
198 Abstract Field Extensions
elements, 0 and 1. We note that f is irreducible, so we may adjoin an element ζ such
that ζ has minimal polynomial f over Z
2
. Then ζ
2
+ ζ + 1 = 0 so that ζ
2
= 1 + ζ
(remember, the characteristic is 2) and the elements 0,1,ζ ,1 + ζ form a field. This
follows from Theorem 5.10 (generalised). It can also be verified directly by working
out addition and multiplication tables:
+
0 1 ζ 1 + ζ
0 0 1 ζ 1 + ζ
1 1 0 1 + ζ ζ
ζ ζ 1 + ζ 0 1
1 + ζ 1 + ζ ζ 1 0
· 0 1 ζ 1 + ζ
0 0 0 0 0
1 0 1 ζ 1 + ζ
ζ 0 ζ 1 + ζ 1
1 + ζ 0 1 + ζ 1 ζ
A typical calculation for the second table runs like this:
ζ (1 + ζ ) = ζ + ζ
2
= ζ + ζ + 1 = 1
Therefore Z
2
(ζ ) is a field with ...