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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: $R/I=(\mathbb Z\times\mathbb Z)/(\mathbb Z\times p\mathbb Z)\cong\mathbb Z/\mathbb Z\times \mathbb Z/p\mathbb Z\cong 0\times \mathbb Z_p\cong \mathbb Z_p$
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    $\begingroup$ Typically, the definition of prime ideal requires it not to be the whole ring. $\endgroup$ – user228113 Jan 10 '18 at 12:52
  • $\begingroup$ Can you elaborate how that is relevant? Or were you just refering to $R$ being a prime ideal is unusual? $\endgroup$ – Buh Jan 10 '18 at 12:54
  • $\begingroup$ The latter, though I am more incline to think that it's a mistake in your notes. $\endgroup$ – user228113 Jan 10 '18 at 13:01
  • $\begingroup$ So everything in 2 is correct and that given $I$ is prime? $\endgroup$ – Buh Jan 10 '18 at 13:04
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    $\begingroup$ $0\times \mathbb Z$ and $\mathbb Z\times 0$ are the minimal primes, maybe that was it? $R$ is never deigned to be a prime ideal of itself, in standard circumstances. $\endgroup$ – rschwieb Jan 10 '18 at 14:48
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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).

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  • $\begingroup$ Alright, I'm a little relieved. This raises another question though: I just asked it here: math.stackexchange.com/questions/2599575/… $\endgroup$ – Buh Jan 10 '18 at 13:17
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    $\begingroup$ $R$ itself not usually being considered a prime ideal is exactly analogous to $1$ usually not being considered a prime number. $\endgroup$ – Arthur Jan 10 '18 at 15:41

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