Finding a nonprime ideal of Z x Z

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Sorry about the formatting, LaTeX doesn't seem to be working, it seems to be giving garbage output.

Homework Statement


Find a nontrivial proper ideal of Z x Z that is not prime.

Homework Equations


Defn: An ideal N in a commutative ring R is prime, if ab in N implies, either a in N, or b in N, for all a, b in R.

Thm: An ideal of N is prime in R iff R/N is an integral domain.

The Attempt at a Solution



The solution at the back of the book is 4Z x {0}, however I don't see how this can be true. If we apply the definition of a prime ideal, we can show this ideal is in fact prime.

The product of 2 typical elements of N = 4Z x {0} is (a,b)(c,d) = (ac, bd), with bd = 0. Since b and d are in Z, which is an integral domain, bd = 0 implies b = 0 or d = 0. So (a,b) or (b,d) is in N. Thus N is prime by definition.

Alternatively, we can compute (Z x Z) / (4Z x {0}) = { (a,b) + 4Z x {0} | (a,b) in Z x Z}. Since, by the division algorithm, we can write a = 4q + r for some r = 0, 1, 2, 3, this simplifies to: (Z x Z) / (4Z x {0}) = { (a,b) + 4Z x {0} | a in {0, 1, 2, 3}, b in Z}, which is clearly isomorphic to {0 ,1 , 2, 3} x Z, which is isomorphic to Z by the map f((a,b)) = 4*a + b, for (a,b) in {0 ,1 , 2, 3} x Z. But Z is a domain, so (Z x Z) / (4Z x {0}) is also a domain and by the theorem above, 4Z x {0} is prime.

So did I make a huge mistake somewhere, or is the solution 4Z x {0} wrong. If so, how do I find the correct solution?
 

Answers and Replies

  • #2
Dick
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You don't want to consider the product of two elements in 4Zx{0}. You want to consider the product of two elements in ZxZ. (2,0)*(2,0)=(4,0). (4,0) is in 4Zx{0}. (2,0) isn't.
 
  • #3
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You don't want to consider the product of two elements in 4Zx{0}. You want to consider the product of two elements in ZxZ. (2,0)*(2,0)=(4,0). (4,0) is in 4Zx{0}. (2,0) isn't.

Ahh, yes. Quite a silly mistake to make.

Hmmm, that would mean by second proof is also wrong. However, I can't see any mistakes there. Any ideas?
 
  • #4
Dick
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Ahh, yes. Quite a silly mistake to make.

Hmmm, that would mean by second proof is also wrong. However, I can't see any mistakes there. Any ideas?

{0,1,2,3}xZ is not isomorphic to Z, and it's not an integral domain. (2,0)*(2,0)=(0,0). f((a,b))=4*a+b is NOT a ring isomorphism.
 

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