Unformatted text preview: ω 2 5 ²1 3 ±43 ω 5 ² 3 Using cubic reciprocity on the latter symbol composed of two PRIMARY primes, this is = ± ω 2 5 ²1 3 ± 543 ω ² 3 Now we must reduce 5 mod43 ω . This is: 543 ω = 5 (43 ω )43 ω (43 ω ) =2015 ω 2 13 Since an element of Z [ ω ] close to this fraction is2ω 2 =1 + ω , we compute 5(43 ω )(2ω 2 ) =22 ω = 2 ω 2 and hence ± 3ω 5 ² 3 = ± ω 2 5 ²1 3 ± 2 ω 243 ω ² 3 = ± ω 2 5 ²1 3 ± ω 243 ω ² 3 ±43 ω 2 ² 3 = ³ ω 5 ´2 3 ± ω43 ω ² 2 3 ±ω 2 ² 3 = ( ω 2 )2 ( ω ) 2 ( ω ) = ω 2 so 3ω is not a cubic residue mod 5....
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This note was uploaded on 04/28/2009 for the course MATH 18.781 taught by Professor Brubaker during the Spring '09 term at MIT.
 Spring '09
 BRUBAKER
 Number Theory, Congruence

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