Find the degree and a basis for $ \scriptstyle\mathbb{Q}(\sqrt{3}+\sqrt{5}) $ over $ \scriptstyle\mathbb{Q}(\sqrt{15}) $. Find the degree and a basis for $ \scriptstyle\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}) $ over $ \scriptstyle\mathbb{Q} $.
$ \scriptstyle[\mathbb{Q}(\sqrt{3}+\sqrt{5}):\mathbb{Q}]\ =\ [\mathbb{Q}(\sqrt{3}+\sqrt{5}):\mathbb{Q}(\sqrt{15})][\mathbb{Q}(\sqrt{15}):\mathbb{Q}] $
We know that $ \scriptstyle[\mathbb{Q}(\sqrt{3}+\sqrt{5}):\mathbb{Q}]=4 $ and $ \scriptstyle[\mathbb{Q}(\sqrt{15}):\mathbb{Q}]=2 $, so $ \scriptstyle[\mathbb{Q}(\sqrt{3}+\sqrt{5}):\mathbb{Q}(\sqrt{15})] $ must be $ \scriptstyle2 $. A basis is given by $ \scriptstyle\{1,\sqrt{3}+\sqrt{5}\} $.
Equivalent bases are given by $ \scriptstyle\{1,\sqrt{3}\} $ and $ \scriptstyle\{1,\sqrt{5}\} $. To see why, consider two elements from $ \scriptstyle\mathbb{Q}(\sqrt{15}) $: $ \scriptstyle a+b\sqrt{15} $ and $ \scriptstyle c+d\sqrt{15} $, where $ \scriptstyle a,b,c,d\in\mathbb{Q} $. Conceptually we know that every element in $ \scriptstyle\mathbb{Q}(\sqrt{3}+\sqrt{5}) $ is some linear combination of $ \scriptstyle1,\ \sqrt{3},\ \sqrt{5}, $ and $ \scriptstyle\sqrt{15} $. Consider the basis $ \scriptstyle\{1,\sqrt{3}+\sqrt{5}\} $ over $ \scriptstyle\mathbb{Q}(\sqrt{15}) $. The span of this basis has elements of the form:
$ \scriptstyle(a+b\sqrt{15})+(c+d\sqrt{15})(\sqrt{3}+\sqrt{5}) $
$ \scriptstyle=\ a\ +\ b\sqrt{15}\ +\ (c+5d)\sqrt{3}\ +\ (c+3d)\sqrt{5} $,
which is the necessary linear combination. Don't be fooled into thinking that the coefficients $ \scriptstyle(c+5d) $ and $ \scriptstyle(c+3d) $are dependent on each other; the system is overdetermined, so given a value for, say, $ \scriptstyle(c+5d) $, there are infinitely many solutions for $ \scriptstyle(c+3d) $.
Now consider the basis $ \scriptstyle\{1,\sqrt{3}\} $ over $ \scriptstyle\mathbb{Q}(\sqrt{15}) $. The span of this basis has elements of the form:
$ \scriptstyle (a+b\sqrt{15})+(c+d\sqrt{15})(\sqrt{3}) $
$ \scriptstyle=\ a\ +\ b\sqrt{15}\ +\ c\sqrt{3}\ +\ 3d\sqrt{5} $.
It's actually easier to see the linearity through this basis! The same can be done with $ \scriptstyle\{1,\sqrt{5}\} $.
$ \scriptstyle[\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}):\mathbb{Q}]\ =\ [\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]\ =\ [\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}):\mathbb{Q}(\sqrt[4]{2})][\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}] $,
so that $ \scriptstyle[\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}):\mathbb{Q}] $ is divided by both $ \scriptstyle[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3 $ and $ \scriptstyle[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]=4 $.
Then, note that $ \scriptstyle x^3-2\ \in\ \mathbb{Q}(\sqrt[4]{2})[x] $, and $ \scriptstyle\sqrt[3]{2} $ is a zero of that polynomial, so $ \scriptstyle[\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}):\mathbb{Q}(\sqrt[4]{2})] $ is at most 3. This connection can be made also because $ \scriptstyle\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2})\ =\ \mathbb{Q}(\sqrt[3]{2},\sqrt[4]{2}) $, since the presence of the $ \scriptstyle\sqrt[4]{2} $ implies the presence of its greater power $ \scriptstyle\sqrt{2} $.
So, $ \scriptstyle[\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}):\mathbb{Q}]\ =\ $ 4,8, or 12, but it must also be divisible by 3, so it must be 12.
The basis for $ \scriptstyle\mathbb{Q}(\sqrt{2},\sqrt[3]{2},\sqrt[4]{2}) $ over $ \scriptstyle\mathbb{Q} $ must include all unique compositions of $ \scriptstyle2^{1/4} $ and $ \scriptstyle2^{1/3} $. That would be the set:
$ \textstyle\{1,2^{\scriptstyle1/4}\textstyle,2^{\scriptstyle2/4}\textstyle,2^{\scriptstyle3/4}\textstyle,2^{\scriptstyle1/3}\textstyle,2^{\scriptstyle1/3+1/4}\textstyle,2^{\scriptstyle1/3+2/4}\textstyle,2^{\scriptstyle1/3+3/4}\textstyle,2^{\scriptstyle2/3}\textstyle,2^{\scriptstyle2/3+1/4}\textstyle,2^{\scriptstyle2/3+2/4}\textstyle,2^{\scriptstyle2/3+3/4}\textstyle\} $
$ \textstyle=\ \{1,\sqrt[4]{2},\sqrt{2},\sqrt[4]{8},\sqrt[3]{2},\sqrt[12]{128},\sqrt[6]{32},\sqrt[12]{8192},\sqrt[3]{4},\sqrt[12]{2048},\sqrt[6]{128},\sqrt[12]{131072}\} $.
- --Nick Rupley 02:05, 22 April 2009 (UTC)
