Saturday, 23 April 2016

Exercise about field extensions




Consider a1,,anZ.



i) Suppose a1,,an are pairwise relatively prime. I have to see by induction on n that [Q(a1,,an):Q]=2n




Once I proved the equality is true for n=1, I suppose it is true for n1, so let's prove it for n:
Applying the tower law:
[Q(a1,,an):Q]=[Q(a1,,an):Q(a1,,an1)][Q(a1,,an1):Q]



By induction, [Q(a1,,an1):Q]=2n1



So we only have to see that [Q(a1,,an):Q(a1,...,an1)]=2



[Q(a1,,an):Q(a1,,an1)]=[Q(a1,,an1)(an):Q(a1,,an1)]=deg(Irr(an,Q(a1,,an))




Irr(an,Q(a1,,an1)=X2an??



I have to see that anQ(a1,an1). By contradiction,



ifanQ(a1,,an1)an=a+ban1 where a,bQ(a1,,an1. How can I get to a contradiction???



ii) Consider P the set of prime numbers and F an extension of Q: F=Q(p,pP) Which is the degree of F/Q? Is it finitely generated?



Could you help me with this problem please?




Thank you for your time and help.


Answer



I think that the most elegant proof of this question, which is a consequence of the fact that



Square roots of different square free positive integers are linearly independent over Q,



can be found in:



http://www.thehcmr.org/issue2_1/mfp.pdf



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