How do I obtain the result that all prime ideals in $R=\mathbb Z\times\mathbb Z$ are $0\times\mathbb Z$, $\mathbb Z\times 0$ and $R$?
- I see that these are all prime ideals and because of $(1,0)\cdot(0,1)=(0,0)\in I$ we have either $(1,0)\in I$ or $(0,1)\in I$, or both. These cases now somehow yield the prime ideals listed above. But I don't understand how $(1,0)\in I\implies I=\mathbb Z\times 0$ if $(0,1)\notin I$. Obviously $\mathbb Z\times 0\subseteq I$ - the other inclusion is what I don't get.
- Why isn't $I=\mathbb Z\times p\mathbb Z$ a prime ideal in $R$ ($p$ prime)? Isn't this correct: $R/I=(\mathbb Z\times\mathbb Z)/(\mathbb Z\times p\mathbb Z)\cong\mathbb Z/\mathbb Z\times \mathbb Z/p\mathbb Z\cong 0\times \mathbb Z_p\cong \mathbb Z_p$