1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.