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 ?