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Unformatted text preview: F p [ x ] x 4 + 1 = ( x 2 + ax + 1 )( x 2 ax + 1 ) or x 4 + 1 = ( x 2 + ax 1 )( x 2 ax 1 ). (iv) Prove that x 4 + 1 factors in F p [ x ] for all primes p . Solution. In part (i), we saw that x 4 + 1 factors in F 2 [ x ] , and in part (iii), we saw that it factors in F p [ x ] for every odd prime p if certain congruences can be solved. As every odd prime (indeed, every odd integer) is congruent to 1 or 3 mod 4, Exercise 3.100 completes the proof....
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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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