Pound–Drever–Hall technique

In mathematics, dimension theory is a branch of commutative algebra studying the notion of the dimension of a commutative ring, and by extension that of a scheme.

The theory is much simpler for an affine ring; i.e., an integral domain that is a finitely generated algebra over a field. By Noether's normalization lemma, the Krull dimension of such a ring is the transcendence degree over the base field and the theory runs in parallel with the counterpart in algebraic geometry; cf. Dimension of an algebraic variety. The general theory tends to be less geometrical; in particular, very little works/is known for non-noetherian rings. (Kaplansky's commutative rings gives a good account of the non-noetherian case.) Today, a standard approach is essentially that of Bourbaki and EGA, which makes essential use of graded modules and, among other things, emphasizes the role of multiplicities, the generalization of the degree of a projective variety. In this approach, Krull's principal ideal theorem appears as a corollary.

Throughout the article, $\dim$ denotes Krull dimension of a ring and $\mathrm{ht}$ the height of a prime ideal (i.e., the Krull dimension of the localization at that prime ideal.)

Basic results

Let R be a noetherian ring or valuation ring. Then

$\dim R[x]=\dim R+1.$

If R is noetherian, this follows from the fundamental theorem below (in particular, Krull's principal ideal theorem.) But it is also a consequence of the more precise result. For any prime ideal $\mathfrak{p}$ in R,

$\mathrm{ht}(\mathfrak{p}R[x])=\mathrm{ht}(\mathfrak{p})$.

$\mathrm{ht}(\mathfrak{q})=\mathrm{ht}(\mathfrak{p})+1$ for any prime ideal $\mathfrak{q}⊋\mathfrak{p}R[x]$ in $R[x]$ that contracts to $\mathfrak{p}$.

This can be shown within basic ring theory (cf. Kaplansky, commutative rings). By the way, it says in particular that in each fiber of $\mathrm{Spec}R[x]\to \mathrm{Spec}R$, one cannot have a chain of primes ideals of length $\ge 2$.

Since an artinian ring (e.g., a field) has dimension zero, by induction, one gets the formula: for an artinian ring R,

$\dim R[x_1,\dots ,x_n]=n.$

Fundamental theorem

Let $(R,\mathfrak{m})$ be a noetherian local ring and I a $\mathfrak{m}$-primary ideal (i.e., it sits between some power of $\mathfrak{m}$ and $\mathfrak{m}$). Let $F(t)$ be the Poincaré series of the associated graded ring ${\mathrm{gr}}_{I}R=\underset{0}{\overset{\mathrm{\infty }}{\oplus }}{I}^n/I^{n+1}$. That is,

$F(t)=\sum _0^{\mathrm{\infty }}\ell (I^n/I^{n+1})t^n$

where $\ell$ refers to the length of a module (over an artinian ring $({\mathrm{gr}}_{I}R{)}_0=R/I$). If $x_1,\dots ,x_s$ generate I, then their image in $I/I^2$ have degree 1 and generate ${\mathrm{gr}}_{I}R$ as $R/I$-algebra. By the Hilbert–Serre theorem, F is a rational function with exactly one pole at $t=1$ of order, say, d. It also says (contained in the proof) that $d\le s$. Since

$(1-t{)}^{-d}=\sum _0^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{d-1+j}{d-1})t^j$,

we find that, for n large, the coefficient of $t^n$ in $F(t)=(1-t{)}^{d}F(t)(1-t{)}^{-d}$ is of the form

$\sum _0^N{a}_k(\genfrac{}{}{0ex}{}{d-1+n-k}{d-1})=(\sum a_k)\frac{n^{d-1}}{d-1!}+O(n^{d-2}).$

That is to say, $\ell (I^n/I^{n+1})$ is a polynomial $P$ in n of degree $d-1$ when n is large. P is called the Hilbert polynomial of ${\mathrm{gr}}_{I}R$.

We set $d(R)=d$. We also set $\delta (R)$ to be the minimum number of elements of R that can generate a $\mathfrak{m}$-primary ideal of R. Our ambition is to prove the fundamental theorem:

$\delta (R)=d(R)=\dim R$.

Since we can take s to be $\delta (R)$, we already have $\delta (R)\ge d(R)$ from the above. Next we prove $d(R)\ge \dim R$ by induction on $d(R)$. Let ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_m$ be a chain of prime ideals in R. Let $D=R/{\mathfrak{p}}_0$ and x a nonzero nonunit element in D. Since x is not a zero-divisor, we have the exact sequence

$0\to D\stackrel{x}{\to }D\to D/xD\to 0$.

The degree bound of the Hilbert-Samuel polynomial now implies that $d(D)>d(D/xD)\ge d(R/{\mathfrak{p}}_1)$. (This essentially follows from the Artin-Rees lemma; see Hilbert-Samuel function for the statement and the proof.) In $R/{\mathfrak{p}}_1$, the chain ${\mathfrak{p}}_i$ becomes a chain of length $m-1$ and so, by inductive hypothesis and again by the degree estimate,

$m-1\le \dim (R/{\mathfrak{p}}_1)\le d(R/{\mathfrak{p}}_1)\le d(D)-1\le d(R)-1$.

The claim follows. It now remains to show $\dim R\ge \delta (R).$ More precisely, we shall show:

Lemma: R contains elements $x_1,\dots ,x_s$ such that, for any i, any prime ideal containing $(x_1,\dots ,x_i)$ has height $\ge i$.

