1.4 Dimension Theory
A dimension theory normally takes the form of a theorem which says that two notions of dimension are the same.
Following the Appendix of Bruns-Herzog, we will use two notions, one related to the lengths a systems of parameters and one related to the heights of prime ideals. We have not defined these terms, so we will do so first, and then prove out theorem.
As our main tool, we use Krull’s Principal Ideal Theorem (aka the Hauptidealsatz). One can also use Hilbert Polynomials, see Atiyah-MacDonald.
Now, we can start to define dimension.
Let \(R\) be a ring, and \(\mathfrak {p}\) be a prime ideal in \(R\). Then the height of \(\mathfrak {p}\) is the sup of the lengths of chains
The height of an ideal \(I \subset R\) is the infimum of the heights of prime ideals containing \(I\).
Sanity lemma: this definition of height agrees with the previous when \(I\) is prime (all by def).
The Krull dimension of the \(R\) is the supremum of heights of prime ideals \(\mathfrak {p} \subset R\).
The following is a nice sanity check lemma.
Let \(R\) a ring and \(\mathfrak {p}\) a prime ideal, then
This is a “straightforward” application of the correspondance theorem for localizations.
An ideal \(I\) is \(\mathfrak {p}\)-primary if it’s radical is equal to \(\mathfrak {p}\).
Let \((R,\mathfrak {m},k)\) be a local (noetherian?) ring. Let \(\nu (R)\) be the minimum of the lengths of generating sets of \(\mathfrak {m}\)-primary ideals.
Now we state Krull’s height theorem and its (partial) converse.
Let \(R\) be a noetherian ring, and \(I\) a prinicpal ideal. Then each minimal prime ideal over \(I\) (i.e. minimal among primes containing I) has height at most 1.
See the Wikipedia page about Krull’s principal ideal theorem (recieved June 27 2023) for a detailed proof, it uses passing to the quotient and the definition of Artinian rings.
Let \(I = (x_{1}, \ldots , x_{n})\) be an ideal. Then each minimal prime over \(I\) has height at most \(n\).
Use Krull’s prinipal ideal and induct on the number of elements, see Wikipedia for details.
Let \(R\) a noetherian ring, and \(I\) a proper ideal of height \(n\). Then for all \(1 \leq i \leq n\), there exist elements \(x_{1}, \ldots , x_{i}\) such that \(\operatorname {height}(x_{1}, \ldots , x_{i}) = i\).
We will successively apply prime avoidance.
Base case: \(I\) is not contained in any minimal prime of \(R\), otherwise \(I\) would have height zero (if I had height zero, we’re already done). By prime avoidance, we can pick an element \(x_{1}\) which is not in any minimal prime. Thus, any prime containing \(x_{1}\) must also contain a minimal prime, so all such primes have height at least 1.
Pick a prime which is minimal among those containing \((x_{1})\) (equivalently, pick one which is minimal in the ring \(R / (x_{1})\)). By the above discussion, it must have height at least one. By Krull’s principal ideal/height theorem, it must have at most height 1, and we conclude it has height 1.
Inductive step: Copy-paste the previous discussion, with "zero" replaced by "\(n-1\)", "1" replaced by "\(n\)", and "minimal prime of \(R\) " replaced by "minimal prime containing \((x_{1}, \ldots , x_{i-1})\)", "\(x_{1}\)" replaced by "\(x_{i}\)", and "\((x_{1})\) replaced by \((x_{1}, \ldots , x_{i})\)".
Let \((R,\mathfrak {m},k)\) be a local noetherian ring. Let \(n \in \mathbb {N}\). Then the following are equivalent
\(\dim R = n\)
\(\operatorname {height}\mathfrak {m} = n\)
\(\nu (R) = n\).
Let \(R\) be a ring, and \(M\) an \(R\)-module. Let \(x\) be an \(M\)-regular element. Then \(\dim M / xM \leq \dim M - 1\).
As \(x\) is \(M\)-regular, we have that \(x \notin \mathfrak {p}\) for all \(p \in \operatorname {Assoc}M\), otherwise there would be \(M\)-torsion.
By definition, \(\dim M = \dim R / \operatorname {Ann}M\). The associated primes of \(M\) are the same as the minimal primes in \(R / \operatorname {Ann}M\), so \(x\) is not contained in any such prime. But then, any chain of prime ideals in \((R / \operatorname {Ann}M) / (x)\) lifts to a chain in \(R / \operatorname {Ann}M\) of ideals which contain \((x)\). But then the smallest ideal in the chain is not minimal since it contains \(x\), so we can extend it to a minimal prime.
Let \(x \in R\) be an element which is not in any minimal prime ideal. Then \(\dim R / (x) \leq \dim R - 1\)
Take \(M = R\) in the previous theorem.
Other dimension theory necessary? Probably?