Under what conditions is an irreducible polynomial in $\operatorname{GF}(p^m)$ also irreducible in $\operatorname{GF}(p^n)$? 
Consider finite fields $\operatorname{GF}(p^m)$ and $\operatorname{GF}(p^n)$. Let $f(x)$ be a monic polynomial with coefficients in a common subfield of the two fields. If $f(x)$ is irreducible in $\operatorname{GF}(p^m)$, under what conditions is it also irreducible in $\operatorname{GF}(p^n)$?

For example, $x^2 + x + 1$ is irreducible in $\operatorname{GF}(2^{2k+1})$ but not in $\operatorname{GF}(2^{2k})$.
As a special case, if $f(x)$ has odd degree and is irreducible in $\operatorname{GF}(2^{2k+1})$, is it also irreducible in $\operatorname{GF}(2^{4k+2})$?
 A: Let $d$ be a degree of the polynomial $f(x)$ and $\operatorname{GF}(p^k)$ be 
a common subfield of $F_1=\operatorname{GF}(p^m)$ and $F_2=\operatorname{GF}(p^n)$ of the form $\operatorname{GF}(p)(f_0,\ldots,f_d)$, (that is $\operatorname{GF}(p)$ extended by the system $(f_0,\ldots,f_d)$) where $f(x) = f_dx^d+\ldots+f_1x+f_0$. For two integers $a,b$ denote via $(a,b)$ the greatest common divisor of $a$ and $b$.
We claim that the implication: $f$ is irreducible over $F1$ $\Rightarrow$ $f$ is irreducible over $F_2$ is valid iff the following implication is valid
$$
(d,\frac{m}{k}) =1 \Rightarrow (d, \frac{n}{k})=1.
$$
Which of course valid in particular case: $d=1$.

Lemma. Let $g(x)$ be an irreducible polynomial of degree $d$ over the field $\operatorname{GF}(q)$. Then $g(x)$ splits in the product of $(d,e)$ irreducible factors of degree $\frac{d}{(d,e)}$ over the field $\operatorname{GF}(q^e)$.

We give a sketch of the proof of the above lemma. Polynomial $g(x)$ has the following canonical decomposition over $\operatorname{GF}(q^d)$
$$
(x-\alpha)(x-\alpha^q)\ldots(x-\alpha^{q^{d-1}}).
$$
Then over $\operatorname{GF}(q^e)$ we have the following required decomposition 
$$
\begin{array}{l}
(x-\alpha)(x-\alpha^{q^{\frac{d}{(d,e)}}})(x-\alpha^{q^{\frac{2d}{(d,e)}}})\ldots (x-\alpha^{q^{((d,e)-1)\frac{d}{(d,e)}}})\cdot \\
(x-\alpha^q)(x-\alpha^{q^{\frac{d}{(d,e)}+1}})(x-\alpha^{q^{\frac{2d}{(d,e)}+1}})\ldots (x-\alpha^{q^{((d,e)-1)\frac{d}{(d,e)}+1}})\cdot\\
\ldots\\
(x-\alpha^{q^{\frac{d}{(d,e)}-1}})(x-\alpha^{q^{2\frac{d}{(d,e)}-1}})(x-\alpha^{q^{3\frac{d}{(d,e)}-1}})\ldots (x-\alpha^{q^{d-1}})\cdot
\end{array}
$$
Using the above lemma and condition $f$ is irreducible over $\operatorname{GF}(q^m)$ we obtain $(d,\frac{m}{k})=1$. And our claim now is trivial.
