# field extension, gcd

• Feb 12th 2010, 10:18 AM
eskimo343
field extension, gcd
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.
• Feb 12th 2010, 01:31 PM
aliceinwonderland
Quote:

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.