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I am struggling with the following:

Let's have the field extension $R/Q$. How can I find $[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q]$? (*)

So far all I have is that:

$[Q(\sqrt 6):Q] = 2$, because $minpoly_{\sqrt6}(X) = X^2 - 6$

Samely $[Q(\sqrt 10):Q] = [Q(\sqrt 15):Q] = 2$.

Another thing I read somewhere (couldn't find it anywhere else), is that:

$[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q] = [Q(\sqrt 2 \cdot \sqrt 3,\sqrt 2 \cdot \sqrt 5,\sqrt 3 \cdot \sqrt 5):Q] = [Q(\sqrt 2,\sqrt 3,\sqrt 5):Q]$

Is that true? Can I conclude that way? And if so, what can I do afterwards to find (*)?

My first idea was to calculate it this way:

$[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q] = [Q(\sqrt 6,\sqrt 10,\sqrt 15):Q(\sqrt 6,\sqrt 10)] \cdot [Q(\sqrt 6, \sqrt 10):Q(\sqrt 6)] \cdot [Q(\sqrt 6):Q]$

but I don't know how to compute the terms $[Q(\sqrt 6,\sqrt 10,\sqrt 15):Q(\sqrt 6,\sqrt 10)]$ and $[Q(\sqrt 6, \sqrt 10):Q(\sqrt 6)]$ either.

What I also found is that $\sqrt 6, \sqrt 10$ and $\sqrt 15$ are linearly dependent over $Q$, since $\sqrt 6 \sqrt 10 - 2 \cdot \sqrt 15 = 0$.

Is there some general way to find $[Q(\alpha, \beta):Q]$ and $[Q(\alpha, \beta, \gamma):Q]$?

Thank you in advance.

K.A.
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  • It would help if you tied this question to Galois theory. In the case of a quadratic extension there is little difficult in determining the degree of extension and the corresponding Galois group. – hardmath Jan 14 '16 at 12:52

3 Answers3

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Hint: We have $\sqrt 6 \ \sqrt{10} = 2 \sqrt{15}$ and so $\mathbb Q(\sqrt 6,\sqrt {10},\sqrt {15})=\mathbb Q(\sqrt 6,\sqrt {10})$, which simplifies things...

lhf
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  • Ok thank you, so in general, if I have $[Q(a_1, ..., a_n,a):Q]$ and $a$ is linearly dependent of $a_1,...,a_n$, then I can say $[Q(a_1, ..., a_n,a):Q]$ = $[Q(a_1, ..., a_n):Q]$?

    That helps a lot, so all I need to calculate now is actually $[Q(\sqrt 6, \sqrt 10):Q(\sqrt 6)]$. I assume I have to show that $\sqrt 10$ can't be expressed by $a + b \sqrt 6$?

     – K.A. Jan 14 '16 at 12:56
  • @J.C., you're on the right track. – lhf Jan 14 '16 at 12:56
  • thank you, I got it now. That helps me a lot with the rest of the question I would have had. – K.A. Jan 14 '16 at 13:05
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$\mathbf Q(\sqrt 6,\sqrt 10,\sqrt 15)=\mathbf Q(\sqrt 6,\sqrt 10)$, since $\sqrt 15=\frac12\sqrt 6 \sqrt 10$, whence $$[\mathbf Q(\sqrt 6,\sqrt 10,\sqrt 15):\mathbf Q(\sqrt 6,\sqrt 10)]=1.$$ $x^2-10$ is irreducible over $\mathbf Q(\sqrt 6)$, sin you can easily check no $\alpha=a+b\sqrt 6$ is a root of $x^2-10$ just writing down the equations for $a$ and $b$.. Thus $$[\mathbf Q(\sqrt 6,\sqrt 10,\sqrt 15):\mathbf Q]=[\mathbf Q(\sqrt 6,\sqrt 10:\mathbf Q]=[\mathbf Q(\sqrt 6,\sqrt 10):\mathbf Q(\sqrt 6)]\cdot[\mathbf Q(\sqrt 6):\mathbf Q]=4.$$

Bernard
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  • Ok, so $x^2-10$ is irreducible over $Q(\sqrt 6)$, because I can't have $a,b \in Q$ such that $(a+b \sqrt 6) - 10 = 0$. I verified it. And therefore $[Q(\sqrt 6, \sqrt 10):Q(\sqrt 6)]$ is the degree of $minpoly_{\sqrt 10}$, which is $x^2 - 10$, hence it's $2$. Is that right? – K.A. Jan 14 '16 at 13:03
  • It's exactly that. – Bernard Jan 14 '16 at 13:51
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$\sqrt6\cdot \sqrt{10}=2\sqrt{15}\Rightarrow \sqrt {15}\in \mathbb Q(\sqrt 6,\sqrt {10})$. Hence the asked degree is the degree of $\mathbb Q(\sqrt 6,\sqrt {10})$. Thus the answer is $\color{red}4$.

Piquito
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