Generator of multiplicative group of the finite field Is it true that if the finite field $\mathbb{F}_q$ have the following realisation:
$$\mathbb{F}_q \simeq \mathbb{F_p[\alpha]}/(f(\alpha))$$
then $\alpha$ is the generator of its multiplicative group $\mathbb{F}^{\times}_q$ ?
I tried the following.
First note that $|\mathbb{F}^{\times}_q| = q - 1$, so if 
$$q - 1 = d_1^{e_1}\dots d_k^{e_k}$$
$\alpha$ is the generator $\iff$ $\forall i=1\dots k(\alpha^{d_i'} \neq 1 \mod f(\alpha)) $ where $d_i' = d_i^{e_i}$
But does there exist an example, where $(\alpha^{d_i'} = 1 \mod f(\alpha))$ for some $i \in \{1\dots k\}$ ?
 A: Answering the first question:

Is it true that if the finite field $\mathbb{F}_q$ have the following realisation:
$$\mathbb{F}_q \simeq \mathbb{F_p[\alpha]}/(f(\alpha))$$
then $\alpha$ is the generator of its multiplicative group $\mathbb{F}^{\times}_q$ ?

No. This is easiest to see in the particular case when $q = p^4$. In this case, each element of $\mathbb F_q \setminus \mathbb F_{p^2}$ generates the $\mathbb F_p$-algebra $\mathbb F_q$ (since otherwise, the $\mathbb F_p$-algebra it would generate would be a proper intermediate field between $\mathbb F_p$ and $\mathbb F_q$; but the only such intermediate field is $\mathbb F_{p^2}$), and thus qualifies as an $\alpha$. Hence, there are $q - p^2 = p^4 - p^2$ many possible $\alpha$s in $\mathbb F_q$. But the number of generators of $\mathbb F_q^\times$ is $\phi\left(q-1\right) = \phi\left(p^4-1\right)$, which is usually smaller than $p^4 - p^2$. (For example, if $p = 2$, then $\phi\left(p^4 - 1\right) = \phi\left(15\right) = 8$ is smaller than $p^4 - p^2 = 12$.)
I'm curious about the general case; is there an infinite family of $q$s for which every generator of the $\mathbb F_p$-algebra $\mathbb F_q$ is also a generator of the multiplicative group $\mathbb F_q^\times$, or is this merely a small-number phenomenon?
