Factoring $ x^n-1$ over $ \mathbb{Q}$

In other words, we shall express $ x^n-1$ as a product of irreducible polynomials having coefficients in $ \mathbb{Q}$.

This is harder - the constraint ``having coefficients in $ \mathbb{Q}$'' is more restrictive than ``having coefficients in $ \mathbb{C}$''. First, let us consider some examples:

$ n=1$
$ x-1$ is already irreducible.

$ n=2$
$ x^2-1=(x-1)(x+1)$ is the factorization into irreducibles over $ \mathbb{Q}$.

$ n=3$
$ x^3-1=(x-1)(x^2+x+1)$ is the factorization into irreducibles over $ \mathbb{Q}$.

Exercise 2.12.7   Show $ x^2+x+1$ is irreducible over $ \mathbb{Q}$.

$ n=4$
$ x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1)$ is the factorization into irreducibles over $ \mathbb{Q}$.

Exercise 2.12.8   Show $ x^2+1$ is irreducible over $ \mathbb{Q}$.

In general, we have a ``formula'' for the irreducible factors of $ x^n-1$. To state this we need a definition.

Definition 2.12.9   Let $ n=p_1^{e_1}...p_k^{e_k}$ be the prime factorization of an integer $ n>1$. Define the Möbius function $ \mu:\mathbb{N}\rightarrow \{0,-1,1\}$ by: $ \mu(1)=1$ and, if $ n>1$ then

\begin{displaymath}
\mu(n)=
\left\{
\begin{array}{cc}
(-1)^k,& {\rm if}\ e_1=...=e_k=1,\\
0, &{\rm if\ some}\ e_i>1.
\end{array}\right.
\end{displaymath}

For each $ m>1$, define the $ m^{th}$ cyclotomic polynomial by

$\displaystyle C_m(x)=\prod_{d\vert m}(x^d-1)^{\mu(m/d)}.
$

Since cyclotomic polynomials are useful for the construction and theory of finite fields, they are also important in algebraic coding theory.

Theorem 2.12.10   For each $ m>1$, $ C_m(x)$ is irreducible over $ \mathbb{Q}$.

For the proof, which goes beyond the scope of this book, see Lang [La], Ch VIII, §3, or Theorem 6.5.5 of Herstein [Her].

Theorem 2.12.11   We have

$\displaystyle x^n-1=\prod_{m\vert n}C_m(x),
$

where $ C_m(x)$ is as above.

This follows from the Möbius inversion formula (see Lidl, Pilz [LP], Ch. 3, §13).

Example 2.12.12   According to this theorem, $ x^4-1=C_1(x)C_2(x)C_4(x)$, where $ C_1(x)=1$, $ C_2(x)=\frac{x^2-1}{x-1}$, and $ C_4(x)=(x-1)^0(x^2-1)^{-1}(x^4-1)^1=x^2+1$. This is the same result as the calculation ``by hand'' above.

Exercise 2.12.13   Show $ C_5(x)=x^4+x^3+x^2+x+1$. More generally, show that for any prime $ p$, $ C_p(x)=x^{p-1}+...+1$.

Exercise 2.12.14   Completely factor $ x^8-1$ into irreducibles over $ \mathbb{Q}$.

Exercise 2.12.15   Completely factor $ x^{10}-1$ into irreducibles over $ \mathbb{Q}$.

Exercise 2.12.16   Completely factor $ x^{12}-1$ into irreducibles over $ \mathbb{Q}$.

Exercise 2.12.17   Compute $ C_{i}(x)$, $ 1\leq i\leq 9$.



David Joyner 2007-09-03