# Non-Zero Rings

• November 19th 2008, 08:57 AM
pkr
Non-Zero Rings
http://i453.photobucket.com/albums/q...y188/math9.jpg

This question confuses me in many ways, by "non-zero" is it meant a ring with no divisors of zero?

As an ID is a commutative ring-with-a-1 no divisors of zero, I guess i'm going to have to show there does exist a divisor of zero in R1 X R2?

Not sure at all how to go about starting this one...
• November 19th 2008, 10:10 AM
JaneBennet
It depends on how you define rings. Some authors insist that a ring must contain a nonzero multiplicative identity. Others do not insist on this condition. If you use the second alternative, then it is possible for a ring to contain just one element, namely the zero element.

A nonzero ring would then be a ring that contains at least one nonzero element.

Here’s a hint to get you started. Let $a_1$ be a nonzero element in $R_1$ and $a_2$ be a nonzero element in $R_2.$ Then $(a_1,0)$ and $(0,a_2)$ are nonzero elements in $R_1\times R_2.$ Now multiply them together. What do you get?
• November 19th 2008, 10:39 AM
pkr
You would get (0,0) in R1 X R2 which would imply this ring has a element, say Q, such that Q multiplied by any element = Q, so it isn't an integral domain?

Obviously i'd write out the proof more concise than, but am I getting at the right idea?
• November 19th 2008, 10:48 AM
JaneBennet
Quote:

Originally Posted by pkr
You would get (0,0) in R1 X R2 which would imply this ring has a element, say Q, such that Q multiplied by any element = Q, so it isn't an integral domain?

What is the definition of an integral domain?
• November 19th 2008, 10:53 AM
pkr
Commutative ring with a 1, no divisors of zero.

So, because a1 and a2, are non-zero elements of R1 and R2, and R1 X R2 = (0, 0) implies R1 X R2 does posses a divisor of zero, and thus cannot be a integral domain.