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Assume that $F$ be a field and $f(x)$ be an irreducible polynomial in $F[x]$. Suppose that $\alpha, \beta$ be two distinct root of $f(x)$. So, there exists an isomorphism $\sigma$ from $F(\alpha)$ to $F(\beta)$. Consider that $h(x)$ be an irreducible polynomial in $F(\alpha)[x]$. So, $\sigma(h(x))$ is an irreducible polynomial in $F(\beta)[x]$. Assume that $a$ be a root of $h(x)$ and $g(x)$ be the minimal polynomial of $a$ over $F$. Is it true that the minimal polynomial of any root of $\sigma(h(x))$ over $F$ is $g(x)$?

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Yes. Consider the following steps.

  1. $h(x)$ is a factor of $g(x)$ in the ring $F(\alpha)[x]$.
  2. Therefore $\sigma(h(x))$ is a factor of $\sigma(g(x))$ in the ring $F(\beta)[x]$.
  3. But $\sigma(g(x))=g(x)$ because $g(x)\in F[x]$.
  4. Therefore any zero of $\sigma(h(x))$ is also a zero of $g(x)$.
  5. But as the minimal polynomial of some element $g(x)$ is irreducible in $F[x]$. Therefore it is automatically the minimal polynomial (over $F$) of any of its zeros in whichever extension field of $F$ that zero happens to lie.
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