Can the root of an irreducible polynomial be the root of a lower degree polynomial as well? Consider the field $F_{p^m}$ for some prime $p$. Let $y(x) \in F_{p}[x]$ be an irreducible polynomial of degree $m$ with root $ \alpha \in F_{p^m}$. My question is can $\alpha$ be a root of some other polynomial of degree less than $m$ in $F_{p}[x]$? If not, then I want to use this fact to prove that $\{1, \alpha, \dots, \alpha^{m-1}\}$ is a basis of $F_{p^m}$.
 A: For any field $F$, let
$y(x) \in F[x] \tag 1$
be an irreducible polynomial with
$\deg y = m, \tag 2$
and let $E$ be a field extension of $F$ with
$\alpha \in E \supsetneq F \tag 3$
a root of $y(x)$:
$y(\alpha) = 0. \tag 4$
Suppose that there exists a non-trivial polynomial
$q(x) \in F[x] \tag 5$
with
$\deg q < m = \deg p, \tag 6$
also taking $\alpha$ as a root:
$q(\alpha) = 0.  \tag 7$
We observe that
$\deg q > 1, \tag 8$
for otherwise we may write
$q(x) = ax + b, \; a, b \in F, \tag 9$
and, by (7),
$a\alpha + b = 0; \tag{10}$
note that
$a \ne 0, \tag{11}$
lest (7) become
$b = q(\alpha) = 0, \tag{12}$
and with
$a = b = 0, \tag{13}$
we have
$q(x) = 0, \tag{14}$
the prohibited trivial case.  Now by virtue of (10) we may write
$\alpha = -\dfrac{b}{a} \in F, \tag{15}$
and the root $\alpha$ of $y(x)$ is in fact an element of $F$, and hence $y(x)$ may be written as
$y(x) = (x - \alpha)p(x), \tag{16}$
where
$p(x) \in F[x] \tag{17}$
is a polynomial of degree $m - 1$; but this shows that $y(x)$ is in fact reducible in $F[x]$, contrary to assumption, and thus $q(x)$ cannot take the form (9), and hence (8) binds.
Now of all polynomials satisfying the hypotheses (5)-(7), we may select one, $\mu(x)$, of least degree.  Then via Euclidean division of polynomials we may uniquely write
$y(x) = d(x)\mu(x) + r(x) \tag{18}$
for some
$d(x), r(x) \in F[x] \tag{19}$
with $r(x)$ satisfying either
$r(x) = 0 \tag{20}$
or
$\deg r < \deg \mu; \tag{21}$
now if $r(x) = 0$, then (18) reduces to
$y(x) = d(x)\mu(x), \tag{22}$
and since
$\deg \mu < \deg y \tag{23}$
and
$\deg y = \deg \mu + \deg d, \tag{23}$
it follows that
$\deg d \ge 1, \tag{24}$
and thus that $y(x)$ is reducible over the field $F$, again in contradiction to our hypothesis.  Thus we rule out (20), and evaluating (18) at $\alpha$ we find
$0 = y(\alpha) =  d(\alpha)\mu(\alpha) + r(\alpha) = d(\alpha) \cdot 0 + r(\alpha) = r(\alpha); \tag{25}$
but now in light of (21) this contradicts the choice of $\mu(x)$ as a polynomial of minimum degree for which (7) binds; and therefore no $q(x)$ fulfilling (5)-(7) may exist, that is, $\alpha$ is not a zero of any polynomial of degree less than $y(x)$.
Since $\alpha$ is not a zero of any polynomial in $F[x]$ of degree less than $m = \deg y$, it follows that $\{1, \alpha, \alpha^2, \ldots, \alpha^{m- 1}\}$ is a linearly independent set in $E$ over $F$, and furthermore, that the field extension $F(\alpha)/F$ is of degree $m$ over $F$:
$[F(\alpha):F] = m. \tag{26}$
We have derived all these conclusions over a general base field $F$; in case we take
$F = F_p, \tag{27}$
these results readily apply.  Since
$[F_{p^m}:F_p] = m, \tag{28}$
the linear independence of $\{1, \alpha, \alpha^2, \ldots, \alpha^{m- 1}\}$ over $F$ shows that this set is indeed a basis for $F_{p^m}$ over $F_p$.
Note Added in Edit, Thursday 26 August 2021. The astute reader may have noticed that the argument that (8) binds, i.e. that $\deg q >1$ (and which is presented ca. equations (8)-(17) of the text of my answer), is not in fact directly related to the main line argument presented herein, which focuses on the topics presented in the text of the question itself, i.e. the nonexistence of polynomials such as $q(x)$ and the fact that $\{1, \alpha, \alpha^2, \ldots, \alpha^{m - 1}
\}$ forms a basis for $F(\alpha)$.  I am not sure precisely why I included the proof that $\deg q >1$ here, except that it came to me whilst I was rummaging about trying to get the details on my solution in place, and it seems like it clarifies the situation somewhat, so I thought it might be best to make it available to my readers. End of Note.
