Let be some Finite Field extension of and odd
Let
The Cyclic Code of length with defining set is the code:
Note that as the Generator Polynomial of divides ,
we only need to ensure that for all .
Remark
If we let such that and set
we find that
and so has elements.
As is cyclic, so is so we find:
Thus any defining set is .
Example
We find
where and have no linear factors,
so they are irreducible.
We also find that Splitting Field of is
In fact, all elements of are roots of
Say is a root of .
Then and also are
So , , must be roots of
The defining set
gives
The defining set
gives
If combines roots from these two sets then
etc.