On page thirty of Matsumura, it says that $\Bbb{Z}$ has krull dimension 1 because every prime ideal is maximal. I understand this because for any prime p you have $0 \subset p$.
However, for artinian rings, it says that the dimesion is zero because every prime ideal is maximal. This is where I'm confused. As above, we have $0 \subset p$ for any nonzero prime ideal...but then why isn't he dimension 1?
On
Every prime ideal $\mathfrak p$ in an Artinian ring $R$ is maximal. To see this, let $x \in R / \mathfrak p$, $x \ne 0$. Consider the descending chain $(x) \supset (x^2) \supset \cdots$ to show that $x$ is invertible.
On the other hand, $\mathbb Z$ is a principal ideal domain. Every nonzero prime ideal in a PID is maximal. The word "nonzero" here is the crucial difference.
It follows that $\mathbb Z$ has dimension $1$ whereas Artinian rings have dimension $0$.
Hint: Is $0$ a prime ideal in all rings? If $0$ is a prime ideal, and it is maximal, is there any other prime ideal?