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I have a proof of the primitive element theorem for subfields $K$ of $\mathbb{C}$ which relies on there being infinitely many elements in $K$. In a book I've been reading it states that every finite separable extension is simple and that this follows from the earlier version the primitive element theorem.

Surely this doesn't work if I'm in a finite field? Is there another proof of the primitive element theorem that works in this case? Or is there a more general version that doesn't required infinitely many elements in $K$?

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    The multiplicative group of a finite field is cyclic.2012-05-17
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    I'm not that sure Eagle ... Edit : Oh didn't see "field" sorry.2012-05-17
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    Also be advised that in the context of finite fields *a primitive element* usually means a generator of the multiplicative group. This differs markedly from the context of arbitrary field extensions, where a primitive element is any element not contained in any intermediate field. The finite fields usage is more restrictive (even though a primitive element in that sense is also a primitive element in the more general sense).2012-05-17
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    But Chris' comment answers the question in the case of finite fields. Whenever I have seen this result stated, separate proofs have been stated for the two cases: infinite and finite.2012-05-17
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    @ChrisEagle: oh of course. So separate proofs in each case then.2012-05-17

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A common way of handling this is to deal with the infinite fields using the general theorem, and to deal with the finite fields using the cyclicity of the multiplicative group. A finite degree extension of a finite field is itself finite, and the generator of its multiplicative group then is a primitive element in both senses of the word.