Primitive element theorem: Difference between revisions

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== Example ==

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 of a field extension|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^&nbsp;'', ''α''², ''α''³ 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>

== The theorems ==