Proving a Sequence of Extension Fields for $\sqrt{1+\sqrt{2}+\sqrt{3}+\sqrt{5}}$

saadsarfraz
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Homework Statement



Find a sequence of extension fields (i.e. tower)
Q= F_{0}\subseteq...\subseteqF_{n}.

where \sqrt{1+\sqrt{2}+\sqrt{3}+\sqrt{5}} \in F_{n}

Prove that all the steps are non-trivial. except the last one. btw Q is the set of rational number. and 0 and n on F were meant to be subscripts not superscripts (i don't know how to do that)

Homework Equations





The Attempt at a Solution



I'm a bit confused as to what to do in this question? I don't think I understand the question.\sqrt{}
 
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The "trivial" step is F_1= Q(\sqrt{1}) since \sqrt{1}= 1 which already is a rational number. Take F_2= F_1(\sqrt{2})= Q_(\sqrt{2}), F_3= F_2(\sqrt{3}), etc.
 
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ok i think i got it, can anyone please check my answer

K_0 = 2 which corresponds to F_0
K_1= 1 + \sqrt{2} for F_1
K_2= 1 + \sqrt{2} + \sqrt{3} for F_2
K_3= 1 + \sqrt{2} + \sqrt{3} + \sqrt{5} for F_3
K_4= \sqrt{1+\sqrt{2}+\sqrt{3}+\sqrt{5}} for F_4except the last one is supposed to go on forever? can anyone help me in this.
 
Your original question did not "go on forever", it stopped at \sqrt{5}.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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