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

  • $\begingroup$ To find the roots, you may use that in $\mathbb{F}_{p^n}$ we have $x^{p^n} = x$ (and any non zero element satisfies $x^{p^n-1} = 1$). So for example, in for $x \in \mathbb{F}_4$ we have $p(x) = x^4+x^3+1=x^3+x+1$, and if $x \ne 0$, then $p(x)=x$. So $p$ clearly has no root in $\mathbb{F}_4$. $\endgroup$ – Joel Cohen Nov 18 '19 at 0:12
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    $\begingroup$ Note that $p(x+1) = x^4+x^3+x^2+x+1$, whose roots are exactly the fifth roots of unity, excluding 1. $p(x)$ will have a root in your field if and only if $p(x+1)$ does. Now $\mathbb{F}_{p^k}$ has $\gcd (n, p^k-1)$ $n$-th roots of unity in it (consider a generator of the multiplicative group). From this we see that $\mathbb{F}_{16}$ has all the fifth roots of unity in it, so $p(x)$ splits completely here. I'm not quite sure what to do about the other fields. $\endgroup$ – Liam Nov 18 '19 at 0:47
  • $\begingroup$ Hints 1) You know that any of the zeros of $p(x)$ generates $\Bbb{F}_{16}$. Because $\Bbb{F}_{16}$ is a quadratic extension of $\Bbb{F}_4$, the minimal polynomials of those zeros must be quadratic over $\Bbb{F}_4$. But those minimal polynomials must also be factors of $p(x)$, so.... 2) $\Bbb{F}_{64}$ contains $\Bbb{F}_4$, so... For odd degree extensions, see DonAntonio's answer. $\endgroup$ – Jyrki Lahtonen Nov 18 '19 at 17:39

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 $\;4\,\nmid\,6\;$...

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  • $\begingroup$ What I am thinking is that since $p(x)$ factors linearly in $\mathbb{F}_{16}$, but $\mathbb{F}_{16}$ is not a subfield of $\mathbb{F}_{64}$ by the divisibility argument, so it has no roots in $\mathbb{F}_{64}$. But is it irreducible in $\mathbb{F}_{64}$? I think so because if $ p(x) = \prod_{i=1}^{4} (x-a_{i})$, then $ (x-a_{i})(x-a_{j}) \notin \mathbb{F}_{64}[x]$. Also, could a similar argument be applied for $p(x)$ is $\mathbb{F}_{32}$. That is, it is irreducible in $\mathbb{F}_{32}$ by a similar argument to $\mathbb{F}_{64}$? $\endgroup$ – user100101212 Nov 18 '19 at 15:13
  • $\begingroup$ Your explanation about $\Bbb{F}_8$ is on the money. But the hint about $4\nmid6$ feels strange. The gcd of the degree of the polynomial and of the degree of the field extension is more relevant. May be you had a different idea in mind, but I found that point a bit opaque. $\endgroup$ – Jyrki Lahtonen Nov 18 '19 at 17:42

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