Dimension and Height

Section 14.1 Dimension and Height

“I don’t like heights. This is why I stopped growing at fifth grade.”
―Billy Crystal

Remark 14.1.

We will now spend a while discussing the notion of dimension of a ring and dimension of a variety. To motivate the definition, let’s first think in terms of varieties.

We take our inspiration from the fundamental setting of dimension theory: vector spaces. The notion of basis doesn’t make sense for varieties (What does it mean to span? Where is zero?), but one relevant thing we do have for both vector spaces and for varieties is subobjects.

One way to characterize the dimension of a vector space $V$ is the the largest number $d$ such that there is a proper chain of subspaces

\begin{equation*} \{0\} = V_0 \subsetneq V_1 \subsetneq V_2 \subsetneq \cdots \subsetneq V_d = V. \end{equation*}

We can try something similar for varieties, but for a reducible variety, this is not a very good notion. For example, for a union of $m$ points, we can cook up a chain of $m$ proper subvarieties by adding one more point each time, but a point should be zero-dimensional by any reasonable measure. So, if we want this approach to work, we should stick to chains of irreducible subvarieties.

Definition 14.2. Dimension of Variety.

The dimension of an affine variety$X$ is defined as

\begin{equation*} \sup \{ d \ | \ \exists \ \text{a strictly decreasing chain of irreducible subvarieties of $X$:} \ \ X_d \supsetneq X_{d-1} \supsetneq \cdots \supsetneq X_0\}. \end{equation*}

Definition 14.3. Krull Dimension of a Ring.

The Krull dimension of a ring $R$ is

\begin{equation*} \dim(R) := \sup \{d \mid \exists \text{ a strictly increasing chain of prime ideals } P_0 \subsetneq \cdots \subsetneq P_d\}. \end{equation*}

We often call it simply the dimension of $R\text{.}$ Alternatively, $\dim(R)$ is the supremum of the lengths of saturated chains of primes of $R\text{.}$

Definition 14.4. Dimension of a Module.

The dimension of an $R$-module $M$ is defined as $\dim(R/\ann_R(M))\text{.}$

Example 14.5.

The dimension of any field is zero.
A ring is zero-dimensional if and only if every minimal prime of $R$ is a maximal ideal.
The ring of integers $\Z$ has dimension $1\text{,}$ since there is one minimal prime $(0)$ and every other prime is maximal. Likewise, any PID that is not a field has dimension one.

Lemma 14.6. UFD height 1.

Let $R$ be a ring and $I$ be an ideal in $R\text{.}$ If $R$ is a UFD, $I$ is a prime of height one if and only if $I=(f)$ for a prime element $f\text{.}$

Proof.

If $I=(f)$ with $f$ irreducible, and $0 \subsetneq P \subseteq I\text{,}$ then $P$ contains some nonzero multiple of $f\text{,}$ say $af^n$ with $a$ and $f$ coprime. Since $a \notin I\text{,}$ $a \notin P\text{,}$ so we must have $f \in P\text{,}$ so $P = (f)\text{.}$ Thus, $I$ has height one. On the other hand, if $I$ is a prime of height one, we claim $I$ contains an irreducible element. Indeed, $I$ is nonzero, so contains some $f \neq 0\text{,}$ and primeness implies one of the prime factors of $f$ is contained in $I\text{.}$ Thus, any nonzero prime contains a prime ideal of the form $(f)\text{,}$ so a height one prime must be of this form.

Example 14.7. Dim of $k[x_1,x_2,\dots]/(x_1^2,x_2^2,\dots)$ is Zero.

For a field $k\text{,}$ the ring $R = k[x_1,x_2,\dots]/(x_1^2,x_2^2,\dots)$ is not noetherian, as

\begin{equation*} (x_1) \subsetneq (x_1,x_2) \subsetneq (x_1,x_2,x_3) \subsetneq \cdots \end{equation*}

is an infinite ascending chain, so $R$ is not noetherian. However, $\sqrt{(0)} = (x_1,x_2,\dots)$ is a maximal ideal, and hence $(x_1,x_2,\dots)$ is the unique minimal prime of $R\text{,}$ which is also maximal. Therefore, $\dim(R) = 0\text{.}$

Example 14.8. Nagata.

