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Find the degree and a basis for $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$.

My attempt:

I first proved that $\mathbb{Q}(\sqrt{3} + \sqrt{5}) = \mathbb{Q}(\sqrt{3}, \sqrt{5})$, Then using tower law its fairly straight forward to establish that $[\mathbb{Q}(\sqrt{3} + \sqrt{5}) : \mathbb{Q}(\sqrt{15})] = 2$.

But I am having difficulty determining the basis for this extension.

I can only take a guess that the answer is $\{1, \sqrt{3}\}$ or $\{1, \sqrt{5}\}$.

Is there a better method than just trail and error to determine the basis in this case.

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    $\begingroup$ Both of your guesses are correct. $\endgroup$ – Starfall Apr 8 '17 at 18:41
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One way to prove this is :

$a=(\sqrt{3}+\sqrt{5})$

$a^2=(\sqrt{3}+\sqrt{5})^2$

Thus $a^2-8+2\sqrt{15}=0$ and the polynomial $f(X)=X^2-8+2\sqrt{15}$ is irreducible over $\mathbb{Q}(\sqrt{15})$ because its roots do not lie in $\mathbb{Q}(\sqrt{15})$.

Thus a basis is the set $\{1,\sqrt{3}+\sqrt{5}\}$

Also $\{1,\sqrt{5}\}$ is indeed a basis for the extension because an element $x \in \mathbb{Q}(\sqrt{3},\sqrt{5})$ is of the form $x=a+b \sqrt{3} +c\sqrt{5}+d\sqrt{15}=a+\frac{b}{5} \sqrt{15} \sqrt{5} +c \sqrt{5}+d \sqrt{15}=(a+d \sqrt{15})+(c+ \frac{b}{5} \sqrt{15}) \sqrt{5}$.

In the same way you can prove that $\{1,\sqrt{3}\}$ is a basis of the extension you ask in your question.

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  • $\begingroup$ I guess $a^2 = {( \sqrt 3 + \sqrt 5)}^2 \implies a^2 - 8 - 2{\sqrt {15}} = 0$ and since that polynomial is irreducible over $\Bbb Q(\sqrt {15})$ can {1, $ \sqrt{8+2\sqrt{15}}$} be a choice of a basis? $\endgroup$ – Utsav Dewan Feb 24 '18 at 7:55
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There's a very obvious way to do this when you know a primitive element: clearly, we have that $ \mathbf Q(\sqrt{3} + \sqrt{5}) = \mathbf Q(\sqrt{15})(\sqrt{3} + \sqrt{5}) $, that is, the extension is generated over $ \mathbf Q(\sqrt{15}) $ by the same primitive element. Since we know that the extension degree is $ 2 $, we can easily surmise that a basis for the extension is simply $ \{ 1, \sqrt{3} + \sqrt{5} \} $, which is readily verified. Of course, the extension has many bases...

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