Corrections:
- The pattern identified in the question is wrong. The core reason is this. $(0)$ is not a prime ideal of $\mathbb{Z}^2$ because $\mathbb{Z}^2$ is not an integral domain. For proof, consider $(0x+1y)(1x+0y) = 0x+0y$, so $\mathbb{Z}^2$ has nonzero zero divisors. Nonzero zero divisors, more generally, demonstrate the failure of the complement of $(0)$ in $\mathbb{Z}^2$ to be multiplicatively closed. The other parts of the setup build on this mistake and infer a pattern that doesn't hold.
I was reading about the definition of the going-up property and wondering what chains of prime ideals typically look like.
Can we use chains of prime ideals to define a notion of dimension? If so, how do we do it?.
I'll define a maximal chain of prime ideals as a strict ascending chain of prime ideals that cannot be extended into a larger chain, and, additionally, has the longest possible length of any such chain. So, these chains satisfy both a local optimality property and a global one.
I tried looking around for maximal examples in easy to understand rings: $\mathbb{Z}^n$.
The maximal strict chains of prime ideals of $\mathbb{Z}$ look like:
$$ (0) < (p) < (1) $$
A maximal strict chain of prime ideals of $\mathbb{Z}^2$ looks like the following, where a pair $(a, b)$ will be denoted as a formal sum $ax+by$.
The chain below is written with redundant generators to make the pattern more obvious.
$$ (0, 0) < (0, py) < (0, y) < (qx, y) < (x, y) $$
Let's also note that because multiplication in each dimension is componentwise, every ideal of $\mathbb{Z}^n$ looks like a product of ideals of $\mathbb{Z}$. For example, in the two-dimensional case:
$$ (ax+by, cx+dy) = (a,c)x + (b,d)y $$
This follows by the Chinese Remainder Theorem and the fact that we can freely and independently choose the coordinates of the coefficient of $ax+by$ and likewise for $cx+dy$.
For $\mathbb{Z}^3$ we can come up with a similar chain of length $7$.
Conceptually, we're laying out chains of ideals for each dimension on top of each other and then subtracting the overlap.
The ideal $(0)$ is also a bit of an odd duck, since its presence as a prime ideal basically just tells us that we're working with an integral domain.
So, if we tentatively define $D(R)$ where $R$ is a ring as the length of a maximum strict ascending chain of prime ideals that does not contain $(0)$, then we get the following nice formula:
$$ D(\mathbb{Z}^n) = 2n $$
For the field $\mathbb{R}$, we get the even nicer result:
$$ D(\mathbb{R}^n) = n $$
Is there a real notion of dimension that this is similar to?
Best Answer
That length $n$, if it is finite, defines Krull dimension of a ring. It enjoys several sensible properties that you can find in most commutative algebra books.
Why would you leave out $\{0\}$ if it's prime? Oh well. Anyhow, $\mathbb Z^n$ still only has Krull dimension $1$ for any $n$. The Krull dimension of a finite product of rings is just the maximum of the rings involved in the product. (See this for example). A maximal chain of prime ideals for $n=2$ is given by $\{0\}\times\mathbb Z\subseteq (p)\times \mathbb Z$. The dimension is given by the number of links: $1$.