cubicrecip

cubicrecip - 2 5 -1 3 -4-3 5 3 Using cubic reciprocity on...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 = ±
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 5 -1 3 -4-3 5 3 Using cubic reciprocity on the latter symbol composed of two PRIMARY primes, this is = 2 5 -1 3 5-4-3 3 Now we must reduce 5 mod-4-3 . This is: 5-4-3 = 5 (-4-3 )-4-3 (-4-3 ) =-20-15 2 13 Since an element of Z [ ] close to this fraction is-2- 2 =-1 + , we compute 5-(-4-3 )(-2- 2 ) =-2-2 = 2 2 and hence 3- 5 3 = 2 5 -1 3 2 2-4-3 3 = 2 5 -1 3 2-4-3 3 -4-3 2 3 = 5 -2 3 -4-3 2 3 - 2 3 = ( 2 )-2 ( ) 2 ( ) = 2 so 3- is not a cubic residue mod 5....
View Full Document

Ask a homework question - tutors are online