Let $R = k[x_{11},x_{21},x_{22}, x_{31}, x_{32}, x_{33},\ldots]$ be a polynomial ring in infinitely many variables, which we are thinking of as arranged in an infinite triangle. The ring $R$ is clearly infinite-dimensional and not noetherian. Let

\begin{equation*} W = R \setminus \big((x_{11}) \cup (x_{21},x_{22}) \cup (x_{31},x_{32},x_{33}) \cup \cdots \big) = \bigcap_{n \geqslant 1} \big(R \setminus (x_{n1},\dots,x_{nn})\big) \end{equation*}

and $S=W^{-1}R\text{.}$ Note that $W$ is an intersection of multiplicatively closed subsets, so this is a valid localization of $R\text{.}$

For any $n\text{,}$ we have a chain of primes

\begin{equation*} (x_{n1}) \subsetneq (x_{n1},x_{n2}) \subsetneq \cdots \subsetneq (x_{n1},\ldots, x_{nn}) \end{equation*}

in $R\text{.}$ As these primes are all contained in $(x_{n1},\ldots, x_{nn})\text{,}$ none of these intersects $W\text{,}$ so the expansion to $S$ of the chain above yields a proper chain of primes in $S\text{.}$ It follows that $\dim(S) \geqslant n$ for all $n \geqslant 1\text{,}$ so $S$ is infinite-dimensional.

It turns out that $S$ is noetherian, which is not at all obvious and a bit technical, so we will not prove it.

Lemma 14.9. Properties of Dimension and Height.

Let $R$ be a ring.

A prime has height zero if and only if it is a minimal prime of $R\text{.}$
An ideal has height zero if and only if it is contained in a minimal prime of $R\text{.}$ In particular, in a domain, every nonzero ideal has positive height.
$\dim(R) = \sup \{ \dim(R/P) \mid P \in \Spec(R) \} = \sup\{ \dim(R/P) \mid P \in \Min(R) \}\text{.}$
$\dim(R) = \sup \{ \ht(P) \mid P \in \Spec(R) \}=\sup\{ \mathrm{height}(Q) \mid Q \in \mSpec(R) \}\text{.}$
If $I$ is an ideal, then $\dim(R/I) = \sup \{ n \mid \exists P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_n, P_i \in V(I) \}\text{.}$
If $P$ is prime,

\begin{equation*} \dim(R/P) + \ht(P) \leqslant \dim(R). \end{equation*}
If $I$ is an ideal,

\begin{equation*} \dim(R/I) + \ht(I) \leqslant \dim(R). \end{equation*}
If $W$ is a multiplicative set, then $\dim(W^{-1}R) \leqslant \dim(R)\text{.}$
If $P$ is prime, then $\ht(P)=\dim(R_P)\text{.}$
If $P \subseteq Q$ are primes, then $\dim(R_Q/ P R_Q)$ is the supremum of the lengths of saturated chains of primes in $R$ of the form $P = P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_n = Q\text{.}$

Proof.

We will prove some of these and leave the rest as an exercise.

A prime has height $0$ if and only if it contains no other prime, which is equivalent being a minimal prime of $R\text{.}$

If $I$ is contained in a minimal prime $Q$ of $R\text{,}$ then $\ht(Q)=0$ by (1), so $\ht(I)=0$ by definition. Conversely, if $\ht(I)=0\text{,}$ by definition, there is a minimal prime of $I$ of height $0\text{,}$ so some minimal prime of $I$ is a minimal prime of $R\text{,}$ so $I$ is contained in a minimal prime of $R\text{.}$

