The first claim follows from the fact that the Galois group of Fpn/Fp is generated by the Frobenius map x↦xp. Thus, from Galois theory, the subfield of Fpn fixed by G(Fpn/Fp) is exactly Fp. For your second statement, recall that Tr(yp)=Tr(y), now apply Tr on both sides of the equation.
The equation yp−y=x in finite fields.
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manpreet![Tuteehub forum best answer]()
Best Answer
2 years ago
Define the trace by
Now define yet another mapping:
I know that this is a linear mapping. My syllabus stated the following I could't prove: