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Let $x^n-b ∈ F[x]$ be an irreducible polynomial . Prove that $x^m−α^m∈F(α^m)[x]$ is the minimal polynomial of α over $F(α^m)$.

It's so simple and apparently easy yet I can't say riguously why. I know that is equivalent to prove that $x^m-α^m$ is irreducible over the field $F(α^m)$.

Jaqawa
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  • Is $\alpha$ assumed to be transcendental? A simple algebraic counter-example is $\alpha=\sqrt[3]{2}$, $m=2$. – Mor A. Apr 03 '21 at 20:29
  • Why not? take $n=3$, $b=2$, $m=2$ and you get my example. (With $F=\mathbb Q$ and $K=\mathbb R$, for example) – Mor A. Apr 03 '21 at 20:55
  • Probably we need the condition that $m$ is a divisor of $n$. – Berci Apr 03 '21 at 20:58

1 Answers1

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Note that $[F(\alpha):F]=n$. Also, from Dedekind’s Product Theorem, $$ n=[F(\alpha):F]=[F(\alpha):F(\alpha^m)][F(\alpha^m):F].\tag{1}$$

Write $n=mk$. If $g(x)$ is the minimal polynomial for $\alpha^m$ over $F$, and has degree $r$, then $g(x^m)$ is a polynomial that is satisfied by $\alpha$, and hence a multiple of $x^n-b$. That means that $n\leq\deg(g(x^m)) = m\deg(g) = mr$. Therefore, $r\geq k$. Thus, the minimal polynomial of $\alpha^m$ is of degree at least $k$. But clearly, $\alpha^m$ satisfies the polynomial $x^k - b$, so this is the minimal polynomial for $\alpha^m$ over $F$. Thus, $[F(\alpha^m):F]=k = \frac{n}{m}$.

Plugging into (1), we get $$ n = [F(\alpha):F(\alpha^m)]k,$$ so $[F(\alpha):F(\alpha^m)] = m$.

Since $\alpha$ clearly satisfies $x^m-\alpha^m$, and this is monic of the correct degree, it must be the minimal polynomial for $\alpha$ over $F(\alpha^m)$.

Arturo Magidin
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