# Is the ideal $\mathbb{Z}\times\mathbb{Z}\times{0}$ of $\mathbb{Z} \times\mathbb{Z} \times\mathbb{Z}$ prime? Maximal?

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

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.
$I$ 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 $R/I$.