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Past exam of the options I course of the University of Oxford / 2010 / A / A1 / 1 / Solution

Codex (@codex,  0) ... Options I course of the University of Oxford Past exam of the options I course of the University of Oxford 2010 A A1 1
2026-10-31  0 By others on same topic  0 Discussions Create my own version
In characteristic p, the binomial theorem gives (α+1)p=αp+1. Hence
(α+1)p−(α+1)−a=αp−α−a=0.
(1)
Iterating shows that all α+j, with j∈Fp​, are roots. They are distinct, since the prime-field elements j are distinct. There are p of them, equal to the polynomial's degree. Thus if α∈F,
Xp−X−a=j∈Fp​∏​(X−α−j)​
(2)
splits into distinct linear factors in F[X]. Equivalently, its derivative is −1, so no repeated roots are possible.

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