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In some text book of Commutative Algebra, the authors defined the height of an ideal $I$ of a commutative ring $R$ is the maximal of length of a prime ideal chains : $\mathfrak{p}_{0}\subset \mathfrak{p}_{1}\subset...\subset \mathfrak{p}_{k}=\mathfrak{p}$

The Krull dimension of a ring $R$ is defined to be the sup ht$\mathfrak{p}$ for $\mathfrak{p}$ is a prime ideal of $R$.

My first question is : 1- Does $\mathfrak{p}_0$ must be different from the zero ideal ?

In this link the author claimed that $\mathbb{Z}/n\mathbb{Z}$ has Krull dimension 0. Now, take $n=4$, then in $\mathbb{Z}/4\mathbb{Z}$ we have : $\bar{0}\subset \bar{2}$, then I think the Krull dimension is 1.

The second question is : 2- What is wrong in my argument ?

Update Thanks to Sanchez for the answer. My question now is what is the prime ideal in $\mathbb{Z}/n\mathbb{Z}$? Can we list them to conclude that $\mathbb{Z}/n\mathbb{Z}$ has dimension 0 ?

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4  
1. No. 2. 0 is not a prime ideal in $\mathbb{Z}/4\mathbb{Z}$, for 2 is a zero divisor. – user27126 Dec 4 '12 at 4:26
    
@morse: what you wrote in the first paragraph is the height of a prime ideal. – user18119 Dec 4 '12 at 7:57
1  
Your questions show not much effort to do things by yourself. Do you know the ideals of $\mathbb{Z}/n\mathbb{Z}$? Do you know the fundamental theorem of isomorphism for rings? How look the quotient ring of $\mathbb{Z}/n\mathbb{Z}$ by an arbitrary ideal? Which one of these quotients are integral domains (=fields)? – user26857 Dec 4 '12 at 8:57

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