# Z2xZ2

• Nov 24th 2012, 12:13 PM
seventhson
Z2xZ2
Hi,

Is Z2xZ2 a field?
• Nov 24th 2012, 12:20 PM
Plato
Re: Z2xZ2
Quote:

Originally Posted by seventhson
Is Z2xZ2 a field?

What are the field axioms and are they satisfied?
• Nov 24th 2012, 01:08 PM
Deveno
Re: Z2xZ2
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).