n]$, so that clearing out denominators in the expression $$0=\sum_{i=0}^{n-1}\alpha_i(s^n)\cdot s^i=\sum_{i=0}^{n-1}\frac{p_i(s^n)}{q_i(s^n)}\cdot s^i,$$ yields an expression $\sum_{i=0}^{n-1}r_i(s^n)\cdot s^i=0$ with $r_i(s^n)\in K[s^n]$ given by $$r_i(s^n):=p_i(s^n)\cdot\prod_{\substack{j=0\\j\neq i}}^{n-1}q_j(s^n).$$ Now consider the polynomial $f(X)\in K[X]$ given by $$f(X):=\sum_{i=0}^{n-1}r_i(X^n)\cdot X^i.$$ It is a polynomial over $K$ which has $s$ as a root, contradicting the fact that $s$ is an indeterminate. Hence $\{1,s,\ldots,s^{n-1}\}$ is linearly independent.
Exercise: Why doesn't this argument work for $\{1,s,\ldots,s^n\}$?