Primitive element theorem: Difference between revisions

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== Proof ==

Suppose first that <math>F</math> is infinite. By induction, it suffices to prove that any finite extension <math>E=F(\beta,\gamma) </math> is simple. For <math>c\in F</math>, suppose <math>\alpha = \beta+ c\gamma </math> fails to be a primitveprimitive element, <math>F(\alpha)\subsetneq F(\beta,\gamma)</math>. Then <math>\gamma\notin F(\alpha)</math>, since otherwise <math>\beta = \alpha-c\gamma\in F(\alpha)=F(\beta,\gamma)</math>. Consider the minimal polynomials of <math>\beta,\gamma </math> over <math>F(\alpha) </math>, respectively <math>f(X), g(X) \in F(\alpha)[X]</math>, and take a splitting field <math>L </math> containing all roots <math>\beta,\beta',\ldots</math> of <math>f(X) </math> and <math>\gamma,\gamma',\ldots </math> of <math>g(X) </math>. Since <math>\gamma\notin F(\alpha)</math>, there is another root <math>\gamma'\neq \gamma</math>, and a field automorphism <math>\sigma:L\to L </math> which fixes <math>F(\alpha) </math> and takes <math>\sigma(\gamma)=\gamma' </math>. We then have <math>\sigma(\alpha) =\alpha </math>, and:

:<math>\beta + c \gamma = \sigma(\beta + c \gamma) = \sigma(\beta) + c \, \sigma(\gamma) </math>, and therefore <math>c = \frac{\sigma(\beta) - \beta}{\gamma - \sigma(\gamma)}</math>.

Since there are only finitely many possibilities for <math>\sigma(\beta)=\beta'</math> and <math>\sigma(\gamma)=\gamma'</math>, only finitely many <math>c\in F</math> fail to give a primitive element <math>\alpha=\beta+c\gamma</math>. All other values give <math>F(\alpha)=F(\beta,\gamma) </math>.