Shouldn't it be α^18, which in GF(16) simplifies to α^3 since α^15 = 1 in GF16? Same is also confusing for the 3rd and 4th conjugate of α^7 while finding the minimal polynomial in GF (16), how they get to have α^13 and α^11? To make matter easy I am also attaching. an image for conjugacy classes different roots of GF16
2026-03-24 19:08:57.1774379337
When finding the minimal polynomials in GF(16), Why the 4th conjugate of α³ becomes α^9?
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The conjugates, apart from $\newcommand{\al}{\alpha}\al^3$ are $\al^6$, $\al^{12}$ and $\al^{24}=\al^9$. I got these by repeatedly applying the Frobenius map $x\mapsto x^2$ to $\al^3$.