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Let $R = \mathbb{Z} \times\mathbb{Z} \times\mathbb{Z}$ and $I = \mathbb{Z}\times\mathbb{Z}\times{0}$. Then which of the following statements is correct?

  1. $I$ is a maximal ideal but not a prime ideal of $R$.
  2. $I$ is a prime ideal but not a maximal ideal of $R$.
  3. $I$ is both maximal ideal as well as a prime ideal of $R$.
  4. $I$ is neither a maximal ideal nor a prime ideal of $R$.

I am stuck on this problem. Can anyone help me please?

What is the general method of checking maximal ideal and prime ideal?

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Proceed by definition. – Calvin Lin Jan 3 '13 at 7:03
I often look at the ring $R/I$. – Hurkyl Jan 3 '13 at 7:03
check if $R/I$ is field or integral domain. – Ram Jan 3 '13 at 7:05
General comments about your question: an ellipsis consists of 3 dots, not 15, and each sentence should end with at most one question mark. Also, remember: to get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people are much more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Jan 3 '13 at 7:08
up vote 6 down vote accepted

If $R=R_1\times\cdots\times R_n$ and $I=I_1\times\cdots\times I_n$, where $I_i$ is an ideal of $R_i$, then $$R/I\simeq R_1/I_1\times\cdots\times R_n/I_n.$$ In your case $R/I\simeq\mathbb{Z}/\mathbb Z \times\mathbb{Z}/\mathbb Z \times\mathbb{Z}/(0)\simeq \mathbb Z$. Since $\mathbb Z$ is an integral domain, but not a field, your ideal is prime, but not maximal.

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$I$ is a prime ideal since it satisfies the definition of a prime ideal. Alternatively, since $1 \in R$, and $R / I$ is an integral domain, hence $I$ is a prime ideal

$I$ is not a maximal ideal, since (by definition) it is contained in another prime ideal. Alternatively, since $1\in R$, and $R/I$ is not a field, hence $I$ is not a maximal ideal.

Generally speaking, it is better (and often easier) to check the properties of $R/I$.

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