Prime ideals in $R=\mathbb Z\times\mathbb Z$

Question

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 $R$?

1. 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.
2. Why isn't $I=\mathbb Z\times p\mathbb Z$ a prime ideal in $R$ ($p$ prime)? Isn't this correct: $R/I=(\mathbb Z\times\mathbb Z)/(\mathbb Z\times p\mathbb Z)\cong\mathbb Z/\mathbb Z\times \mathbb Z/p\mathbb Z\cong 0\times \mathbb Z_p\cong \mathbb Z_p$

Answer 1

The text of the assignment is wrong. In fact:

1. Usually (as in "this fact pops out all the time"), the definition of prime ideal of $R$ is: an ideal $P$ such that $P\ne R$ and, for all $a,b\in R$, $ab\in P$ only if $a\in P$ or $b\in P$. So $R$ should not be considered a prime ideal.

2. The prime ideals in $A\times B$ are actually $\mathfrak p\times B$ or $A\times \mathfrak q$, where $\mathfrak p,\mathfrak q$ are primes of $A$ or $B$ respectively. The reason is essentially the one you've given in (2).