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Fraleigh Section 23

12.Give an example to show that a factor ring of an integral domain may be a field

13.Give an example to show that a factor ring of an integral domain may have divisors of $0$

14.Give an example to show that a factor ring of a ring with divisors of 0 may be an integral domain.

For 12, $Z/2Z$ works

For 13, $Z/4Z$ works

For 14, $4Z/8Z$ ( isomorphic to Z$_2$?) works. -> I just figured out 4Z is not a ring with divisors of 0...

I'm unsure about my answer for 14. Is it correct? And any other examples for 12,13,14?

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InfimumMaximum
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    I don't understand your notation in 14. 4Z is an ideal not a ring, and even if you allow rings without identities so 4Z is a ring then it still wouldn't have zero divisors. – Noah Snyder Aug 05 '13 at 02:33
  • @NoahSnyder I just edited before seeing your comment! Thank you noah! – InfimumMaximum Aug 05 '13 at 02:34
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    Literally any ring with zero divisors works for 14 by considering a maximal ideal. – Karl Kroningfeld Aug 05 '13 at 02:36
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    Indeed 4Z/8Z is iso to Z/2Z as an abelian group, but as a ring it isn't even unital (it's just two elements, 0 and 4, and every multiplication yields 0). You're looking for (Z/8Z)/(2), which is indeed iso to Z/2Z as a ring. – anon Aug 05 '13 at 02:45
  • @anon Oh! So Z/8Z is a ring with a zero divisor since, for example, 2*4 = 0 and (Z/8Z)/(2) is an integral domain since Z/2Z is an integral domain! Right?! – InfimumMaximum Aug 05 '13 at 02:57
  • Yes. Note (Z/8Z)/(2) is iso to Z/2Z since the only cosets of (2) in Z/8Z are 0+(2) and 1+(2), and it is easy to verify this has the same addition and multiplication table as Z/2Z. Also, the ideal (2) in Z/8Z has four elements (necessary since to get a factor ring with two elements out of a ring with eight the ideal would need to have size four), which may be why you were thinking of the number four in your proposed example. – anon Aug 05 '13 at 03:08

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The problem with trying to use $4\Bbb Z$ as a ring is that it is a ring without identity. The ring $4\Bbb Z/8\Bbb Z$ is not isomorphic to $\Bbb Z_2$. In fact, all of its elements are zero divisors, so it is definitely not a domain.

Your example is certainly a start, but you're going to have to make sure you're looking at a ring with identity (as I suspect your book wants you to.)

Hint: $R/I$ is a domain iff $I$ is a prime ideal. Consider $R=\Bbb Z/4\Bbb Z$.

rschwieb
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