Krull Dimension

Definition of Dimension

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

Definition 1 The Krull dimension of a ring \(A\) is defined, over descending chains of prime ideals of \(A\)

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

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

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

More generally, we define the following.

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

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

In \(A\), for a prime ideal \(\mathfrak{p}\), the codimension of \(\mathfrak{p}\), \(\codim \mathfrak{p}\), 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 an arbitrary \(A\)-module \(M\), the dimension and codimension of \(M\) are defined as the dimension and codimension of \(\ann(M)\), respectively.

As a point of caution, under the definition above, \(\mathfrak{a}\) has two definitions of dimension: its dimension as an ideal defined first, and its dimension when viewed as an \(A\)-module; the values given by these two definitions may differ. Therefore, when using the notation \(\dim \mathfrak{a}\), we mean only the dimension as an ideal of \(A\), as defined first in Definition 2.

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

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

Therefore, the following inequality

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

holds. Contrary to what its name suggests, the reverse inequality does not hold in general.

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

Computing Dimensions

In general, when dealing with dimension, one mostly considers the case where the ring \(A\) is Noetherian. One of the main 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 precisely what \(0\)-dimensional Noetherian rings are.

Corollary 3 For a Noetherian ring \(A\), \(\dim A =0\) if and only if \(A\) is Artinian.

Meanwhile, by the following proposition, we know that in general, if \(\phi:A \rightarrow B\) is integral, then the dimension of an ideal of \(B\) is equal to the dimension of its preimage.

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

Proof

The first statement is simply §Integral Extensions and Ideals, ⁋Proposition 1. For the second, since \(\phi^{-1}\mathfrak{b}\) is the kernel of the composition \(A \rightarrow B \rightarrow B/\mathfrak{b}\), \(A/\phi^{-1}\mathfrak{b}\hookrightarrow B/\mathfrak{b}\) is also an integral extension; the inequality \(\dim \mathfrak{b}\geq \dim \phi^{-1}\mathfrak{b}\) holds by the second statement of §Integral Extensions and Ideals, ⁋Proposition 1, and the reverse inequality holds by §Integral Extensions and Ideals, ⁋Corollary 4.

We now turn our attention to what happens in dimension 1. Before that, we make the following somewhat technical definition.

Definition 5 For a prime ideal \(\mathfrak{p}\subseteq A\), the \(n\)th symbolic power of \(\mathfrak{p}\), \(\mathfrak{p}^{(n)}\), is defined by the following equation:

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

By definition, \(\mathfrak{p}^{(n)}\) is obtained via the localization \(A \rightarrow A_\mathfrak{p}\) by transferring the ideal \((\mathfrak{p}A_\mathfrak{p})^n\) to \(A\). Then elements outside \(\mathfrak{p}\) become non-zerodivisors modulo \(\mathfrak{p}^{(n)}\), and it is clear that \(\mathfrak{p}^{(n)}A_\mathfrak{p}=(\mathfrak{p}A_\mathfrak{p})^n\). There is also a descending chain of symbolic powers

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

We can now prove the following theorem.

Theorem 6 (Codimension one Principal Ideal Theorem) Let \(A\) be a Noetherian ring and let \(a\in A\) be arbitrary. Suppose \(\mathfrak{p}\) is 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 every prime ideal \(\mathfrak{q}\subsetneq \mathfrak{p}\), which in turn, by §Localization, ⁋Proposition 8, reduces to showing that \(\dim A_\mathfrak{q}=0\).

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

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

Meanwhile, since \(\mathfrak{p}A_\mathfrak{p}\) is minimal among the prime ideals containing \(\mathfrak{a}A_\mathfrak{p}\), the ring \(A_\mathfrak{p}/\mathfrak{a}A_\mathfrak{p}\) is Artinian by §The Jordan-Hölder Theorem, ⁋Corollary 8. From this, we know that 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 stabilize. Thus, let \((\mathfrak{q}A_\mathfrak{p})^{(n)}+\mathfrak{a}A_\mathfrak{p}= (\mathfrak{q}A_\mathfrak{p})^{(n+1)}+\mathfrak{a}A_\mathfrak{p}\). Then, since

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

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

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

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

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

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

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

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

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

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

Using this, we can now prove the following by induction.

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

That is, any prime ideal of a Noetherian ring satisfies the descending chain condition, and the length of a chain starting at \(\mathfrak{p}\) is less than or equal to the number of generators of \(\mathfrak{p}\). Nevertheless, there exist Noetherian rings of infinite dimension. ([Nag, Appendix, Example 1])

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

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

Proof

As claimed, suppose that \(\mathfrak{p}\) has codimension \(c\). Starting from the ideal generated by \(0\) elements, that is, the zero ideal \((0)\), we will inductively choose elements \(x_1,\ldots, x_r\) to construct the desired ideal. Now, given \(0\leq r< c\), suppose that for \(r\) we have constructed the ideal generated by \(x_1,\ldots, x_r\). Then, so that it does not belong to any of the minimal prime ideals containing the ideal \((x_1,\ldots, x_r)\), we must choose a suitable \(x_{r+1}\in \mathfrak{p}\).