It suffices to show that if $\dim(R/P) \geqslant a$ and $\ht(P) \geqslant b$ then $\dim(R) \geqslant a+b\text{.}$ By definition, $\ht(P) \geqslant b$ means that there is a chain of primes

\begin{equation*} Q_0 \subsetneq Q_1 \subsetneq \cdots \subsetneq Q_b = P. \end{equation*}

By (5), $\dim(R/P) \geqslant a$ means that there is a chain of primes

\begin{equation*} \mathfrak{a}_0 \subsetneq \mathfrak{a}_1 \subsetneq \cdots \subsetneq \mathfrak{a}_a \end{equation*}

with $\mathfrak{a}_0 \supseteq P\text{.}$ We can assume without loss of generality that $\mathfrak{a}_0 = P\text{,}$ since if not, we can add it to the bottom of the chain. Putting these chains together, we get a chain of length $a+b$ in $\Spec(R)\text{,}$ so $\dim(R) \geqslant a+b\text{.}$

Let $\dim(R/I) \geqslant a$ and $\ht(I) \geqslant b\text{.}$ The inequality $\ht(I) \geqslant b$ means that for every minimal prime $P$ of $I\text{,}$ $\ht(P) \geqslant b\text{.}$ The inequality $\dim(R/I) \geqslant a$ implies that there exists a minimal prime of $P$ of $I$ such that $\dim(R/P) \geqslant a\text{.}$ For such a minimal prime as in the latter statement, using (5), we get the desired conclusion.

Remark 14.10.

We know that in noetherian rings, there can be arbitrarily long chains of primes, since the dimension can be infinite as in Nagata’s. On the other hand, any ascending proper chain of primes is finite, as a consequence of the definition. Does this imply that every prime has finite height? This does not prevent that possibility that there could be an infinite descending chain of primes, which would then give any of the primes in the chain infinite height. (This seems strange in conjunction with the fact that in any ring, any prime contains a minimal prime, but it does not contradict this.) Another possible problem is there being two primes $P \subseteq Q$ such that for all $n \geqslant 1$ there exists a chain of primes of length $n$ from $P$ to $Q\text{.}$

However, we will later show that the height of any ideal in a noetherian ring is finite.

Example 14.11. Int Extension of Domains Nonzero Multiple.

Let $R \subseteq S$ be an integral extension of domains. Show that every nonzero $s \in S$ has a nonzero multiple in $R\text{,}$ meaning that there exists $t \in S$ such that $ts \neq 0$ and $ts \in R\text{.}$

Example 14.12. .

Let $k$ be an infinite field and $R = k[t^3,t^4,t^5]\text{.}$ We have shown that

\begin{equation*} \displaystyle R\cong \frac{k[x,y,z]}{(x^3-yz,y^2-xz,z^2-x^2y)}, \end{equation*}

and that $R$ is the coordinate ring of the curve

\begin{equation*} X = \{ (t^3,t^4,t^5) \ | t\in k\}. \end{equation*}

As $X$ is a curve (meaning it is parameterized by a single parameter), we should expect the dimension of the variety $X\text{,}$ and equivalently of the ring $R\text{,}$ to be $1\text{.}$ Let’s prove this.

On the one hand, $R$ is a domain, so $(0)$ is the unique minimal prime. To show that $\dim(R) = 1\text{,}$ we need to show that any nonzero prime ideal is maximal.

Set $S = k[t^3]\subseteq R\text{.}$ We note that $t^3$ does not satisfy any algebraic relation over $k\text{,}$ so $S$ is isomorphic to a polynomial ring in one variable, where that one variable corresponds to $t^3\text{.}$ Moreover, note that the inclusion $S \subseteq R$ is integral, since $t^4$ satisfies the monic polynomial $(T^4)^3- (t^3)^4=0$ and $t^5$ satisfies the monic polynomial $(T^5)^3 - (t^3)^5=0\text{.}$ Polynomial rings in one variable have dimension $1\text{,}$ so $\dim(S) = 1\text{.}$

