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I am trying to figure out the prime ideals in $\mathbb{Z}_{30}$. My understanding is a little shaky so I would just like to test my understanding of the conditions for a prime ideal.

So for a prime ideal $P$ in $R$ and $a, b \in R$, if $ab \in P$ then $a \in P$ or $b \in P$. Also, $P \neq R$.

So clearly (1) is not a prime ideal as it is equal to $R$.

$P = (0)$ is not prime since we have $ab = 0 \in P$, however $a = 2, b = 15$ is a solution where neither are elements of $P$.

Is this a suitable explanation?

Thanks again for your help as always.

Inazuma
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  • Yes, you are right. In addition, there are some non-trivial prime ideals in $\mathbb{Z}_{30}$ – Focus Feb 11 '18 at 03:55
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    I have previously also been looking at maximal ideals, where any divisor of 30 generates a maximal ideal. However, the "opposite" seems to be true for prime ideals, in that for divisors of 30, we have $ab = 0$, so we can choose $a$ and $b$ to be any divisors of 30 and so they are not prime ideals. Is this the right train of thought, and should I then be looking at non-divisors? – Inazuma Feb 11 '18 at 04:03
  • Nvm I think I may have jumped the gun and need to think about this a bit more. – Inazuma Feb 11 '18 at 04:07
  • In,fact, in $\mathbb{Z}_m$,any maximal ideal is prime ideal. Because finite imtegral domain is field. – Focus Feb 11 '18 at 04:11
  • Thank you for reminding me of that theorem! – Inazuma Feb 11 '18 at 04:15

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