Some more funky facts about minimal

polynomials:

Let F ⊆ E be fields, and α ∈ E. We denote F[α] to be the smallest subring of E containing both F and α, and F(α) to be the smallest subfield containing both F and α. Let n(x) ∈ F[x] be the minimal polynomial of α over F.

Obviously F(α) ⊆ F[α], since all subfields are subrings. Now F(α) = F[α] iff α is algebraic over F (i.e. n(x) exists).

Also, F(α) ≅ F[x] / n(x)F[x].