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Verify that the following numbers are constructible by determining the sequence of their extension fields:

$$ \sqrt{1+\sqrt2+\sqrt3+\sqrt5}, \frac{\sqrt5+\sqrt{11}}{1+\sqrt{7-\sqrt3}},\frac{\sqrt{2+\sqrt3}}{\sqrt[3]{2}+\sqrt{1+\sqrt{2+\sqrt{5}}+\sqrt{3-\sqrt{7}}}} $$

Taking the first number, we can take the rational numbers as $F_0$, where $k_0=2$ and $k_1=\sqrt2$, where $k_1$ is not in $F_0$. Then:

$$ k_2=1+\sqrt2 \\ k_3=3+\sqrt{1+\sqrt2} \\ k_3=5+\sqrt{3+\sqrt{1+\sqrt2}} $$

which eventually gives $\sqrt{5+\sqrt{3+\sqrt{1+\sqrt2}}}$. But that doesn't match the initial number. Is it possible to take multiple $k$ values when extending a field? What steps do I need to take verify the above numbers?

hohner
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  • How about the intermediate fields $\Bbb Q\subseteq \Bbb Q(\sqrt2)\subseteq \Bbb Q(\sqrt2,\sqrt3)$? What would be the next one? – Arthur Aug 02 '15 at 11:53
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    let's take $k=\sqrt{1+\sqrt 2+\sqrt 3+\sqrt 5}$ then: $$\mathbb Q\left({\sqrt 2+\sqrt 3+\sqrt 5}\right)=\mathbb{Q}(k^2)\subseteq \mathbb{Q}(k) $$

    and the quetient is $2$ so the idea is to follow squaring and this way you will get what you want : $$ \mathbb Q \subseteq \mathbb Q \left(\sqrt {30}\right)\subseteq \mathbb Q \left(\sqrt 6,\sqrt 5\right)\subseteq \mathbb Q \left(\sqrt 2,\sqrt 3,\sqrt 5\right)=\mathbb Q\left({\sqrt 2+\sqrt 3+\sqrt 5}\right)\subseteq \mathbb{Q}(k) $$ and the degree of over $\mathbb{Q}(k)$ over $\mathbb{Q}$ is $16=2^4$

     – Elaqqad Aug 02 '15 at 12:03
  • @Elaqqad Maybe I'm missing something, but why not just take $$\Bbb{Q} \subseteq \Bbb{Q}(\sqrt{2}) \subseteq \Bbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \Bbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}) \subseteq \Bbb{Q}(\sqrt{1 + \sqrt{2} + \sqrt{3} + \sqrt{5}})$$ ? – A.P. Aug 02 '15 at 16:15
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    @A.P. you're absolutely right , I was thinking very far. – Elaqqad Aug 02 '15 at 18:36
  • I haven't used the $\subseteq \Bbb Q$ syntax before so I don't fully understand everyone's comments. Would you mind explicitly linking it to the idea of extending subfields with $k$ as above? – hohner Aug 02 '15 at 19:32

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