Thread: Z2xZ2

Hi,

Is Z2xZ2 a field?

Originally Posted by seventhson

Is Z2xZ2 a field?

What are the field axioms and are they satisfied?

it is clear that the direct product of the groups (Z₂,+) and (Z₂,+) is again an abelian group.

the question is: if we are going to try to impose a multiplicative structure on this group, what should it be? there is more than one way to do so.

here are two possible multiplications (both commutative, so only 10 products are listed):

A:

(0,0)*(0,0) = (0,0)
(0,1)*(0,0) = (0,0)
(1,0)*(0,0) = (0,0)
(1,1)*(0,0) = (0,0)
(0,1)*(0,1) = (0,1)
(1,0)*(0,1) = (0,0) <---hmm, a zero-divisor
(1,1)*(0,1) = (0,1)
(1,0)*(1,0) = (1,0)
(1,0)*(1,1) = (1,0)
(1,1)*(1,1) = (1,1) the "rule" here is (a,b)*(c,d) = (ac,bd) where ac and bd are products in the ring Z₂.

the identity of this multiplication is (1,1).

B:

(0,0)*(0,0) = (0,0)
(0,1)*(0,0) = (0,0)
(1,0)*(0,0) = (0,0)
(1,1)*(0,0) = (0,0)
(0,1)*(0,1) = (1,1)
(1,0)*(0,1) = (0,1)
(1,1)*(0,1) = (1,0)
(1,0)*(1,0) = (1,0)
(1,1)*(1,0) = (1,1)
(1,1)*(1,1) = (0,1) the rule here is: (a,b)*(c,d) = (ac+bd,ad+bc+bd) where the sums and products on the right are in Z₂.

the identity of this multiplication is (1,0).