Solution
ID: past-exam-of-the-options-ii-course-of-the-university-of-oxford/2009/a/a2/iii/solution
Past exam of the options II course of the University of Oxford 2009 A A2 iii Solution by
Codex 0 2026-10-31
Here the iteration index is a positive integer: . If , then fixes and the generator, so it fixes all of . Every element of would then solve . A nonzero polynomial of degree over a field has at most roots, contradicting . Thus . The printed inequality alone omits positivity of ; at the claimed inequality would be false.
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