p(x) = x6 + x5 + 1
is primitive (hence,
irreducible) over GF(2) . Use p(x)
to construct a log/antilog table for GF(26) .
p(x) = x2 + x + 2
is primitive (hence,
irreducible) over GF(3) . Use p(x)
to construct a log/antilog table for GF(32)
.
ab = 0
Prove
that R is not a field.
DEGREE 4 ... 3 37D ...
o
Since p(x) is irreducible
and of degree 3, it follows that
GF(24) =
GF(2)[x] mod p(x)
List
all the elements of GF(24) in the above representation, i.e., in terms of
ksi = x mod p(x).
o
Let ksi = x mod p(x).
Why is { ksik} not a complete list of all the non-zero elements of GF(24)?
Last Modified: March 1, 2001