Determine irreducibility of polynomial over finite field

Question

My questions are the following:
How does the reducibility a polynomial $p(x)$ in $\mathbb{F}_{p}[x]$ relate to its reducibility in $\mathbb{F}_{p^{n}}$, for some $ n \in \mathbb{N}$? I was thinking that if $p(x)$ is reducible in $\mathbb{F}_{p}[x]$, then it is reducible in $\mathbb{F}_{p^{n}}[x]$. But irreducibility in $\mathbb{F}_{p}[x]$ does not imply irreducibility in $\mathbb{F}_{p^{n}}[x]$.
Also, I was wondering about the relationship of roots of a polynomial in $\mathbb{F}_{p^{n}}[x]$. In order to find a root of a polynomial in $\mathbb{F}_{p^{n}}$, does one have to construct the field of that order by finding irreducible polynomial of degree $n$ and consider the quotient field?
Specifically, how does one find the roots and determine the irreducibility of $p(x) = x^{4} + x^{3} + 1$ in $\mathbb{F}_{4}, \mathbb{F}_{8}, \mathbb{F}_{16}, \mathbb{F}_{64}$. I know that it is irreducible in $\mathbb{F}_{2}$ because the irreducible polynomials in that field do not divide it. But I don't know how to proceed on the higher order fields

Answer 1

Observe that $\;p(x)=x^4+x^3+1\;$ has no linear factor in $\;\Bbb F_2\;$ since none of the elements in this field is a root, and also the unique irreducible quadratic polynomial over $\;\Bbb F_2\;$ , namely $\;x^2+x+1\;$ , divides it. Thus, $\;\Bbb F_2[x]/\langle p(x)\rangle\cong\Bbb F_{2^4=16}\;$ is the splitting field of $\;p(x)\;$, and from here that $\;p(x)\;$ remains irreducible over $\;\Bbb F_{2^3=8}\;$, otherwise:

We would be able to write $\;p(x)=g(x)h(x)\in\Bbb F_8[x]\;$, with $\;g,h\;$ of degree exactly two (as none of the roots of $\;p\;$ in $\;\Bbb F_8\;$, otherwise it would split on this field...), and since all the roots of $\;p\;$ are in $\;\Bbb F_{16}\;$, this last field would would be a splitting field of $\;g,\,h\;$ over $\;\Bbb F_8\;$ , which is absurd since this last field is not even a subfield of $\;\Bbb F_{16}\;$ (because $\;3\,\nmid\,4\;$).

Try now to deduce the situation in $\;\Bbb F_{2^6=64}\;$, taking into account that $\;4\,\nmid\,6\;$...