Let $P \in \Spec(R)$ be nonzero. Note first that $P \cap S \neq 0\text{,}$ since if $f \in P\text{,}$ then there is some nonzero multiple of $f$ in $S$ by . Since $\dim(S)=1\text{,}$ $P \cap S$ is maximal. The inclusion

\begin{equation*} \frac{S}{P \cap S} \hookrightarrow \frac{R}{P} \end{equation*}

is integral: a dependence relation for any representative yields a dependence relation. Since $\frac{S}{P \cap S}$ is a field, $\frac{R}{P}$ is a domain, and the inclusion is integral, by we can conclude that $R/P$ is a field, so $P$ is a maximal ideal. This shows that every nonzero prime in $R$ is maximal, and thus $\dim(R) = 1\text{.}$

Definition 14.13. Catenary Ring.

A ring is catenary if for every pair of primes $Q \supseteq P\text{,}$ every saturated chain of primes

\begin{equation*} P = P_0 \subsetneq P_1 \subsetneq \cdots \subsetneq P_n = Q \end{equation*}

has the same length.

Definition 14.14. Equidimensional Ring.

A ring is equidimensional if every maximal ideal has the same finite height, and every minimal prime has the same dimension.

Example 14.15. .

Consider the ring

\begin{equation*} R = \displaystyle \frac{k[x,y,z]}{(xy,xz)}. \end{equation*}

We can find the minimal primes of $R$ by computing $\Min((xy,xz))$ in $k[x,y,z]\text{.}$ The prime ideals $(x)$ and $(y,z)$ are incomparable, and $(x)\cap(y,z)=(xy,xz)\text{,}$ so $\Min(R)=\{(x),(y,z)\}\text{.}$ We claim that the height of $(x-1,y,z)$ is one: it contains the minimal prime $(y,z)\text{,}$ and any saturated chain from $(y,z)$ to $(x-1,y,z)$ corresponds to a saturated chain from $(0)$ to $(x-1)$ in $k[x]\text{,}$ which must have length 1 since this is a PID. The height of $(x,y-1,z)$ is at least 2, as witnessed by the chain $(x) \subseteq (x,y-1) \subseteq (x,y-1,z)\text{.}$ So $R$ is not equidimensional.

Example 14.16. .

The ring $\Z_{(2)}[x]$ is a domain that is not equidimensional. On the one hand, the maximal ideal $(2,x)$ has height at least two, which we see from the chain

\begin{equation*} (0) \subsetneq (x) \subseteq (x,2). \end{equation*}

Thus $\dim(R) \geqslant 2\text{.}$ On the other hand, we will show later that the prime ideal $P = (2x-1)$ has height 1, and it is maximal since $R/P \cong \Q\text{.}$ Therefore, $\dim(R/P)=0\text{,}$ and thus

\begin{equation*} \dim(R/P)+ \ht(P)=1 \end{equation*}

whereas $\dim R \geqslant 2\text{.}$ In fact, $\dim R=2\text{,}$ but I won’t justify this.

The ring $R$ in this example is in fact a catenary domain, which we will not justify. Notice that there are maximal ideals of distinct heights in this ring, for example the ideal $P$ given above is a prime of height 1 whereas and the maximal ideal $\fm=(2,x)$ has height 2. Thus this ring is not equidimensional.

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Postmodern Algebra

Section 14.1 Dimension and Height

Remark 14.1.

Definition 14.2. Dimension of Variety.

Definition 14.3. Krull Dimension of a Ring.

Definition 14.4. Dimension of a Module.

Example 14.5.

Lemma 14.6. UFD height 1.

Proof.

Example 14.7. Dim of \(k[x_1,x_2,\dots]/(x_1^2,x_2^2,\dots)\) is Zero.

Example 14.8. Nagata.

Lemma 14.9. Properties of Dimension and Height.

Proof.

Remark 14.10.

Example 14.11. Int Extension of Domains Nonzero Multiple.

Example 14.12. .

Definition 14.13. Catenary Ring.

Definition 14.14. Equidimensional Ring.

Example 14.15. .

Example 14.16. .