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Unformatted text preview: f elds k and k with k ∼ = Q and k ∼ = F p for some prime p . Solution. Suppose such sub f elds k and k exist, and consider their intersection k = k ∩ k . The one element 1 lies in k ; since k ≤ k , we have 0 = p · 1 = 1 + 1 + ··· + 1, where there are p summands equal to 1. On the other hand, in k , we have p · 1 ±= 0. This is a contradiction. (ii) Prove that a f eld K cannot have sub f elds k and k with k ∼ = F p and k ∼ = F q , where p ±= q . Solution. The argument in part (i) also works here: if p < q , then p · 1 = 0 in k = k ∩ k , because k ≤ k , but p · 1 ±= 0 in k because k ≤ k ....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.
 Fall '11
 KeithCornell

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