cubicrecip

# cubicrecip - ω 2 5 ²-1 3 ±-4-3 ω 5 ² 3 Using cubic...

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18.781, Fall 2007 Performing Cubic Reciprocity In this note, we expand our in-class discussion of solutions to the congru- ence: x 3 (3 - ω ) (mod 5) Recall that we are working in the ring Z [ ω ] / 5 Z [ ω ] for this congruence, where ω = - 1+ i 3 2 . Then it suﬃces to compute: ± 3 - ω 5 ² 3 , the cubic residue symbol mod 5, which is 1 if and only if 3 - ω is a cubic residue mod 5. We’d like to apply cubic reciprocity to this symbol, but unfor- tunately, the statement of cubic reciprocity we gave requires both primes to be “primary” – that is, both primes must be congruent to 2 mod 3. (Recall that we know 3 - ω is prime, as it has norm 13.) So we seek a unit u ∈ {± 1 , ± ω, ± ω 2 } with u (3 - ω ) 2 (3) . Taking u = ω 2 , we get ω 2 (3 - ω ) = - 1 + 3 ω 2 = - 1 - 3 - 3 ω = - 4 - 3 ω 2 (3) So now, ± 3 - ω 5 ² 3 = ±
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Unformatted text preview: ω 2 5 ²-1 3 ±-4-3 ω 5 ² 3 Using cubic reciprocity on the latter symbol composed of two PRIMARY primes, this is = ± ω 2 5 ²-1 3 ± 5-4-3 ω ² 3 Now we must reduce 5 mod-4-3 ω . This is: 5-4-3 ω = 5 (-4-3 ω )-4-3 ω (-4-3 ω ) =-20-15 ω 2 13 Since an element of Z [ ω ] close to this fraction is-2-ω 2 =-1 + ω , we compute 5-(-4-3 ω )(-2-ω 2 ) =-2-2 ω = 2 ω 2 and hence ± 3-ω 5 ² 3 = ± ω 2 5 ²-1 3 ± 2 ω 2-4-3 ω ² 3 = ± ω 2 5 ²-1 3 ± ω 2-4-3 ω ² 3 ±-4-3 ω 2 ² 3 = ³ ω 5 ´-2 3 ± ω-4-3 ω ² 2 3 ±-ω 2 ² 3 = ( ω 2 )-2 ( ω ) 2 ( ω ) = ω 2 so 3-ω 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.

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