The krull dimension of $\Bbb{Z}$ and artinian rings 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?
 A: 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?
A: 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 $0$.
A: Fields are obviously artinian, as they have only trivial ideals.
Suppose $R$ is an artinian ring which is not artinian and let $x\in R$ an element which is neither $0$ nor invertible. Consider the descending chain of ideals
$$
(x)\supset(x^2)\supset\cdots\supset(x^n)\supset(x^{n+1})\supset \cdots.
$$
The chain stabilizes by assumption.
If $(x^n)=(0)$ eventually, $x$ is a nilpotent element.
If $(x^n)=(x^{n+1})\neq(0)$ eventually, we can write $x^n=yx^{n+1}$ for some $y\in R$ from which we get $x^n(xy-1)=0$, i.e. $x$ is a $0$-divisor.
In either case $R$ is not a domain, hence the ideal $(0)$ isn't prime.
