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 November 11th, 2013, 05:39 AM #1 Newbie   Joined: Nov 2013 Posts: 3 Thanks: 0 extension fields over Q I need to find the degree and a basis for the extension-field $Q(\sqrt{2},i\sqrt{5},\sqrt{2}+\sqrt{7})$... I had no Problem finding a basis and the degree of $Q(\sqrt{3},\sqrt{5}$ ... but now i don´t know what to do with the $\sqrt{2}+\sqrt{7}$ - part my first idea would be : $Q(\sqrt{2},i\sqrt{5},\sqrt{2}+\sqrt{7})=\{a_0+a_1\ sqrt{2}+a_2i\sqrt{5}+a_3(\sqrt{2}+\sqrt{7})+a_4i\s qrt{10}+a_5(\sqrt{2}\cdot( \sqrt{2}+\sqrt{7}))+a_6(i\sqrt{5}\cdot(\sqrt{2}+\s qrt{7})+a_7\cdot(i\sqrt{10}\cdot(\sqrt{2}+\sqrt{7} )\}$ and so the degree would be 8 ... is that correct or am i wrong?
 November 12th, 2013, 06:06 PM #2 Member   Joined: Mar 2013 Posts: 90 Thanks: 0 Re: Extension fields over Q A simplier basis would be $\{1,\sqrt2,\,\sqrt7,\,\sqrt{14},\,i\sqrt5,\,i\sqrt {10},\,i\sqrt{35},\,i\sqrt{70}\}$; the degree is indeed 8.
November 13th, 2013, 05:20 AM   #3
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Re: extension fields over Q

Quote:
 Originally Posted by Nehushtan A simplier basis would be $\{1,\sqrt2,\,\sqrt7,\,\sqrt{14},\,i\sqrt5,\,i\sqrt {10},\,i\sqrt{35},\,i\sqrt{70}\}$; the degree is indeed 8.
Very nice.

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