may02 - a Suppose 2 = αβ in Z √ 13 Then 4 = N α N β...

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Math 4124 Monday, May 2 May 2, Ungraded Homework Prove that 2 is irreducible but not prime in Z [ 13 ] . Here Z [ 13 ] = { a + b 13 | a , b Z } . Define N : Z [ 13 ] Z by N ( a + b 13 ) = a 2 - 13 b 2 for a , b Z . Then N ( αβ ) = N ( α ) N ( β ) for α , β Z [ 13 ] ; to see this write α = a + b 13 and β = c + d 13. Then N ( αβ ) = N ( ac + 13 bd +( ad + bc ) 13 ) = ( ac + 13 bd ) 2 - 13 ( ad + bc ) 2 , N ( α ) N ( β ) = ( a 2 - 13 b 2 )( c 2 - 13 d 2 ) and ( ac + 13 bd ) 2 - 13 ( ad + bc ) 2 = ( a 2 - 13 b 2 )( c 2 - 13 d 2 ) . Since N is a multiplicative homomorphism (but not a ring homomorphism), we see that if α is a unit in Z [ 13 ] , then N ( α ) is a unit in Z , i.e. ± 1. Conversely suppose N ( a + b 13 ) = ± 1. Then ( a + b 13 )( a - b 13 ) = ± 1 and we see that a + b 13 is a unit in Z [ 13 ] . Next we note that there is no α Z [ 13 ] such that N ( α ) = ± 2. Indeed if N ( a + b 13 ) = ± 2, then a 2 - 13 b 2 = ± 2, so a 2 = ± 2 mod 13. However there is no such a (just go through the 13 possibilities for
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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 divides-1 + √ 13 or 1 + √ 13. This is not the case, because the resulting quotients would be-1 / 2 + √ 13 / 2 and 1 / 2 + √ 13 / 2, neither of which are in Z [ √ 13 ] . The result follows....
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