Dimension

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Definition of Dimension

In the next few posts we define the dimension of a ring and study its properties. The definition itself is not difficult.

Definition 1 The Krull dimension of a ring \(A\) is defined as the supremum of the lengths \(r\) of descending chains of prime ideals of \(A\)

\[\mathfrak{p}_r\supseteq \mathfrak{p}_{r-1}\supseteq\cdots\supseteq \mathfrak{p}_1\supseteq \mathfrak{p}_0\]

and is denoted by \(\dim A\). If no such \(r\) exists, we define \(\dim A=\infty\).

For short, we call the Krull dimension of a ring \(A\) simply the dimension of \(A\). For example, a field \(\mathbb{K}\) has only the prime ideal \((0)\), so \(\mathbb{K}\) is always \(0\)-dimensional.

More generally, we make the following definition.

Definition 2 For an ideal \(\mathfrak{a}\) of a ring \(A\), the dimension of \(\mathfrak{a}\) is defined by

\[\dim \mathfrak{a}=\dim A/\mathfrak{a}\]

For a prime ideal \(\mathfrak{p}\) of \(A\), the codimension \(\codim \mathfrak{p}\) of \(\mathfrak{p}\) is defined as the dimension of \(A_\mathfrak{p}\); for a general ideal \(\mathfrak{a}\), \(\codim \mathfrak{a}\) is defined as the minimum of the codimensions of the prime ideals containing \(\mathfrak{a}\).

Finally, for any \(A\)-module \(M\), the dimension and codimension of \(M\) are defined respectively as the dimension and codimension of \(\ann(M)\).

One point requiring some care: according to the definition above, \(\mathfrak{a}\) acquires two notions of dimension—the dimension as an ideal defined first, and the dimension when regarded as an \(A\)-module—and these two may give different values. Therefore, when we use the notation \(\dim \mathfrak{a}\), we agree that it refers only to the dimension as an ideal of \(A\), as defined first in Definition 2.

Then by §Localization, ⁋Proposition 8, \(\codim \mathfrak{p}\) equals the supremum of the lengths of decreasing chains starting from the prime ideal \(\mathfrak{p}\)

\[\mathfrak{p}=\mathfrak{p}_r\supseteq \mathfrak{p}_{r-1}\supseteq\cdots\supseteq \mathfrak{p}_1\supseteq \mathfrak{p}_0\]

Hence the inequality

\[\dim \mathfrak{a}+\codim \mathfrak{a}\leq \dim A\]

holds. Despite its name, the reverse inequality does not hold in general.

On the other hand, for a local ring \((A, \mathfrak{m})\), we can always prepend \(\mathfrak{m}\) to the beginning of any chain of prime ideals giving \(\dim A\), so necessarily \(\dim A=\codim \mathfrak{m}\).

Computation of Dimension

In general, when we treat dimension we mainly consider the case where the ring \(A\) is noetherian. One of the principal reasons is that Theorem 7 (Principal Ideal Theorem) holds only for noetherian rings. Before computing dimensions in earnest, let us first look at a simple example.

First, through the equivalence between the first and third conditions of §The Jordan-Hölder Theorem, ⁋Theorem 4, we know exactly what \(0\)-dimensional noetherian rings are.

Corollary 3 For a noetherian ring \(A\), \(\dim A =0\) is equivalent to \(A\) being artinian.

On the other hand, by the following proposition we know that in general, if \(\phi:A \rightarrow B\) is integral, then dimension does not change.

Proposition 4 Let \(\phi: A \rightarrow B\) be integral. Then for any prime ideal \(\mathfrak{p}\) of \(A\) containing \(\ker\phi\), there exists a prime ideal \(\mathfrak{q}\) of \(B\) such that \(\mathfrak{p}=\phi^{-1} \mathfrak{q}\). Moreover, for any ideal \(\mathfrak{b}\) of \(B\), \(\dim \mathfrak{b}=\dim \phi^{-1} \mathfrak{b}\).

