# Maximal and prime ideals of $\mathbb{Z} \times \mathbb{Z}$

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??

• Hint: $(1,0) * (0,1) = 0$. – Brandon Carter Mar 12 '14 at 3:35
• 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 – zcn Mar 12 '14 at 3:50
• what do you mean by a "nontrivial ring"? – terrible at math Mar 12 '14 at 3:54
• I mean a ring that is not the zero ring. A quotient $R/I$ is trivial iff $I = R$ is the unit ideal – zcn Mar 12 '14 at 3:56
• Ah okay, thank you. That makes sense. – 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.
• 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? – terrible at math Mar 12 '14 at 3:38