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Currently, I have that I can do:

$$ x^5-1 = (x-1)(x^4+x^3+x^2+x+1) $$

Then I thought I could do:

$$ x^5-1 = (x-1)(x(x+1)(x^2+1)+1) $$

However I want to use the factorization to find all cyclic codes in $[5, k]$. For that, I need to find all irreducible factors of $ x^5-1 $. I am not seeing if I can actually reduce $x^4+x^3+x^2+x+1$ any further

user840091
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    Does this mean you are you factorising over $\mathbb Z_5$? – Luke Collins Dec 13 '20 at 00:45
  • @LukeCollins not it is in $Z_2$ – user840091 Dec 13 '20 at 00:47
  • If you want to factor over the reals...$(x^5 - 1)=(x-1)(x^2 + \frac {1-\sqrt 5}{2} x + 1)(x^2 + \frac {1-\sqrt 5}{2} x + 1) = (x+\cos \pi)(x^2 + 2\cos\frac {2\pi}{5} x + 1)(x^2 +2\cos\frac {4\pi}{5} x + 1)\$ Over the Compex numbers $(x - e^{2\pi i})(x - e^{\frac {2\pi}{5} i})(x - e^{\frac {4\pi}{5} i})(x - e^{\frac {6\pi}{5} i})(x - e^{\frac {8\pi}{5} i})$ – Doug M Dec 13 '20 at 00:47
  • @DougM this is over a galois field $GF_2$. So unfortunately, in this case I do not think that factorization is valid for this problem? – user840091 Dec 13 '20 at 00:55
  • It ends here. $x^4+x^3+x^2+x+1$ is irreducible over $GF(2)$. Meaning that there are relatively few binary cyclic codes of length five. The repetition code, the overall parity code and the trivial codes :-) – Jyrki Lahtonen Dec 13 '20 at 06:16
  • A general result relevant here is that the cyclotomic polynomial $\Phi_n(x)$, $n$ odd, is irreducible over $GF(2)$ if and only if $2$ generates the multiplicative residue class group $\Bbb{Z}_n^$. Here $$\Bbb{Z}_5^={2^0=1,2^1=2, 2^2=4, 2^3=3}$$ so this holds. The proof uses Galois theory of finite fields. For example here. – Jyrki Lahtonen Dec 13 '20 at 06:21

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Over GF(2)$=\Bbb F_2$ we have $1=-1$ and $2=0$. The factorization (using prime factors in $\Bbb F_2[x]$) of the given polynomial is already $$ x^5+1 =(x+1)(x^4+x^3+x^2+x+1)\ . $$ To see that $f=(x^4+x^3+x^2+x+1)$ is irreducible (= prime in $\Bbb F_2[x]$), it is enough to check there is no factor of degree one or two. The irreducible factors in degrees one and two are $x$, $(x+1)$, and $(x^2+x+1)$. The factors in degree one are easily excluded. (Since $f(0)=f(1)=1$.) And the rest by division with rest of $f$ by $(x^2+x+1)$ is $x+1$, so the factor in degree two is also excluded.

A computer algebra system is useful in such cases, for instance sage delivers the factorization:

sage: R.<x> = PolynomialRing(GF(2))
sage: factor(x^5+1)
(x + 1) * (x^4 + x^3 + x^2 + x + 1)

sage: # note that x^2 + 1 is reducible... 
sage: factor(x^2+1)
(x + 1)^2
dan_fulea
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