Proof

The first statement is simply §Integral Extensions and Ideals, ⁋Proposition 1. For the second, \(\dim \mathfrak{b}\geq \dim \phi^{-1}\mathfrak{b}\) holds by the second statement of §Integral Extensions and Ideals, ⁋Proposition 1, and the reverse inequality holds by §Integral Extensions and Ideals, ⁋Corollary 4.

Now we turn our attention to what happens in dimension one. Before that, we give the following somewhat technical definition.

Definition 5 For a prime ideal \(\mathfrak{p}\subseteq A\), we define the \(n\)th symbolic power \(\mathfrak{p}^{(n)}\) of \(\mathfrak{p}\) by the formula

\[\mathfrak{p}^{(n)}=\{a\in A\mid\text{$ba\in \mathfrak{p}^n$ for some $b\in A\setminus \mathfrak{p}$}\}\]

By definition, \(\mathfrak{p}^{(n)}\) is the inverse image in \(A\) of the ideal \((\mathfrak{p}A_\mathfrak{p})^n\) under the localization \(A \rightarrow A_\mathfrak{p}\). Then elements outside \(\mathfrak{p}\) become non-zerodivisors modulo \(\mathfrak{p}^({n})\), and it is clear that \(\mathfrak{p}^{(n)}A_\mathfrak{p}=(\mathfrak{p}A_\mathfrak{p})^n\). Moreover, there is a descending chain of symbolic powers

\[A=\mathfrak{p}^{(0)}\supseteq \mathfrak{p}=\mathfrak{p}^{(1)}\supseteq \mathfrak{p}^{(2)}\supseteq \mathfrak{p}^{(3)}\supseteq\cdots\]

Now we can prove the following theorem.

Theorem 6 (Codimension one Principal Ideal Theorem) Let \(A\) be a noetherian ring and let \(a\in A\) be arbitrary. Let \(\mathfrak{p}\) be minimal among the prime ideals containing the principal ideal \(\mathfrak{a}=(a)\). Then \(\codim \mathfrak{p}\leq 1\).

Proof

It suffices to show that \(\codim \mathfrak{q}=0\) for any prime ideal \(\mathfrak{q}\subsetneq \mathfrak{p}\), and by §Localization, ⁋Proposition 8 this is equivalent to showing that \(\dim A_\mathfrak{q}=0\).

Now in \(A_\mathfrak{p}\), \(\mathfrak{p}A_\mathfrak{p}\) is the unique maximal ideal, so \(\mathfrak{p}\) contains the ideals \(\mathfrak{q}A_\mathfrak{p}\), \((\mathfrak{q}A_\mathfrak{p})^{(n)}\), and \(\mathfrak{a}A_\mathfrak{p}\). In particular, we obtain the two chains

\[\mathfrak{a}A_\mathfrak{p}\subseteq (\mathfrak{q}A_\mathfrak{p})^{(n)}+\mathfrak{a}A_\mathfrak{p}\subseteq \mathfrak{p}A_\mathfrak{p},\qquad \mathfrak{q}A_\mathfrak{p}\subseteq \mathfrak{p}A_\mathfrak{p}\]

On the other hand, since \(\mathfrak{p}A_\mathfrak{p}\) is minimal among prime ideals containing \(\mathfrak{a}A_\mathfrak{p}\), by §The Jordan-Hölder Theorem, ⁋Corollary 8 the quotient \(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p}\) is artinian. Hence the descending chain of symbolic powers

\[(\mathfrak{q}A_\mathfrak{p})^{(1)}+\mathfrak{a}A_\mathfrak{p}\supseteq (\mathfrak{q}A_\mathfrak{p})^{(2)}+\mathfrak{a}A_\mathfrak{p}\supseteq\cdots\]

must terminate. Suppose that \((\mathfrak{q}A_\mathfrak{p})^{(n)}+\mathfrak{a}A_\mathfrak{p}= (\mathfrak{q}A_\mathfrak{p})^{(n+1)}+\mathfrak{a}A_\mathfrak{p}\). Then

