Abstract algebra: I have a problem, I need to show that if F is a field then F is an integral domain.

easy enough. Let a,b be in F. Assume ab=0 where a is not 0, then since F is a field, we have a^-1ab = b = 0, so every element is a non zero divisor, and so F is an integral domain.

my problem RxR is a Field, and (0,1)(1,0) = (0,0), so (1,0) and (0,1) are zero divisors.

what am I not seeing?

Question

Abstract algebra: I have a problem, I need to show that if F is a field then F is an integral domain.

easy enough. Let a,b be in F. Assume ab=0 where a is not 0, then since F is a field, we have a^-1ab = b = 0, so every element is a non zero divisor, and so F is an integral domain. 

my problem RxR is a Field, and (0,1)(1,0) = (0,0), so (1,0) and (0,1) are zero divisors.

what am I not seeing?

anonymous · Answer

Your proof for the first part is correct. However, what you showed afterwards was that $\Bbb{R}\times\Bbb{R}$ with component wise addition and multiplication has zero divisors and hence $\Bbb{R}\times\Bbb{R}$ not a field even though $\Bbb{R}$ itself is a field.  Are you sure the question was really claiming that $\Bbb{R}\times\Bbb{R}$ is a field?

anonymous · Answer

$\mathbb{R}^2$ is not a field with componentwise multiplication... if you equip it with $(a,b)\cdot(c,d)=(ac-bd,ad+bc)$ it is however

anonymous · Answer

^^ that's called the complex numbers $\mathbb{C}$

zzr0ck3r · Answer

word, yeah I thought I had seen a proof that RxR was a field, but of course it cant be
(1,0)*(0,1) = (0,0)