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I got the following question. The polynom $x^5-1$ should be factorized over $F_{11}$.

I have this as a first solution: $x^5-1=(x-1)(x+1)(x^2+1)(x-1)$ could this be possible?

I don't know which irred. polynomials I can use for factorization. Can I use polynomials of higher degree than 1 in $F_{11}$?

Another question is regarded to the factorization of $x^5-1$ over $F_{19}$. But this polynomial should split into 3 factors.

I'm not totally sure which irred. polynomials I can use for factorization. I thought that I only can use polynomials those degree divides n in $F_p^n$

Thanks for your help!

Ovomaltine
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1 Answers1

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Over any finite field, there are irreducible polynomials of every degree $n \geq 1$, so any degree of irreducible polynomial is possible as a factor of a polynomial.

However, in this case, the polynomial $x^5-1$ has a very special form: its roots satisfy $x^5=1$, which means that they are elements of order 5 inside the multiplicative group of the field in which they reside. The multiplicative group of units of $F_{11}$ is of order 10 and therefore has a subgroup of order $5$; all of the elements of this subgroup will satisfy $x^5=1$ (by Lagrange's theorem), hence they will be roots of $x^5-1$. This means that $x^5-1$ has five distinct roots in $F_{11}$. Therefore, you must have made an error in your factorization; notice, in particular, that $x+1$ cannot be a factor, because then $-1$ would be a root of $x^5-1$, which it is not.

Over $F_{19}$, the situation is different, because its multiplicative group has order 18, which is not divisible by 5. Therefore, the multiplicative group of $F_{19}$ does not contain any elements of order 5, which means that $F_{19}$ contains no roots of $x^5-1$ except for $x=1$. Therefore, $x^5-1$ does not split completely into linear factors over $F_{19}$.

Brent Kerby
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  • Brent Kerby: Thank you for your great explanation. The hint with the subgroups was good. But regarding $(x^5-1)$ in $F_{19}$: This is an task of an exercise where the additional hint was given, that this polynomial $(x^5-1)$ should be factorized by three factors. So is this hint a deception? Thank you! – Ovomaltine Feb 25 '15 at 07:20
  • The hint is good. Over $F_{19}$ it does factor into three irreducible polynomials. – Brent Kerby Feb 25 '15 at 19:04
  • Ok :-) Can you give me a short hint how to find the right polynomials? I started with division by $(x-1)$. Then I tried to find some irred. polyn. for the rest $(x^4+x^3+x2+x+1)$ like $(x^2+x+1)$ and $(x^3+x+1)$ but they were all not right. Which polynomials should I consider? – Ovomaltine Feb 25 '15 at 19:14
  • You can just set $x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)$, expand the right side, and equate coefficients. This will give you a system of 4 equations in the four unknowns $a,b, c$, and $d$. – Brent Kerby Feb 25 '15 at 21:57