Consider a1,…,an∈Z.
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 n−1, so let's prove it for n:
Applying the tower law:
[Q(√a1,…,√an):Q]=[Q(√a1,…,√an):Q(√a1,…,√an−1)][Q(√a1,…,√an−1):Q]
By induction, [Q(√a1,…,√an−1):Q]=2n−1
So we only have to see that [Q(√a1,…,√an):Q(√a1,...,√an−1)]=2
[Q(√a1,…,√an):Q(√a1,…,√an−1)]=[Q(√a1,…,√an−1)(an):Q(√a1,…,√an−1)]=deg(Irr(√an,Q(√a1,…,√an))
Irr(√an,Q(√a1,…,√an−1)=X2−an??
I have to see that √an∉Q(√a1…,√an−1). By contradiction,
if√an∈Q(√a1,…,√an−1)→√an=a+b√an−1 where a,b∈Q(√a1,…,√an−1. How can I get to a contradiction???
ii) Consider P the set of prime numbers and F an extension of Q: F=Q(√p,p∈P) 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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