tag:blogger.com,1999:blog-5195188167565410449.post7447389483205642313..comments2014-03-09T17:09:45.698ZComments on Haskell for Maths: Extension fieldsDavidAhttp://www.blogger.com/profile/16359932006803389458noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5195188167565410449.post-49852924528972031312010-03-14T18:14:40.242Z2010-03-14T18:14:40.242ZThe claim is generally known as "Primitive El...The claim is generally known as "Primitive Element Theorem". It has a half-way constructive proof.<br /><br />http://en.wikipedia.org/wiki/Primitive_element_theoremRoberthttp://www.blogger.com/profile/10644674458756425098noreply@blogger.comtag:blogger.com,1999:blog-5195188167565410449.post-24531283113093299922009-09-16T21:46:03.290+01:002009-09-16T21:46:03.290+01:00Shawn,
Unfortunately I can't find my referenc...Shawn,<br /><br />Unfortunately I can't find my reference for this claim just now. If I could, the proof would probably be constructive. In the case of Q(sqrt2, sqrt3), Q(sqrt2 + sqrt3) does the trick. If you work through that example, it might all become clear.<br /><br />(Clearly sqrt2 + sqrt3 is in Q(sqrt2, sqrt3). We have to show the other way round, that sqrt2, sqrt3 are in Q(sqrt2 + sqrt3). Well, consider (sqrt2 + sqrt3)^2, ^3, etc.)DavidAhttp://www.blogger.com/profile/16359932006803389458noreply@blogger.comtag:blogger.com,1999:blog-5195188167565410449.post-18075570636924495282009-09-16T12:39:57.281+01:002009-09-16T12:39:57.281+01:00Hi,
Luckily, this doesn't matter, as any algeb...Hi,<br />Luckily, this doesn't matter, as any algebraic extension Q(a,b) is equal to Q(c) for some c.<br /><br />How does one compute such a c?<br /><br /><a href="http://www.zoombits.fr/bluetooth/" rel="nofollow">casque bluetooth</a>Shawnhttp://www.blogger.com/profile/15206346233530013399noreply@blogger.com