# Thread: galois theory

1. ## galois theory

Can someone please help me with this question finding it very difficult

Thanks Edgar

Let K be a 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 satisifies a polynomial of degree n over K(s^n); this gives <=. Then show that {1,s,...,s^(n-1)} is linearly independent over K(s^n); this gives you >=.

2. Originally Posted by edgar davids
Can someone please help me with this question finding it very difficult

Thanks Edgar

Let K be a 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 satisifies a polynomial of degree n over K(s^n); this gives <=. Then show that {1,s,...,s^(n-1)} is linearly independent over K(s^n); this gives you >=.
Let $f(x) = x^n - s^n \in K(s^n)$ then clearly $f(s) = 0$ so if $p(x)$ is the minimal polynomial for $s$ over $K(s^n)$ we have that $p(x) | f(x) \implies k= \deg p(x) \leq \deg f(x) = n \implies k\leq n$. Now that means that $\{ 1, s,s^2, ... ,s^{k-1} \}$ is a basis for $K(s)$ over $K(s^n)$. If we can show that $\{1,s,s^2 , ... s^{n-1} \}$ is linearly independent then it would means $k\geq n$ because the dimension of a linearly independent set cannot the dimension of a basis. To show that $\{1,s,s^2,...,s^{n-1} \}$ is linearly independent over $K(s^n)$ it is equivalent to showing that no element in this set can be expressed as a linear combination of the other elements. Take for example, $1$, we cannot express $1$ as a linear combination $a_1 s+a_2s^n + ... + a_{n-1}s^{n-1}$ where $a_i \in K(s^n)$ because $a_i = b^{(i)}_0 + b^{(i)}_1 s^n + b^{(i)}_2 s^{2n} + ...$ so it is impossible to get intermediately exponents between $1$ and $n$. Thus, $k=n$ and this means $p(x) = x^n - s^n$.

This is Mine 87th Post!!!