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I would like to know of some examples of a prime ideal that is not maximal in some commutative ring with unity.

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sub-rings of $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ and sub-rings in $\mathbb{Z}_n$, but I never thought to check $\{0\}$ –  tmpys Feb 28 at 9:15
    
Would still love some more interesting ones.... –  tmpys Feb 28 at 9:15
    
yes I think $k \mathbb{Z}$ for $k\in\mathbb{Z}$ is an ideal. Am I wrong? I am confused by your question;why would I say it if I did not think it...If you can't help, please just don't comment. Or, in the vein of your question,do you think you are helping with your questions? –  tmpys Feb 28 at 9:38
    
@tmpys You do appear to be confused about the difference between subrings and ideals, and surely it is not unhelpful to point this out to you? (Although I agree that the sentence "I can't confusing you more than you are" is confusing.) –  Derek Holt Feb 28 at 10:27

2 Answers 2

Let $R$ be an integral domain and consider $R[x]/(x) \cong R$. It's not a field (unless $R$ is), so $(x)$ is not maximal. Since $R$ has no zero divisors, $(x)$ is a prime ideal.

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this is how I convinced myself that there were such things, but I still cant think of good clean examples. –  tmpys Feb 28 at 9:20
    
@tmpys: You're going to have to use example rings which contain elements that are neither zero divisors nor units (to make it possible to discriminate between prime and maximal ideals). How many such rings do you know? –  Eric Towers Feb 28 at 9:30
    
Note here that $R$ could itself be a suitable polynomial ring, say $K[y]$ for some suitable $K$. –  Mark Bennet Feb 28 at 9:53

Take $(0)$, the zero ideal, in $\mathbb{Z}$, which is prime as the integers are an integral domain, but not maximal as it is contained in any other ideal.

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The same example works if $\mathbb Z$ is replaced by any other integral domain, which is not a field. –  Martin Sleziak Apr 10 at 7:12

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