Factorization of $x^5-1$ over $F_{11}$ and $F_{19}$ 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!
 A: 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 $F_{19}$.
