I have to find a maximal ideal of $\mathbb{Z} \times \mathbb{Z}$ , and a prime ideal that is NOT maximal.

Or, essentially, I want $I$ such that $\mathbb{Z} \times \mathbb{Z} / I$ is a field,

and I want $P$ such that $\mathbb{Z} \times \mathbb{Z} / P $ is a domain but NOT a field.

What's my general strategy for something like this??

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    $\begingroup$ Hint: $(1,0) * (0,1) = 0$. $\endgroup$ – Brandon Carter Mar 12 '14 at 3:35
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    $\begingroup$ More generally, if $R_1$ and $R_2$ are rings, then any ideal of $R_1 \times R_2$ is of the form $I_1 \times I_2$, where $I_1$ is an ideal of $R_1$ and $I_2$ is an ideal of $R_2$. Furthermore, $(R_1 \times R_2)/(I_1 \times I_2) \cong (R_1/I_1) \times (R_2/I_2)$, and a product of nontrivial rings is never a domain $\endgroup$ – zcn Mar 12 '14 at 3:50
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    $\begingroup$ what do you mean by a "nontrivial ring"? $\endgroup$ – terrible at math Mar 12 '14 at 3:54
  • $\begingroup$ I mean a ring that is not the zero ring. A quotient $R/I$ is trivial iff $I = R$ is the unit ideal $\endgroup$ – zcn Mar 12 '14 at 3:56
  • $\begingroup$ Ah okay, thank you. That makes sense. $\endgroup$ – terrible at math Mar 12 '14 at 3:57

Hint 1: What are the maximal ideals in $\mathbb{Z}$? What about the prime ideals in $\mathbb{Z}$? Use this to help you find your answer.

Hint 2: Maximal ideals are always prime ideals (as you seem to already know), so if you have an idea of what the maximal ideals are, the prime ideals that are not maximal should be slightly smaller in some sense...

  • $\begingroup$ The maximal ideals in $\mathbb{Z}$ are all principal, and the prime ideals are exactly the principal ideals $n\mathbb{Z}$ such that $n$ is prime. Is it true then, that any cross product set of $\mathbb{Z}$ is going to have all principal ideas as well? $\endgroup$ – terrible at math Mar 12 '14 at 3:38
  • $\begingroup$ actually I need to correct this a bit. I meant to say that all ideals are principal, and all maximal ideals are nZ, but the same is true for prime with the addition of the zero ideal. $\endgroup$ – terrible at math Mar 12 '14 at 3:41

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