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I have dealt with a number of algebraic field extensions $\mathbb{Q}[\alpha_1, \alpha_2, \ldots]/\mathbb{Q}$ and the primitive element was always $\alpha_1 + \alpha_2 + \cdots$. Is this generally true (provided that there exists a primitive element in the first place) or are there counter examples?

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    $\begingroup$ Certainly not in general, I mean $\mathbb{Q}(\sqrt 2, -\sqrt 2)$ is a counterexample. There are probably less trivial ones too. $\endgroup$ – Noah Olander Jun 9 '15 at 20:09
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    $\begingroup$ @NoahOlander Slightly less trivial: $\mathbb Q[\sqrt 3-\sqrt 2,\sqrt 2]$ $\endgroup$ – Hagen von Eitzen Jun 9 '15 at 20:11
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    $\begingroup$ In general, you can construct counterexamples based on the constructive proof of the primitive element theorem. See for example, planetmath.org/ProofOfPrimitiveElementTheorem and note that you just need to gimmick your generators so that the sum is one of the exceptions. $\endgroup$ – Adam Hughes Jun 9 '15 at 20:15
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    $\begingroup$ It is true that, for finitely many algebraic $\alpha_i$'s, there are primitive elements among their linear combinations. $\endgroup$ – Berci Jun 9 '15 at 20:16
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    $\begingroup$ As Hagen's answer mentions, there is no such thing as "the" primitive element; there will be many equally-good choices. $\endgroup$ – Zev Chonoles Jun 9 '15 at 20:23
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As it is easy to construct explicit counterexamples, one might ask why it happens so often in naturally occuring situations that the sum is primitive. The reason is that almost all linear combinations of the $\alpha_k$ are primitive; recall from the proof that one uses that the inifinitely many fields $\mathbb Q[\alpha_1+c\alpha_2]$ are really just finitely many, and ultimately almost all choices of $c$ lead to a primitive element. Hence a case where $c=1$ does not work can be assumed to be "bad luck" or artificially constructed to have this property (such as $\mathbb Q[\sqrt 3-\sqrt 2,\sqrt 2]$)

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    $\begingroup$ In this case $\mathbb{Q}[\sqrt{3}-\sqrt{2},\sqrt{2}] = \mathbb{Q}[\sqrt{3}-\sqrt{2}]$. How about when we have $\mathbb{Q}[S]$ with $S$ a minimal set? $\endgroup$ – Darth Geek Jun 9 '15 at 20:23

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