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This question is generalized based on my previous question: Is $x^5 + x^3 + 1$ irreducible in $\mathbb{F}_{32}$ and $\mathbb{F}_8$?

Problem:

Consider an irreducible polynomial $f = x^4 + x^3 + 1$ in $\mathbb{F}_2 [x]$. I am interested in its solvability and reducibility in a "chain" of fields. Namely, does $f$ have root in the following fields? Is $f$ reducible in the following fields?

$$ \color{red}{ \mathbb{F}_2},\mathbb{F}_4,\mathbb{F}_8,\color{red}{\mathbb{F}_{16}},\mathbb{F}_{32},\cdots,\color{red}{\mathbb{F}_{256}}$$

My progress:

$\mathbb{F}_2,\mathbb{F}_{16},\mathbb{F}_{256}$:

It is easy to verify that $f$ has no root in $\mathbb{F}_2$ and it is irreducible as well. As a result, $f$ has a root in $\mathbb{F}_{16} \simeq \mathbb{F}_2/f(x)$ and thus is reducible in $\mathbb{F}_{16}$. Since $\mathbb{F}_{256} = \mathbb{F}_{2^{4\times 2}}$, $\mathbb{F}_{16}$ is a subfield of $\mathbb{F}_{256}$, $f$ has root and is reducible in $\mathbb{F}_{256}$ as well.

$\mathbb{F}_{32},\mathbb{F}_{64},\cdots$:

For the fields "between" $\mathbb{F}_{16}$ and $\mathbb{F}_{256}$, I verified that $f$ cannot have root in them. Take $\mathbb{F}_{64}$ for example. Suppose $f$ actually has root in it, call it $\zeta$. Then apparently $\mathbb{F}_2 (\zeta)$ is a subfield of $\mathbb{F}_{64}$. But since $\mathbb{F}_2 (\zeta) \simeq \mathbb{F}_2 [x]/f(x)$, this would imply that $\mathbb{F}_{16}$ is a subfield of $\mathbb{F}_{64}$, which is impossible. But I haven't figured out how to deal with the reducibility part?

$\mathbb{F}_4,\mathbb{F}_8$:

For the fields "between" $\mathbb{F}_2$ and $\mathbb{F}_{16}$, I am not sure what's the best strategy here. For the existence of root, I am attempted to say that $f$ has no root in both of them since "the smallest" field extension that has a root is $\mathbb{F}_{16}$. I forgot the proof but I think it is correct.

With the operation tables of $\mathbb{F}_4$ from WolframAlpha, I was able to show that $f$ has no root in $\mathbb{F}_4$, and it is irreducible in $\mathbb{F}_4$ as well. But is there a better way?

For $\mathbb{F}_8$, I mimicked the answer in my previous question and managed to show that $f$ is irreducible in $\mathbb{F}_8$. But I don't know how I should deal with the existence of root?

I understand that this is a long post and I appreciate your help. Thanks.

Community
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3x89g2
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  • Because any zero of this polynomial generates the field $\Bbb{F}{16}$ it has a zero in a field $K$, if and only if $\Bbb{F}{16}\subseteq K$. It does factor of over $\Bbb{F}4$ because the minimal polynomials of any element of $\Bbb{F}{16}$ over $\Bbb{F}_4$ are at most quadratic. In an answer of this Q&A I show how the reciprocal of this polynomial, $x^4+x+1$, factors over $\Bbb{F}_4$. The factors of this polynomial are thus reciprocals of the two quadratics listed there. – Jyrki Lahtonen Jul 15 '16 at 08:43
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    It will help you in the future to think of "betweenness" of finite fields much the same way as "intermediate" fields between, say $\Bbb{Q}$ and its extensions. The multiplication rule of the extension degree, if $K\subset L\subset F$, then $[F:K]=[F:L]\cdot [L:K]$ is extremely useful. So $\Bbb{F}4$ and $\Bbb{F}_8$ are between $\Bbb{F}{64}$ and $\Bbb{F}2$, but $\Bbb{F}{16}$ is not (but does intersect $\Bbb{F}_{64}$ at $\Bbb{F}_4$). – Jyrki Lahtonen Jul 15 '16 at 08:46

1 Answers1

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Here is a simple way to answer all these questions, just from the fact that $f$ is irreducible over $\mathbb{F}_2$. Since $f$ is irreducible over $\mathbb{F}_2$, $\mathbb{F}_2[x]/(f)$ is a field. Since $f$ has degree $4$, this field is $4$-dimensional over $\mathbb{F}_2$, and hence is isomorphic to $\mathbb{F}_{16}$. But $f$ has a root in a field $K$ iff $\mathbb{F}_2[x]/(f)$ embeds in $K$ (by sending $x$ to the root in $K$). This means $f$ has a root in $\mathbb{F}_{2^n}$ iff $4\mid n$.

In particular, then, an extension $\mathbb{F}_{2^m}$ of $\mathbb{F}_{2^n}$ will contain a root (and hence all the roots) of $f$ iff $4\mid m$. Also, $\mathbb{F}_{2^m}$ is an extension of $\mathbb{F}_{2^n}$ iff $n\mid m$. This means that adjoining any root of $f$ to $\mathbb{F}_{2^n}$ gives $\mathbb{F}_{2^{\operatorname{lcm}(4,n)}}$. Thus each irreducible factor of $f$ over $\mathbb{F}_{2^n}$ must have degree $\frac{\operatorname{lcm}(4,n)}{n}=\frac{4}{\gcd(4,n)}$. Explicitly, this means that $f$ has a root in $\mathbb{F}_{2^n}$ iff $4\mid n$, is reducible but has no root iff $2\mid n$ and $4\not\mid n$, and is irreducible iff $n$ is odd.

More generally, a similar analysis shows that if $f(x)\in\mathbb{F}_q[x]$ is irreducible of degree $d$, then over $\mathbb{F}_{q^n}$ $f$ splits as a product of irreducible factors of degree $\frac{d}{\gcd(d,n)}$.

Eric Wofsey
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