field extension, gcd

**eskimo343** · February 12th 2010, 10:18 AM

Suppose $K/F$ is a field extension of degree $m$ and that $\alpha \in K$ . Prove that for any integer $n$ such that $\text{gcd}(m, n)=1$ , $F(\alpha)=F(\alpha^n).$

I was thinking initially in this problem to use the tower lemma but nothing seemed to work out after that. I do not know how to show that for any integer $n$ such that $\text{gcd}(m, n)=1$ , $F(\alpha)=F(\alpha^n)$ . Thanks in advance.

**aliceinwonderland** · February 12th 2010, 01:31 PM

Originally Posted by eskimo343

Suppose $K/F$ is a field extension of degree $m$ and that $\alpha \in K$ . Prove that for any integer $n$ such that $\text{gcd}(m, n)=1$ , $F(\alpha)=F(\alpha^n).$

I was thinking initially in this problem to use the tower lemma but nothing seemed to work out after that. I do not know how to show that for any integer $n$ such that $\text{gcd}(m, n)=1$ , $F(\alpha)=F(\alpha^n)$ . Thanks in advance.

This is the sketch of the proof. I'll assume $K=F(\alpha)$ where $[F(\alpha):F]=m$ . Since the degree of extension is finite, it is an algebraic extension. Then, $1, \alpha, \alpha^2, \cdots, \alpha^m$ is a basis for K as a vector space over F. If gcd(m,n)=1, then you see that $F(\alpha^n)$ has the same basis set with $1, \alpha, \alpha^2, \cdots, \alpha^m$ because the field generated by $\alpha$ over F and the field generated by $\alpha^n$ over F is the same.

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