how can we find factorization of $x^{15}-1$ in $\mathbb{Z}/11\mathbb{Z}$?
I already factorize $x^{15}-1=(x-1)(x^2+x+1)(x^4+x^3+x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$ in $\mathbb{Z}[x]$.
it was relatively easy factorizing in $\mathbb{Z}/2\mathbb{Z}$ $x^{15}-1=(x-1)(x^2+x+1)(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$ in $(\mathbb{Z}/2\mathbb{Z})[x]$.
thank you very much in advance.
This can be done by hand with the aid of a bit of ad hoc -trickery.
As John Hughes and Gerry Myerson explained, the polynomial $x^5-1$ splits into linear factors over the field $F=\Bbb{Z}_{11}$. The multiplicative group $K^*$ is cyclic of order ten (we can easily verify that $2$ is a generator), so all the non-zero squares in $K$ are zeros of $x^5-1$. This gives us the linear factors $$ x-1,x-4,x-9,x-16=x-5,\ \text{and}\ x-25= x-3 $$ of $x^{15}-1$.
As $3\mid 15$, the third cyclotomic polynomial $x^2+x+1=(x^3-1)/(x-1)$ is also a factor. The fun part is to factor the degree eight cyclotomic polynomial $$ \Phi_{15}(x)=x^8-x^7+x^5-x^4+x^3-x+1. $$ Its zeros (in some extension field of $K$) are roots of unity of order $15$, and the factors are the minimal polynomials of the said roots of unity. We recall that the multiplicative group of the quadratic extension field $K=\Bbb{F}_{11^2}$ is also cyclic. More precisely, $K^*\simeq C_{120}$. As $15\mid 120$, all the fifteenth roots of unity are in $K$. Because $[K:F]=2$, the minimal polynomials are all quadratic. So we know that $\Phi_{15}(x)$ factors into a product of four irreducible quadratics.
Let $\alpha$ be one of the zeros of $\Phi_{15}$ in $K$. By the Galois theory of extensions of finite fields (Frobenius automorphism) we know that the other zero of the minimal polynomial $m_\alpha(x)$ over the prime field is $\alpha^{11}$. So $$ m_\alpha(x)=(x-\alpha)(x-\alpha^{11})=x^2-[\alpha+\alpha^{11}]x+\alpha^{12}\in F[x]. $$ Next we observe that the constant term $\beta=\alpha^{12}$ is a root of unity of order five. As $\alpha$ ranges over the set of roots of unity of order fifteen, $\beta$ will range over the roots of unity of order five (if $\alpha$ corresponds to $\beta=\alpha^{12}$ then $\alpha^k, k=2,4,8$, corresponds to $\beta^k$). So we have deduced that the constant terms of the four quadratic factors are $4,9,3$ and $5$ in some order.
The remaining task is to figure out the relation between the coefficient $\gamma=\alpha+\alpha^{11}$ of the linear term and that of the constant term $\beta=\alpha^{12}$. Up to this point we have been running on an autopilot, but now I needed trickery (but there may be a more general clever way that I have missed). Let us write $\omega=\alpha^5$. It is a third root of unity in $K$. The other third root of unity is $\omega^2=\alpha^{10}$. They are roots of $\Phi_3(x)=1+x+x^2$, so we have the relation $1+\omega+\omega^2=0$. This comes handy as it implies that $$ \begin{aligned} \gamma^2&=(\alpha+\alpha^{11})^2\\ &=\alpha^2+2\alpha^{12}+\alpha^{22}\\ &=\alpha^{12}+\alpha^2(1+\alpha^{10}+\alpha^{20})\\ &=\beta+\alpha^2(1+\omega^2+\omega)\\ &=\beta. \end{aligned} $$ So we know that $\gamma$ is one of the square roots of $\beta$. As $\beta^5=1$ we know that $(\beta^3)^2=\beta$, so the alternatives are $\gamma=\pm\beta^3$. Which one is it? Of the candidates $\beta^3$ has order five but $-\beta^3$ is a tenth root of unity. We can settle this by calculating $\gamma^5$: $$ \begin{aligned} \gamma^5&=(\alpha+\alpha^{11})^5\\ &=\alpha^5+\binom 51\alpha^{15}+\binom 52\alpha^{25}+\binom53\alpha^{35}+\binom54\alpha^{45}+\alpha^{55}\\ &=\omega+5+10\omega^2+10\omega+5+\omega^2\\ &=10+11\omega+11\omega^2\\ &=-1. \end{aligned} $$ Therefore $\gamma$ always has order ten, and we can deduce that $\gamma=-\beta^3$. Thus $$ m_\alpha(x)=x^2+\beta^3x+\beta. $$ Plugging in $\beta=4,9,5,3$ gives the respective factors $$ x^2+9x+4,x^2+3x+9,x^2+4x+5,\quad\text{and}\quad x^2+5x+3. $$
In such cases i recommend using a computer algebra system. For instance sage.
We have the factorizations in the polynomial rings $S=\Bbb Q[x]$, and respectively $R=\Bbb F_{11}[x]$:
sage: S.<x> = PolynomialRing(QQ)
sage: factor( x^15 - 1 )
(x - 1) * (x^2 + x + 1) * (x^4 + x^3 + x^2 + x + 1)
* (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
sage: R.<x> = PolynomialRing(GF(11))
sage: factor( x^15 - 1 )
(x + 2) * (x + 6) * (x + 7) * (x + 8) * (x + 10)
* (x^2 + x + 1)
* (x^2 + 3*x + 9) * (x^2 + 4*x + 5) * (x^2 + 5*x + 3) * (x^2 + 9*x + 4)
sage: factor(x^4 + x^3 + x^2 + x + 1)
(x + 2) * (x + 6) * (x + 7) * (x + 8)
sage: factor(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
(x^2 + 3*x + 9) * (x^2 + 4*x + 5) * (x^2 + 5*x + 3) * (x^2 + 9*x + 4)
Now humanly, using the above as a substitute for typing.
Modulo eleven, the first factor over $\Bbb Z$ becomes $x-1=x+10$ over $F=\Bbb F_{11}$. The factor $x^2+x+1$ cannot be factored over $F$, no root in $F$. The factor $x^4+x^3+x^2+x+1$ (cyclotomic polynomial $\Phi_5\in\Bbb Q[x]$) splits, this is related to the polynomial $x^5-1$, and to the fact that $F^\times$ has order ten. (And is cyclic.) The remained polynomial of degree $8$ is the work, but observe that it is a reciprocal polynomial.
> Factorization((PolynomialRing(GF(11)))!(x^15-1));
[
<x + 2, 1>,
<x + 6, 1>,
<x + 7, 1>,
<x + 8, 1>,
<x + 10, 1>,
<x^2 + x + 1, 1>,
<x^2 + 3*x + 9, 1>,
<x^2 + 4*x + 5, 1>,
<x^2 + 5*x + 3, 1>,
<x^2 + 9*x + 4, 1>
]
Run the above mamga program, and the answer is:
x^15-1=(x+2)(x+6)(x+7)(x+8)(x+10)(x^2+x+1)(x^2+3x+9)(x^2+4x+5)(x^2+5x+3)(x^2+9x+4).