\[(\mathfrak{q}A_\mathfrak{p})^{(n)}\subseteq (\mathfrak{q}A_\mathfrak{p})^{(n)}+\mathfrak{a}A_\mathfrak{p}= (\mathfrak{q}A_\mathfrak{p})^{(n+1)}+\mathfrak{a}A_\mathfrak{p}\]

so any \(f\in (\mathfrak{q}A_\mathfrak{p})^{(n)}\) can be written in the form

\[f=\alpha a+g,\qquad g\in (\mathfrak{q}A_\mathfrak{p})^{(n+1)}=(\mathfrak{q}A_\mathfrak{p})^{(n)}\]

and from this \(\alpha a\in (\mathfrak{q}A_\mathfrak{p})^{(n)}\). But in this expression, since \(\mathfrak{p}\) is minimal among primes containing \(\mathfrak{a}\), we have \(a\not\in \mathfrak{q}\) and therefore \(\alpha\in (\mathfrak{q}A_\mathfrak{p})^{(n)}\). That is,

\[(\mathfrak{q}A_\mathfrak{p})^{(n)}=\mathfrak{a}(\mathfrak{q}A_\mathfrak{p})^{(n)}+(\mathfrak{q}A_\mathfrak{p})^{(n+1)}\]

holds. Now passing to \(A_\mathfrak{p}/(\mathfrak{q}A_\mathfrak{p})^{(n+1)}\),

\[(\mathfrak{q}A_\mathfrak{p})^{(n)}=\mathfrak{a}(\mathfrak{q}A_\mathfrak{p})^{(n)}\pmod{\mathfrak{q}^{(n+1)}}\]

and since \(a\in \mathfrak{p}A_\mathfrak{p}=J(A_\mathfrak{p})\), by §Integral Extensions, ⁋Lemma 8 (Nakayama) we have \((\mathfrak{q}A_\mathfrak{p})^{(n)}=0\pmod{(\mathfrak{q}A_\mathfrak{p})^{(n+1)}}\). That is, \((\mathfrak{q}A_\mathfrak{p})^{(n)}=(\mathfrak{q}A_\mathfrak{p})^{(n+1)}\). Localizing this equation at \(\mathfrak{q}\),

\[(\mathfrak{q}A_\mathfrak{q})^{n+1}=(\mathfrak{q}A_\mathfrak{q})^{n}\]

and since \(\mathfrak{q}A_\mathfrak{q}=J(A_\mathfrak{q})\), we have \((\mathfrak{q}A_\mathfrak{q})^{n}=0\). From the equivalence of the second and third conditions of §The Jordan-Hölder Theorem, ⁋Corollary 8, \(A_\mathfrak{q}=A_\mathfrak{q}/(0)\) is artinian, and therefore by Corollary 3 we know that \(\dim A_\mathfrak{q}=0\).

Now using this, we can inductively prove the following.

Theorem 7 (Principal Ideal Theorem) Let \(A\) be a noetherian ring and let \(a_1,\ldots, a_c\in A\) be arbitrary. Let \(\mathfrak{p}\) be minimal among the prime ideals containing \(a_1,\ldots, a_c\). Then \(\codim \mathfrak{p}\leq c\).

That is, in a noetherian ring any prime ideal satisfies the descending chain condition, and the length of a chain starting from \(\mathfrak{p}\) is at most the number of generators of the ideal it minimally contains. Nevertheless, there exist noetherian rings of infinite dimension. ([Nag, Appendix, Example 1])

On the other hand, Theorem 7 (Principal Ideal Theorem) also admits the following converse.

Corollary 8 In a noetherian ring \(A\), a prime ideal \(\mathfrak{p}\) of codimension \(c\) is minimal among the prime ideals containing some ideal generated by \(c\) elements.

Proof

Suppose that \(\mathfrak{p}\) has codimension \(c\), as claimed. We will construct the desired ideal by inductively choosing elements \(x_1,\ldots, x_r\), starting from the zero ideal \((0)\) (generated by \(0\) elements). Now suppose that for some \(r\) satisfying \(0\leq r< c\), we have constructed the ideal generated by \(x_1,\ldots, x_r\). Then we must choose a suitable \(x_{r+1}\in \mathfrak{p}\) not belonging to any of the prime ideals containing the ideal \((x_1,\ldots, x_r)\).

