# R^2 a field?

1. Sep 18, 2010

### quasar987

Hello,

Am I missing something or is R^2 a field with the obvious component wise addition and multiplication (a,b)*(c,d)=(ac,bd)?

2. Sep 18, 2010

### rasmhop

EDIT: Completely ignore this. Didn't think it through. For an example of why it's false see Office Shredder's reply.

Yes if R is a field, then R^2 is a field (clearly commutative, and (a,b) has inverse (1/a,1/b) ).

Last edited: Sep 18, 2010
3. Sep 18, 2010

### Office_Shredder

Staff Emeritus
What's the inverse of (3,0)?

4. Sep 19, 2010

### quasar987

Ok, that's what I was missing :)

5. Sep 19, 2010

### Landau

If R, S are non-trivial integral domains, then their direct product $R\times S$ is never an integral domain, because it always has zero-divisors: (a,0).(0,b)=0. In particular, this holds for fields.