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Let $K/F$ be a field extension. If $\alpha \in F(\alpha^m)$, $m > 1$, then $\alpha$ is algebraic in $F$.

A proof provided in the book is as follows:

Proof: Since $\alpha \in F(\alpha^m)$, there exists $f$ and $g$ in $F[x]$ such that $\alpha = \frac{f(a^m)}{g(a^m)}$. Therefore, $\alpha$ is the solution to the polynomial $h(x) := xg(x^m) - f(x^m)$. Suppose $\deg(f(x)) = s, \deg(g(x))=t$. Then $\deg(f(x^m)) = sm$ and $\deg(g(x^m)) = tm$. Since $m>1$, $ms \neq mt + 1$. Therefore, $h \neq 0$, showing that $\alpha \in F(\alpha^m)$.

My question: Why we have those $f$ and $g$ satisfies $a = \frac{f(a^m)}{g(a^m)}$? Does it imply that $s \neq t$?

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  • $\begingroup$ Equivalently with $x$ transcendental then $x \not \in F(x^m)$ for $m > 1$. $\endgroup$ – reuns Aug 17 at 4:23
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    $\begingroup$ What is the form of the elements in $F (\alpha^m)$? $\endgroup$ – xbh Aug 17 at 4:25
  • $\begingroup$ @xbh I am not quite sure, I just thought that $F(\alpha^m)$ is just the smallest subfield of $K$ joining $\alpha^m$. Could you further explain? $\endgroup$ – mathdoge Aug 17 at 4:54
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    $\begingroup$ @mathdoge Same as what Robert Shore says. $F(\alpha) \cong$ the fraction field of $F[\alpha]$. $\endgroup$ – xbh Aug 17 at 5:00
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The definition of $F(\alpha^m)$ is the set of quotients of polynomials in $\alpha^m$. We’re choosing two specific polynomials, $f$ and $g$, so that their quotient witnesses that $\alpha \in F(\alpha^m)$. That means $\alpha=\frac{f(\alpha^m)}{g(\alpha^m)}$. Your question uses $a$ here instead of $\alpha$. Perhaps that’s the source of your confusion.

It’s possible that $s=t$; that won’t affect the proof, which demonstrates that there’s a non-zero polynomial $h(x)=xg(x^m)-f(x^m)$ with the property $h(\alpha)=0$. The existence of that polynomial means (by the definition of algebraic) that $\alpha$ is algebraic over $F$.

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  • $\begingroup$ Thank you for your answer! I indeed made a typo ($\alpha$ and $a$). However I am still not sure, I thought $F(\alpha^m)$ is just the smallest subfields of $K$ containing $\alpha^m$, and $K$ is algebraic over $F$ says that we could find a polynomial, say $p(x) \in F[x]$ such that $p(\alpha^m) = 0$. Could you further explain your first paragraph? $\endgroup$ – mathdoge Aug 17 at 4:52
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    $\begingroup$ The smallest subfield containing $F$ and $\alpha^m$ has to contain all $F$-polynomials in $\alpha^m$ and all fractions of such polynomials. If you haven’t already done so, you should prove that the set of such fractions (with non-zero denominators) is in fact a field, which must therefore be the smallest such field. $\endgroup$ – Robert Shore Aug 17 at 4:58

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