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If one adjoins to the [[rational number]]s <math>F = \mathbb{Q}</math> the two irrational numbers <math>\sqrt{2}</math> and <math>\sqrt{3}</math> to get the extension field <math>E=\mathbb{Q}(\sqrt{2},\sqrt{3})</math> of degree 4, one can show this extension is simple, meaning <math>E=\mathbb{Q}(\alpha)</math> for a single <math>\alpha\in E</math>. Taking <math>\alpha = \sqrt{2} + \sqrt{3} </math>, the powers 1, ''α'', ''α''<sup>2</sup>, ''α''<sup>3</sup> can be expanded as [[linear combination]]s of 1, <math>\sqrt{2}</math>, <math>\sqrt{3}</math>, <math>\sqrt{6}</math> with integer coefficients. One can solve this [[system of linear equations]] for <math>\sqrt{2}</math> and <math>\sqrt{3}</math> over <math>\mathbb{Q}(\alpha)</math>, to obtain <math>\sqrt{2} = \tfrac12(\alpha^3-9\alpha)</math> and <math>\sqrt{3} = -\tfrac12(\alpha^3-11\alpha)</math>. This shows that ''α'' is indeed a primitive element:
:<math>\mathbb{Q}(\sqrt 2, \sqrt 3)=\mathbb{Q}(\sqrt2 + \sqrt3).</math>
One may also use the following more general argument.<ref>{{Cite book |last=Lang |first=Serge |url=http://link.springer.com/10.1007/978-1-4613-0041-0 |title=Algebra |date=2002 |publisher=Springer New York |isbn=978-1-4612-6551-1 |series=Graduate Texts in Mathematics |volume=211 |___location=New York, NY |pages=243 |doi=10.1007/978-1-4613-0041-0}}</ref> The field <math>E=\Q(\sqrt 2,\sqrt 3) </math> clearly has four field automorphisms <math>\sigma_1,\sigma_2,\sigma_3,\sigma_4: E\to E </math> defined by <math>\sigma_i(\sqrt 2)=\pm\sqrt 2 </math> and <math>\sigma_i(\sqrt 3)=\pm\sqrt 3 </math> for each choice of signs. The
== Theorem statement ==
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