next up previous
Next: Ideals in Up: lecture4 Previous: Minimal polynomials and conjugate

Factoring $ x^n-1$

Recall that every element $ \beta \in GF(q^m)$ has an order that divides $ q^m-1$; thus every element is a root of $ x^{q^m-1}-1$. Put another way, the elements of $ GF(q^m)$ are the $ (q^m-1)$st roots of unity, and these are all the nonzero elements of the field.

Given a field, we divide it into conjugacy classes, taking the minimal polynomial from each. Then based on our observation, we must have $ x^{q^m-1}$ as a product of the minimal polynomials of the nonzero elements.
\begin{example}
\begin{displaymath}x^7-1 = (x+1)(x^3+x+1)(x^3+x^2+1).
\end{displaymath}\end{example}

We can now pursue a more general problem, roots of $ x^n-1$ for other values of $ n$. These $ n$ roots of unity must exist in some field. We find the field, then find the minimal polynomials of the conjugacy classes in the field.

Suppose we have an element $ \beta$ of order $ n$ in some field $ GF(p^m)$. Then $ \beta$ is a root of $ x^n-1$ in that field, and so are the elements $ \beta^2, \beta^3, \ldots, \beta^{n-1}$. We can find such a $ \beta$ if we form the field correctly.

Before doing so, it is interesting to pause a moment and suggest an application of this. Consider the DFT:

$\displaystyle X[k] = \sum_{n=0}^{N-1} x[n] e^{-j2\pi nk/N}.
$

Note that the number $ e^{-j2\pi/N}$ is an $ N$th root of unity. We can define a Fourier transform in a finite field $ GF(p^m)$ of length $ N$, provided that we can find a field in which $ N$th roots of unity exist.

Recall that if $ n\vert p^m-1$, then there are $ \phi(n)$ elements of order $ n$ in $ GF(p^m)$.
\begin{definition}
The smallest positive integer $m$ such that $n\vert q^m-1$ is called the
{\bf order of $q$ modulo $n$}.
\end{definition}
If $ m$ is the order of $ q$ modulo $ n$, then $ GF(q^m)$ is the smallest extension field of $ GF(q)$ in which $ n$th roots of unity exist.
\begin{example}
We are looking for an extension of $GF(2)$ in which 5th roots ...
... it has order 15. Then $\beta
= \alpha^3$ has order 5, as needed.
\end{example}

\begin{example}
We want 13th roots of unity in an extension of $GF(3)$. Note tha...
...laymath}13 \vert 3^3-1
\end{displaymath}so they live in $GF(3^3)$.
\end{example}

\begin{example}
We want to find a field $GF(2^m)$ which has 25th roots of unity...
...}(x)
\end{displaymath}(where the subscript represents the degree).
\end{example}

\begin{example}
Let us find a field $GF(7^m)$ in which $x^{15}-1$ has roots. ...
...s $x^{15}-1$ factors into six irreducible polynomials in $GF(7)$.
\end{example}
If we list the exponents of the primitive roots of unity, we get what are called the cylotomic cosets. For example, for the last example we have the following:

Conjugacy class   Cyclotomic cosets
$ \{1 \}$ $ \leftrightarrow$ {0 }
$ \{ \gamma, \gamma^7, \gamma^{4}, \gamma^{13} \}$ $ \leftrightarrow$ { 1,7,4,13}
$ \{ \gamma^2, \gamma^{14}, \gamma^{8}, \gamma^{11} \}$ $ \leftrightarrow$ {2,14,8,11}
$ \{ \gamma^{3}, \gamma^{6}, \gamma^{12}, \gamma^{9} \}$ $ \leftrightarrow$ {3,6,12,9}
$ \{ \gamma^5 \}$ $ \leftrightarrow$ {5}
$ \{ \gamma^{10} \}$ $ \leftrightarrow$ {10}

The cyclotomic cosets modulo $ n$ with respect to $ GF(q)$ contain the exponents of the $ n$ distinct powers of a primitive $ n$th root of unity with respect to $ GF(q)$, each coset corresponding to a conjugacy class. These cosets provide a shorthand representation for the conjugacy class.


next up previous
Next: Ideals in Up: lecture4 Previous: Minimal polynomials and conjugate
Todd Moon 2009-11-06