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In an integral domain, $\{0\}$ is always a prime ideal. What about maximal ideal? $(0)$ is a prime ideal in $\mathbb{Z}$ which is an I.D but it is not maximal in $\mathbb{Z}$. So I can not conclude that $(0)$ is always maximal ideal in I.D, right? Then what about an example of ideal which is prime in $\mathbb{R}$ ( real numbers) but not maximal? As every field is an integral domain, can I conclude that $(0)$ is not always maximal ideal of field?

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    $\begingroup$ Waiting for suitable answer. $\endgroup$ – Kavita Sahu Mar 15 '16 at 6:20
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    $\begingroup$ If you have questions on my answer, you are welcome to post comments. It is indeed a complete answer to your question, however, so I'm not sure what you mean by 'suitable'. $\endgroup$ – Alex Wertheim Mar 15 '16 at 6:43
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Let $R$ be a commutative ring with $1$. You can see in many books related to the topic ( Atiyah-Macdonald, 1.2 and if you want more expression, Lemma 3.1 of the book Steps in Commutative Algebra by Sharp ):

$R$ is a field if and only if $R$ has exactly two ideals: $0$ and $R$ itself.

Case 1. $R$ is a field. (e.g $\mathbb{R}$)
In this case $0$ is the only proper ideal (Hence maximal ideal) of $R$.

Case 2. The domain $R$ is not a field.
In this case $R$ has an ideal other than $0$, say $I$. Since $0\subsetneq I$, you know that $0$ is not maximal.

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  • $\begingroup$ Alex Wertheim 's answer is complete and suitable +1. but i add references with a little more expression. $\endgroup$ – user 1 Mar 15 '16 at 7:25
  • $\begingroup$ Good answer! Thanks for fleshing out my hints. $\endgroup$ – Alex Wertheim Mar 17 '16 at 19:38
  • $\begingroup$ you are welcome. $\endgroup$ – user 1 Mar 18 '16 at 6:56
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Hint: what are the ideals in a field? More generally, do you know the fact that for any commutative ring $R$ and ideal $I$, $R/I$ is a field if and only if $I$ is maximal?

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  • $\begingroup$ This statement is correct for commutative rings with 1, but not for arbitrary rings. $\endgroup$ – Loic Aug 21 '20 at 15:19
  • $\begingroup$ @Loic: yes. I've made this explicit in my answer now. (For me, ring always refers to ring with unity; like many others, I call a ring without $1$ a 'rng'.) $\endgroup$ – Alex Wertheim Aug 27 '20 at 1:41

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