Now let \(\mathfrak{q}_1,\ldots, \mathfrak{q}_s\) be the minimal prime ideals containing \((x_1,\ldots, x_r)\). By Theorem 7 (Principal Ideal Theorem), the codimension of each \(\mathfrak{q}_i\) is at most \(r\), and since \(r< c\), their codimensions are all strictly less than \(c\). Therefore \(\mathfrak{p}\) cannot equal any of them, and in particular \(\mathfrak{p}\not\subseteq \bigcup_{i=1}^s \mathfrak{q}_i\). Thus we can choose \(x_{r+1}\in \mathfrak{p}\setminus \bigcup_{i=1}^s \mathfrak{q}_i\).

By induction we now obtain \(c\) elements \(x_1,\ldots, x_c\) belonging to \(\mathfrak{p}\). Then choosing a minimal prime ideal \(\mathfrak{q}\) containing the ideal \((x_1,\ldots, x_c)\), by Theorem 7 (Principal Ideal Theorem) we have \(\codim \mathfrak{q}\leq c\). On the other hand, since \(\mathfrak{q}\subseteq \mathfrak{p}\) and \(\codim \mathfrak{p}=c\), we must have \(\mathfrak{q}=\mathfrak{p}\).

If in the above corollary \(\codim \mathfrak{p}=0\), then \(\mathfrak{p}\) is a minimal prime containing the ideal generated by \(0\) elements, that is, the zero ideal. Now by §Associated Primes, ⁋Theorem 7, such a prime ideal consists entirely of zerodivisors. Combining this with Theorem 6 (Codimension one Principal Ideal Theorem), we see that if \(\mathfrak{p}\) is a minimal prime ideal containing a non-zerodivisor \(a\), then necessarily \(\codim \mathfrak{p}=1\).

In particular, for a minimal prime \(\mathfrak{p}\) containing a non-zerodivisor \(a\),

\[\dim A/\mathfrak{p}A+\codim \mathfrak{p}=\dim \mathfrak{p}+\codim \mathfrak{p}\leq \dim A\]

and since \(\codim \mathfrak{p}=1\),

\[\dim A/\mathfrak{p}A\leq\dim A-1\]

holds.

In particular, we have seen that for a noetherian local ring \((A, \mathfrak{m})\), \(\dim A=\codim \mathfrak{m}\) holds. Therefore, since \(\codim \mathfrak{m}=d\), by Corollary 8 the maximal ideal \(\mathfrak{m}\) must be generated by at least \(d\) elements.

Dimension in Graded Rings

Let us examine some useful properties for computing dimension in a graded ring \(R = \bigoplus_{d \ge 0} R_d\). First, recall the following. (§Localization of Graded Rings)

Definition 9 An ideal \(\mathfrak{a}\) of a graded ring \(R\) is called homogeneous if it is generated by homogeneous elements. A prime ideal \(\mathfrak{p}\) is called a homogeneous prime ideal if it is a homogeneous ideal and is prime.

The key observation in a graded ring is that prime ideals containing the irrelevant ideal \(\mathfrak{m} = \bigoplus_{d > 0} R_d\) are always homogeneous.

Proposition 10 In a graded ring \(R\), any prime ideal \(\mathfrak{p}\) containing the irrelevant ideal \(\mathfrak{m} = \bigoplus_{d > 0} R_d\) is homogeneous.

Proof

Let \(\mathfrak{p}\) be a prime ideal with \(\mathfrak{m} \subseteq \mathfrak{p}\). Consider the homogenization of \(\mathfrak{p}\)

\[\mathfrak{p}^\ast = \{x \in \mathfrak{p} \mid x \text{ homogeneous}\}\]

That this is a graded prime ideal follows immediately from the definition. The key point is that \(\mathfrak{p} = \mathfrak{p}^\ast\). First, \(\mathfrak{m} \subseteq \mathfrak{p}^\ast \subseteq \mathfrak{p}\) is obvious, so assume \(x \in \mathfrak{p}\) and let us show that \(x\in \mathfrak{p}^\ast\).

