“…Using this k, Theorem 5.1 implies there exists t satisfying 2) and such that t = nmod(p i -l)(p 2 -l) 9 t^n. There exist positive integers m, m' such that mp i + m'p 2 = 2 n+i for some positive integer n. Since m is even if and only if m' is even, we can divide through by a suitable power of 2 to assume m, m' are both odd.…”
Section: Resultsmentioning
confidence: 98%
“…Then since ζ e E' and K'/E' is cyclic of degree /, we have K' = E'(u ill ) 9 Suppose F p = F for all primes p. Then F is a perfect field and all finite extensions £ of F are not Euclidean since F 2 = F implies -l e F 2 g E 2 .…”
Section: Proposition 2 2 Lei E/f Be a Finite Algebraic Extensionmentioning
confidence: 99%
“…Since /=±(2fc / -k) 9 it follows (fc, /) appears in the tree. Set k' = -(k±l) where again the sign is chosen so that fe±/ = 2mod4.…”
Section: A) Every Odd Integer Occurs In the Treementioning
“…Using this k, Theorem 5.1 implies there exists t satisfying 2) and such that t = nmod(p i -l)(p 2 -l) 9 t^n. There exist positive integers m, m' such that mp i + m'p 2 = 2 n+i for some positive integer n. Since m is even if and only if m' is even, we can divide through by a suitable power of 2 to assume m, m' are both odd.…”
Section: Resultsmentioning
confidence: 98%
“…Then since ζ e E' and K'/E' is cyclic of degree /, we have K' = E'(u ill ) 9 Suppose F p = F for all primes p. Then F is a perfect field and all finite extensions £ of F are not Euclidean since F 2 = F implies -l e F 2 g E 2 .…”
Section: Proposition 2 2 Lei E/f Be a Finite Algebraic Extensionmentioning
confidence: 99%
“…Since /=±(2fc / -k) 9 it follows (fc, /) appears in the tree. Set k' = -(k±l) where again the sign is chosen so that fe±/ = 2mod4.…”
Section: A) Every Odd Integer Occurs In the Treementioning
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