# Prime Ideal and Integral Domain

• Feb 3rd 2013, 04:08 PM
spotsymaj
Prime Ideal and Integral Domain
I was given the following question. Let A be a commutative ring with unity, and J an ideal of A. Prove J is a prime ideal if and only if A/J is an integral domain.

This is how I proved it but I dont know if it is complete enough. If anyone has any suggestions please let me know

Suppose that there exists a,b \$in\$ A/J, so that ab = 0. This also means that ab \$in\$ J. We need to prove that there are no zero divisors in A/J. However, since J is a prime ideal, it means that a \$in\$ J or b \$in\$ J. Since a \$in\$ J or b \$in\$ J, this means that a = 0 or b = 0. Thus there means that there are no zero divisors. Hence A/J is an integral domain.

Suppose A/J is an integral domain. Let ab \$in\$ A. Since A/J is an integral domain it has no zero divisors, this means that ab = 0 and either a = 0 or b = 0. Which means that a \$in\$ J or b \$in\$ J and ab \$in\$ J. Thus J is a prime ideal.
• Feb 4th 2013, 05:18 AM
Deveno
Re: Prime Ideal and Integral Domain
you're confusing a and b with the COSETS a+J and b+J. the idea of your proof is correct, but you should be more careful.

suppose J is a prime ideal, and that (a+J)(b+J) = 0+J = J in A/J. this means that ab+J = J, hence ab is in J. Since J is a prime ideal, either a is in J, or b is in J. if a is in J, then a+J = J = 0+J, and similarly for b, so that A/J is an integral domain.

similarly, if we have ab in J, so that ab+J = 0 J = J, then either a+J = 0+J, so that a is in J, or b+J = 0+J, so that b is in J, hence J is a prime ideal.
• Feb 27th 2013, 01:54 AM
arulanandam
Re: Prime Ideal and Integral Domain
Find the number of ideals in a ring Zn where n = product of prime numbers???
I want the answer any body help me
• Feb 27th 2013, 03:09 AM
Deveno
Re: Prime Ideal and Integral Domain
start by looking at the subgroups of the additive group of Zn. are any of these ideals?