Write \(x\) in its homogeneous decomposition \(x = \sum_{d} x_d\). Since \(x_+ = \sum_{d > 0} x_d \in \mathfrak{m} \subseteq \mathfrak{p}\), we have \(x_0 = x - x_+ \in \mathfrak{p}\). Now \(x' = x - x_0 = x_+ \in \mathfrak{p}\), and in the same way we can show that \(x_1 \in \mathfrak{p}\). By induction each \(x_d \in \mathfrak{p}\), and therefore \(x \in \mathfrak{p}^\ast\).

From Proposition 10 we know that prime ideals containing the irrelevant ideal \(\mathfrak{m}\) are homogeneous. Conversely, homogeneous prime ideals not containing \(\mathfrak{m}\) correspond to the points of \(\operatorname{Proj} R\). Now we will show that any prime ideal chain can be refined to a homogeneous prime ideal chain.

Proposition 11 (Refinement by graded prime ideals) Any prime ideal chain in a graded ring \(R\) can be refined to a homogeneous prime ideal chain. That is, for a prime ideal chain \(\mathfrak{p}_0 \supsetneq \cdots \supsetneq \mathfrak{p}_s\), there exists a homogeneous prime ideal chain \(\mathfrak{p}_0^\ast \supsetneq \mathfrak{p}_1^\ast \supsetneq \cdots \supsetneq \mathfrak{p}_s^\ast\).

Proof

Consider a prime ideal chain \(\mathfrak{p}_0 \supsetneq \mathfrak{p}_1 \supsetneq \cdots \supsetneq \mathfrak{p}_s\). For each \(\mathfrak{p}_i\), define the ideal generated by its homogeneous elements

\[\mathfrak{p}_i^\ast = \langle f \in \mathfrak{p}_i : f \text{ is homogeneous} \rangle\]

Each \(\mathfrak{p}_i^\ast\) is a homogeneous prime ideal. Also, if \(\mathfrak{p}_i \supsetneq \mathfrak{p}_{i+1}\) then \(\mathfrak{p}_i^\ast \supseteq \mathfrak{p}_{i+1}^\ast\).

Now let us show that \(\mathfrak{p}_i^\ast \supseteq \mathfrak{p}_{i+1}^\ast\) is a strict inclusion. Since \(\mathfrak{p}_i \supsetneq \mathfrak{p}_{i+1}\), there exists an element \(f\) belonging to \(\mathfrak{p}_i \setminus \mathfrak{p}_{i+1}\). Writing \(f\) as the sum of homogeneous components \(f = f_{d_1} + \cdots + f_{d_k}\), at least one of the \(f_{d_j}\) does not belong to \(\mathfrak{p}_{i+1}\) (otherwise \(f \in \mathfrak{p}_{i+1}\)). Hence this \(f_{d_j} \in \mathfrak{p}_i^\ast \setminus \mathfrak{p}_{i+1}^\ast\).

Consequently, \(\mathfrak{p}_0^\ast \supsetneq \mathfrak{p}_1^\ast \supsetneq \cdots \supsetneq \mathfrak{p}_s^\ast\) is a homogeneous prime ideal chain.

Regular Local Rings

Definition 12 A noetherian local ring \((A, \mathfrak{m})\) is called a regular local ring if \(\mathfrak{m}\) can be generated by exactly \(d\) elements.

Then by §Integral Extensions, ⁋Lemma 8 (Nakayama), for \(a_1,\ldots, a_d\in \mathfrak{m}\), their images in \(\mathfrak{m}/\mathfrak{m}^2\) generate \(\mathfrak{m}/\mathfrak{m}^2\) as an \(A/\mathfrak{m}\)-vector space if and only if \(a_1,\ldots, a_d\) generate \(\mathfrak{m}\) as an \(A\)-module. We will examine further properties of these at the end of the next post.

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References

[Eis] David Eisenbud. Commutative Algebra: with a view toward algebraic geometry. Springer, 1995.
[Nag] Masayoshi Nagata. Local Rings. Interscience publishers, 1962.

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