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

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 $$s \neq t$$?

• Equivalently with $x$ transcendental then $x \not \in F(x^m)$ for $m > 1$. – reuns Aug 17 at 4:23
• What is the form of the elements in $F (\alpha^m)$? – xbh Aug 17 at 4:25
• @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? – mathdoge Aug 17 at 4:54
• @mathdoge Same as what Robert Shore says. $F(\alpha) \cong$ the fraction field of $F[\alpha]$. – xbh Aug 17 at 5:00

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 $$F$$.
• 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? – mathdoge Aug 17 at 4:52
• 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. – Robert Shore Aug 17 at 4:58