We study fields of order 16. The polynomial
Z/2Z[X]
is irreducible. (Verify!)
Put K = Z/2Z[X]/(f) and c = X + (f). The elements of K can be arranged according to the irreducible divisors of f in Z/2Z[X] to which they belong:
| element | zero of | |
| 1 | X + 1 | |
| c, c2, c4, c8 | X4 + X + 1 | |
| c3, c6, c12, c9 | X4 + X3 + X2 + X + 1 | |
| c5, c10 | X2 + X + 1 | |
| c7, c14, c13, c11 | X4 + X2 + 1 |
The multiplicative group is cyclic of order 15, with generator c. The elements of order 3 belong to the subfield {0, 1, c5, c10} of order 4, isomorphic to Z/2Z[X]/(X2 + X + 1).
The elements of order 5 can be recognized by their exponents (having gcd with 15 equal to 3), but also by the corresponding polynomial, which divides X5 - 1.
Each of the three irreducible polynomials of degree 4 can be used to construct a (or the) field of order 16. For verification you can compute the product of all irreducible polynomials occurring in the second column of the table; it should be X16 - X.
In order to carry out addition on K efficiently, you can write
the elements as Z/2Z-linear combinations of
1, c, c2, c3. For
c4, this expression follows from the definition of
K: c4 = c + 1.
For c5
and higher powers, such a relation can be computed by multiplying
with c and expanding the right hand side by use of equations obtained previously.
For example, c5 = c
c4 = c2 + c.
You can also use that squaring in characteristic 2 is easy: c8 = (c4)2 = (c + 1)2 = c2 + 1.
Try and write all elements this way.