TITLE: Is the primitve element of $\mathbb{Q}[\alpha_1, \alpha_2, \ldots]/\mathbb{Q}$ always $\alpha_1 + \alpha_2 + \cdots$?
QUESTION [7 upvotes]: 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?

REPLY [7 votes]: 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 $\mathbb Q[\sqrt 3-\sqrt 2,\sqrt 2]$)