Unformatted text preview: a ). Suppose 2 = αβ in Z [ √ 13 ] . Then 4 = N ( α ) N ( β ) . Since N ( α ) , N ( β ) 6 = ± 2, we see that either N ( α ) = ± 1 or N ( β ) = ± 1. We deduce that α or β is a unit. Also N ( 2 ) = 4 6 = ± 1, so N ( 2 ) is not a unit. Finally 2 6 = 0 and we conclude that 2 is irreducible. Now 2 * 6 = (1 + √ 13 )( 1 + √ 13 ) . Thus if 2 is prime, we have 2 divides1 + √ 13 or 1 + √ 13. This is not the case, because the resulting quotients would be1 / 2 + √ 13 / 2 and 1 / 2 + √ 13 / 2, neither of which are in Z [ √ 13 ] . The result follows....
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This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.
 Spring '08
 Staff
 Math, Algebra

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