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Thread: field extension, gcd

  1. #1
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    field extension, gcd

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

    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 $\displaystyle n$ such that $\displaystyle \text{gcd}(m, n)=1$, $\displaystyle F(\alpha)=F(\alpha^n)$. Thanks in advance.
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  2. #2
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    Quote Originally Posted by eskimo343 View Post
    Suppose $\displaystyle K/F$ is a field extension of degree $\displaystyle m$ and that $\displaystyle \alpha \in K$. Prove that for any integer $\displaystyle n$ such that $\displaystyle \text{gcd}(m, n)=1$, $\displaystyle F(\alpha)=F(\alpha^n).$

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