Krull Dimension

Question

I'm studying Krull dimension and I'm confused about the definition of $\text{ht}(P)$, which is as I understand is the following: let $$P_0\subset P_1\subset\dots\subset P_n=P$$ be a chain of prime ideals we say the length of this chain is $n$, and we define $\text{ht}(P)$ to be the supremum of such lengths.

I'm confused how can a prime ideal $P$ have $0$ as higeht, can't we always write this chain$$(0)\subset P_1=P$$ and there fore $\text{ht}(P)\ge 1$. Thank you for your help.

Answer 1

$(0)$ isn't generally a prime ideal. In fact, it is a prime ideal if and only if the ring is an integral domain.

Now, when $(0)$ is a prime ideal, then it clearly has zero height.

On the other hand, as Marc points out in the comments below, if you're dealing with a ring in which $(0)$ is not a prime ideal, but which does have at least one prime ideal $P$ of finite height (say $n$), then there necessarily exists a prime ideal of height $0$. Why is this? Well, since $P$ has height $n$, then there is a chain of prime ideals $P_0\subset\cdots\subset P_n=P$. Then $P_0$ has height $0$, for if not, then there is a prime ideal $Q\subset P_0$, and putting $Q_0=Q$ and $Q_{k+1}=P_k$ for each $0\le k\le n$, then $Q_0\subset\dots\subset Q_{n+1}=P$ is a chain of prime ideals, which is impossible since $P$ has height $n$.

In the zero ring, there are no prime ideals at all, and in a non-zero ring where every prime ideal has infinite height, there will of course be none of height $0$.