I'm allergic to finite fields, so let's work in characteristic $0$. The primitive element theorem states:

Primitive element theorem: Let $E / F$ be a finite-dimensional field extension, say $n =[E:F]$. Then, there exists an element $\alpha \in E$ such that $E = F[\alpha]$.

The following is not really a "theorem" so much as a basic fact from linear algebra, but anyway:

Cyclic vector theorem: Let $V$ be a finite-dimensional vector space over $F$, say $\mathrm{dim}(V)=n$. Let $T : V \to V$ be a linear transformation with minimum polynomial $m$. If $\mathrm{deg}(m) = \mathrm{dim}(V)$, then there is a vector $v \in V$ such that $v,Tv,\ldots,T^{n-1}v$ is a basis for $V$.

When your extension $E/F$ is singly-generated over $F$ by $\alpha$, you can consider the minimum polynomial (now in the sense of an algebraic element over a field) $m$ of the element $\alpha$, which is irreducible and has degree $n = [E:F]$. It is sort of obvious in this situation that $T : E \to E$ given by $T(\beta) = \alpha \beta$ is an $F$-linear transformation whose minimum polynomial (in the sense of linear algebra) is also $m$. Any nonzero element is a cyclic vector for this transformation. Perhaps $1 \in E$ is a particularly natural choice. So, anyway, there is at least some sort of superficial connection between the boxed results above. But I really want to know if there is a useful connection. I would especially like to know:

Question: Is there some way to cleverly set things up so that the primitive element theorem becomes a consequence of the "cyclic vector theorem"? If not, can the notion of a cyclic vector at least be used to clarify the primitive element theorem somehow?

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    $\begingroup$ I strongly think the answer is no. Of course since the question is not formally stated I can't prove it, but in in order to state something definite let me bet that nobody will post a positive answer before, say, November 30, 23:59:59 GMT :-) $\endgroup$ – Georges Elencwajg Nov 27 '15 at 10:33
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    $\begingroup$ I don't want to spell out all my reasons for my bet, but one of them is that one has to distinguish between finite base field and infinite base field for a theorem whose statement cares only about separable extensions. Separability of the base field plays no role in the cyclic vector theorem, so that a proof of primitivity using pure linear algebra shouldn't have to distinguish cases. $\endgroup$ – Georges Elencwajg Nov 27 '15 at 10:45

We need a minimal polynomial with degree equal to the dimension of $V$ to apply cyclic vector theorem. So if you take $E=V$ you need a polynomal of degree $[E:F]$ and the only good choice for this polynomial I see is the minimal polynomial of $\alpha$.

What you can do is to view the cyclic vector theorem as special case of the primitive element theorem. Use the cyclic vector theorem or the structure theorem for finitely generated modules over a principal ideal domain to identity $V$ with $L=F[X]/(m)$ such that $T$ is represented by the multiplication by $X$. Now if you forget the morphism $T$ you can recover it from $L/F$ using the primitive element theorem and you get the cyclic vector basically for free.

Not sure if it is helpful what I wrote. It looks a bit made up for me. Maybe it is helps anyway.


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