5

$x^{18}-x^3=x^3(x^{15}+1)=x^3(x^{5\cdot3}+1)=x^3(x^5+1)(x^{10}+x^5+1)$

$x^5+1$ has the root $1$ in $\mathbb{F}_2$ so using the factor theorem I got: $x^5+1=(x+1)(x^4+x^3+x^2+x+1)$ (how to prove that the second factor is irreducible over $\mathbb{F}_2$ ?).

Noticing that the multiplicative order is $4$, so the initial polynomial can be factorized only with polynomials of degree $1$, $2$ and $4$ (the divisors of $4$). Since $0$ and $1$ are not roots of $x^{10}+x^5+1$, its factors has to be of degree $2$ and $4$, the only irreducible polynomial deg $2$ over $\mathbb{F}_2$ is $x^2+x+1$, so dividing $x^{10}+x^5+1$ by it, I got $x^8+x^7+x^5+x^4+x^3+x+1$, which it has to decompose in two degree 4 irreducible polynomials. But this method needs too much time.

Another way is to compute the table of $\mathbb{F}_{16}$ (splitting field of the initial polynomial over $\mathbb{F}_2$), the cyclotomic cosets, etc.

Is there any faster method ?

sound wave
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  • To see that $x^4+x^3+x^2+x+1$ is irreducible over $\Bbb{F}2$ just note that $\Bbb{F}{16}$ is the smallest field containing a fifth root of 1 (since 5 must divide the order of the multiplicative group). – sharding4 Jul 04 '18 at 13:13
  • Thanks for reply, but could you explain a bit more ? Why the fact that $\mathbb{F}_{16}$ is the smallest field containing a fifth root of 1 implies that the polynomial is irreducible over $\mathbb{F}_2$ ? – sound wave Jul 04 '18 at 13:19
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    The roots of that polynomial are the $5^{th}$ roots of 1. If the polynomial factored into say 2 quadratics, then the splitting field of one of the quadratics would have 4 elements and contain a $5^{th}$ root of 1 which is impossible since the multiplicative group has order 3. – sharding4 Jul 04 '18 at 13:23
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    The fastest method is to ask WA :-) – lhf Jul 04 '18 at 13:51
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    Because $x^{18}+x^3=x^2(x^{16}+x)$ an even faster way is to search the site we all love. – Jyrki Lahtonen Jul 04 '18 at 14:14
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    Well, I happen to have memorized the irreducible quartics (it's not unlike memorizing lists of small primes when you have gine through the multiplication tables many times over), so it would not occur to me to fire up Mathematica for this. A bit more complicated, sure :-). Anyway, to check whether a quartic or a quintic is irreducible in $\Bbb{F}_2[x]$ all you need to is to check that it has no zero (d'uh) and that it is not divisible by the sole irreducible quadratic $x^2+x+1$. This is not taxing at all, because that quadratic is a factor of $x^3-1$... – Jyrki Lahtonen Jul 04 '18 at 14:19
  • @JyrkiLahtonen yeah I already saw that question, but my question is different from that one, since I want to understand how to factorize polynomials (without using wolframalpha or other tools). – sound wave Jul 04 '18 at 15:05
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    I see. That's great! For that polynomial we can use the general result that over $\Bbb{F}p$ we have the factorization (for all $n$) $$x^{p^n}-x=\prod{q\in\Bbb{F}_p[x], q\ \text{irreducible}, \deg q\mid n}q(x).$$ So in your case of $p=2,n=4$ all the irreducible polynomials of degrees 1,2 or 4 appear as factors with multiplicity one. – Jyrki Lahtonen Jul 04 '18 at 15:34
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    For a quartic to be irreducible its constant term must be $1$ (otherwise $x$ is a factor), it must have an odd number of terms (otherwise $1$ is a zero and $x+1$ is a factor), and it must not be the square of the only irreducible quadratic $x^2+x+1$. You already found that the quartic with five terms is irreducible, so that this leaves the quartics with three terms. $x^4$ and $1$ must be in there, so it is just $x^4+x^3+1,x^4+x^2+1$ and $x^4+x+1$. The last bit is that $(x^2+x+1)^2=x^4+x^2+1$ (freshman's dream). The other two are thus irreducible. – Jyrki Lahtonen Jul 04 '18 at 15:40
  • (cont'd) They are also the factors of the octic you were left with. I don't know if this is "fast". – Jyrki Lahtonen Jul 04 '18 at 15:40
  • I’m hardly an expert at this, @JyrkiLahtonen, but yours is just the method I’d have used. – Lubin Jul 04 '18 at 15:53
  • $x^4+x^3+x^2+x+1$ is also irreducible because trial division by all $\deg\le 2$ candidates $x$, $x+1$, $x^2+x+1$ fails effortlessly. – Hagen von Eitzen Oct 12 '18 at 20:41

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