tag:blogger.com,1999:blog-12535639.post4074186638951146247..comments2024-02-26T06:55:41.876-08:00Comments on Fermat's Last Theorem: Abel's Lemmas on IrreducibilityLarry Freemanhttp://www.blogger.com/profile/06906614246430481533noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-12535639.post-91957283259517152082010-08-17T18:08:19.244-07:002010-08-17T18:08:19.244-07:00In Abel's irreducibility theorem step 3 it is ...In Abel's irreducibility theorem step 3 it is asserted that there exists a common divisor of at least the first order but is that so obvious without some explanation. One way to show this is to assume there is no common divisor and then inspect the equation of step 1 evaluated at r. The RHS is g0 while the LHS is 0. Since we know that g0 is nonzero our assumption must be wrong and a common divisor must exist.david foleyhttps://www.blogger.com/profile/13390635629510293100noreply@blogger.comtag:blogger.com,1999:blog-12535639.post-39801429112230941842010-08-16T17:32:10.252-07:002010-08-16T17:32:10.252-07:00In Abel's lemma step 6 you use an a, b, and c ...In Abel's lemma step 6 you use an a, b, and c step to show C = r^p. From step 3 we know that r is by definition a solution of x^p = C so that r^p = C. It was brought into existence to have the very property that when raised to the pth power it will equal C.david foleyhttps://www.blogger.com/profile/13390635629510293100noreply@blogger.com