Hi,

Is Z2xZ2 a field?

Results 1 to 3 of 3

- November 24th 2012, 11:13 AM #1

- Joined
- Jul 2011
- Posts
- 15

- November 24th 2012, 11:20 AM #2

- November 24th 2012, 12:08 PM #3

- Joined
- Mar 2011
- From
- Tejas
- Posts
- 3,150
- Thanks
- 591

## Re: Z2xZ2

it is clear that the direct product of the groups (Z

_{2},+) and (Z_{2},+) 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 Z_{2}.

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 Z_{2}.

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