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

Now, inductively, in \(\mathfrak{p}\) we obtain \(c\) elements \(x_1,\ldots, x_c\). Then, if we choose a minimal prime ideal containing the ideal \((x_1,\ldots, x_c)\) and contained in \(\mathfrak{p}\), say \(\mathfrak{q}\), 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 corollary above \(\codim \mathfrak{p}=0\), then \(\mathfrak{p}\) is a minimal prime containing an ideal generated by \(0\) elements, that is, the zero ideal. Now, by §Associated Primes, ⁋Theorem 7, such a prime ideal consists only 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 we must have \(\codim \mathfrak{p}=1\).

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

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

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

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

holds.

In particular, for a Noetherian local ring \((A, \mathfrak{m})\), we have seen that \(\dim A=\codim \mathfrak{m}\) holds. Therefore, if we let \(d=\dim A=\codim \mathfrak{m}\), by Theorem 7 (Principal Ideal Theorem), \(\mathfrak{m}\) must be generated by at least \(d\) elements.

Dimension in Graded Rings

Let us examine some properties that are useful when computing dimension in a graded ring \(R = \bigoplus_{d \ge 0} R_d\). We have already defined homogeneous ideals along with their equivalent conditions in [Algebraic Structures] §Graded Rings, ⁋Proposition 6; in particular, an element of a homogeneous ideal has all of its homogeneous components inside that ideal. We give one more name to this.

Definition 9 In a graded ring \(R\), a prime ideal \(\mathfrak{p}\) is called a homogeneous prime ideal if \(\mathfrak{p}\) is a homogeneous ideal.

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

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

Proof

Suppose \(\mathfrak{p}\) is a prime ideal and \(\mathfrak{m} \subseteq \mathfrak{p}\). Consider the homogenization of \(\mathfrak{p}\):

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

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

Write \(x\) as 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 one can show that \(x_1 \in \mathfrak{p}\). Inductively, 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, the homogeneous prime ideals not containing \(\mathfrak{m}\) correspond to the points of \(\operatorname{Proj} R\). We now examine an operation that associates a homogeneous prime ideal to an arbitrary prime ideal.

Proposition 11 For a graded ring \(R\) and a prime ideal \(\mathfrak{p}\), the ideal generated by the homogeneous elements belonging to \(\mathfrak{p}\), \(\mathfrak{p}^\ast\), is the largest homogeneous ideal contained in \(\mathfrak{p}\), and it is also a prime ideal.

Proof

By definition, \(\mathfrak{p}^\ast\) is generated by homogeneous elements, so it is a homogeneous ideal, and since all of its generators belong to \(\mathfrak{p}\), we have \(\mathfrak{p}^\ast \subseteq \mathfrak{p}\). Now if \(\mathfrak{a}\subseteq \mathfrak{p}\) is a homogeneous ideal, then by the third condition of [Algebraic Structures] §Graded Rings, ⁋Proposition 6, \(\mathfrak{a}\) is generated by homogeneous elements, and since these generators are all homogeneous elements belonging to \(\mathfrak{p}\), we have \(\mathfrak{a}\subseteq \mathfrak{p}^\ast\). Thus \(\mathfrak{p}^\ast\) is the largest homogeneous ideal contained in \(\mathfrak{p}\).

Now let us show that \(\mathfrak{p}^\ast\) is prime. First, since \(\mathfrak{p}^\ast \subseteq \mathfrak{p}\subsetneq R\), we have \(\mathfrak{p}^\ast\neq R\). Then by the third condition of §Localization of Graded Rings, ⁋Lemma 2, it suffices to show that whenever \(ab\in \mathfrak{p}^\ast\) for any homogeneous elements \(a,b\in R\), either \(a\in \mathfrak{p}^\ast\) or \(b\in \mathfrak{p}^\ast\). But \(ab\in \mathfrak{p}^\ast\subseteq \mathfrak{p}\) and \(\mathfrak{p}\) is prime, so \(a\in \mathfrak{p}\) or \(b\in \mathfrak{p}\); and since \(a\) and \(b\) are homogeneous, whichever of them lies in \(\mathfrak{p}\) is a generator of \(\mathfrak{p}^\ast\), and in particular an element of \(\mathfrak{p}^\ast\).

For an arbitrary chain of prime ideals \(\mathfrak{p}_0 \supsetneq \cdots \supsetneq \mathfrak{p}_s\), we have \(\mathfrak{p}_0^\ast \supseteq \cdots \supseteq \mathfrak{p}_s^\ast\), but these inclusions need not be strict. When \(R=\mathbb{K}[\x]\) is given the standard grading, every element of \((\x-1)\) vanishes at \(\x=1\), whereas the value of a homogeneous element \(c\x^n\) at \(\x=1\) is \(c\); hence the only homogeneous element belonging to \((\x-1)\) is \(0\). Thus both terms of the chain \((\x-1)\supsetneq (0)\) are sent to \((0)\). On the other hand, in a chain of prime ideals all containing the irrelevant ideal \(\mathfrak{m}\), each term is already homogeneous by Proposition 10, so such a chain is itself a chain of homogeneous prime ideals.

Regular Local Rings

Definition 12 A Noetherian local ring \((A, \mathfrak{m})\) is called a regular local ring if, for \(d=\dim A\), the ideal \(\mathfrak{m}\) can be generated by exactly \(d\) elements.

Then, by §Integral Extensions, ⁋Lemma 8 (Nakayama), the images of \(a_1,\ldots, a_d\in \mathfrak{m}\) 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 study their properties further 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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