Unformatted text preview: 2 x + 1 = 10 or10. Thus, x = 58 or x = 48. GK 1.5 We are trying to solve the congruence x 2 â‰¡ 1 (mod 5 k ) for k = 1 , 2 , Â·Â·Â· , 6. For k = 1, one solutions is x â‰¡ 2 (mod 5). Lift this to a solution for k = 2 using Henselâ€™s lemma. Continue this process ï¬ve times. We have x = ( ... 431212) 5 or x = ( ... 013233) 5 . GK 1.6 Let t âˆˆ F be such that t 2 =1. We claim that 1 + ti âˆˆ F [ i ] is not invertible, and thus F [ i ] is not a ï¬eld. In fact, suppose that (1 + ti )( a + bi ) = 1 for some a,b âˆˆ F . Then ( abt ) + ( at + b ) i = 1 and thus abt = 1 , at + b = 0 . This gives t = t ( abt ) = atbt 2 = at + b = 0 . This is a contradiction. 1...
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 Fall '08
 Shapiro,I
 Math, Number Theory, 5k, Hensel

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