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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 $t^n āˆ’ s^n$.

[Hint: first show that $s$ satisfies a poly of degree $n$ over $K(s^n)$; this gives $\leq$. Then show that $\{1, s, \ldots, s^{nāˆ’1}\}$ is Linearly Independent over $K(s^n)$; this gives $\geq$.]

I would really like some help with this please!

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    What's wrong with the hint you were given? Looks like it does pretty much all the work. – 2011-03-07

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Shwoing that $[K(s):K(s^n)] \leq n$ is easy - in fact you have already written down a polynomial of $s$ over $K(s^n)$, which is $t^n - s^n$.

To show the other direction of the inequality, it suffices to show that $\{1,s, \cdots, s^{n-1}\}$ is linearly independent over $K(s^n)$. Write down a linear dependence relation. Clearing denominators, we can assume that there is a relation $\sum_{i=0}^{n-1} a_i s^i = 0$, with $a_i \in K[s^n]$. Now read off the degree of each term. Can you deduce that $a_i = 0$ for all $i$?