(Notice: $(x_1,\dots ,x_s)$ is then $\mathfrak{m}$-primary.) The proof is omitted. It appears, for example, in Atiyah–MacDonald. But it can also be supplied privately; the idea is to use prime avoidance.

Consequences of the fundamental theorem

Let $(R,\mathfrak{m})$ be a noetherian local ring and put $k=R/\mathfrak{m}$. Then

• $\dim R\le {\dim }_k\mathfrak{m}/{\mathfrak{m}}^2$, since a basis of $\mathfrak{m}/{\mathfrak{m}}^2$ lifts to a generating set of $\mathfrak{m}$ by Nakayama. If the equality holds, then R is called a regular local ring.
• $\dim \hat{R}=\dim R$, since $\mathrm{gr}R=\mathrm{gr}\hat{R}$.

(Krull's principal ideal theorem) The height of the ideal generated by elements $x_1,\dots ,x_s$ in a noetherian ring R is at most s. Conversely, a prime ideal of height s can be generated by s elements.

Proof: Let $\mathfrak{p}$ be a prime ideal minimal over such an ideal. Then $s\ge \dim R_{\mathfrak{p}}=\mathrm{ht}\mathfrak{p}$. The converse was shown in the course of the proof of the fundamental theorem.

If $A\to B$ is a morphism of noetherian local rings, then

$\dim B/{\mathfrak{m}}_{A}B\ge \dim B-\dim A.$

The equality holds if $A\to B$ is flat or more generally if it has the going-down property. (Here, $B/{\mathfrak{m}}_{A}B$ is thought of as a special fiber.)

Proof: Let $x_1,\dots ,x_n$ generate a ${\mathfrak{m}}_A$-primary ideal and $y_1,\dots ,y_m$ be such that their images generate a ${\mathfrak{m}}_B/{\mathfrak{m}}_{A}B$-primary ideal. Then ${{\mathfrak{m}}_B}^s\subset (y_1,\dots ,y_m)+{\mathfrak{m}}_{A}B$ for some s. Raising both sides to higher powers, we see some power of ${\mathfrak{m}}_B$ is contained in $(y_1,\dots ,y_m,x_1,\dots ,x_n)$; i.e., the latter ideal is ${\mathfrak{m}}_B$-primary; thus, $m+n\ge \dim B$. The equality is a straightforward application of the going-down property.

If R is a noetherian local ring, then

$\dim R[x]=\dim R+1$.

Proof: If ${\mathfrak{p}}_0⊊{\mathfrak{p}}_1⊊\cdots ⊊{\mathfrak{p}}_n$ are a chain of prime ideals in R, then ${\mathfrak{p}}_{i}R[x]$ are a chain of prime ideals in $R[x]$ while ${\mathfrak{p}}_{n}R[x]$ is not a maximal ideal. Thus, $\dim R+1\le \dim R[x]$. For the reverse inequality, let $\mathfrak{q}$ be a maximal ideal of $R[x]$ and $\mathfrak{p}=R\cap \mathfrak{q}$. Since $R[x]/\mathfrak{p}R[x]=(R/\mathfrak{p})[x]$ is a principal ideal domain, we get $1+\dim R\ge 1+\dim R_{\mathfrak{p}}\ge \dim R[x{]}_{\mathfrak{q}}$ by the previous inequality. Since $\mathfrak{q}$ is arbitrary, this implies $1+\dim R\ge \dim R[x]$.

Regular rings

Let R be a noetherian ring. The projective dimension of a finite R-module M is the shortest length of any projective resolution of R (possibly infinite) and is denoted by ${\mathrm{pd}}_{R}M$. We set $\mathrm{gl.dim}R=\sup \{{\mathrm{pd}}_{R}M|\text{M is a finite module}\}$; it is called the global dimension of R.

Assume R is local with residue field k.

Proof: We claim: for any finite R-module M,

${\mathrm{pd}}_{R}M\le n\iff {\mathrm{Tor}}_{n+1}^R(M,k)=0$.

By dimension shifting (cf. the proof of Theorem of Serre below), it is enough to prove this for $n=0$. But then, by the local criterion for flatness, ${\mathrm{Tor}}_1^R(M,k)=0\Rightarrow M\text{ flat }\Rightarrow M\text{ free }\Rightarrow {\mathrm{pd}}_R(M)\le 0.$ Now,

$\mathrm{gl.dim}R\le n\Rightarrow {\mathrm{pd}}_{R}k\le n\Rightarrow {\mathrm{Tor}}_{}^{n+1}(-,k)=0\Rightarrow {\mathrm{pd}}_R-\le n\Rightarrow \mathrm{gl.dim}R\le n,$

completing the proof.

Proof: If ${\mathrm{pd}}_{R}M=0$, then M is R-free and thus $M\otimes R_1$ is $R_1$-free. Next suppose ${\mathrm{pd}}_{R}M>0$. Then we have: ${\mathrm{pd}}_{R}K={\mathrm{pd}}_{R}M-1$ when $K$ is the kernel of some surjection from a free module to M. Thus, by induction, it is enough to consider the case ${\mathrm{pd}}_{R}M=1$. Then there is a projective resolution:

$0\to P_1\to P_0\to M\to 0$,

which gives:

${\mathrm{Tor}}_1^R(M,R_1)\to P_1\otimes R_1\to P_0\otimes R_1\to M\otimes R_1\to 0$.

But tensoring $0\to R\stackrel{f}{\to }R\to R_1\to 0$ with M we see the first term vanishes. Hence, ${\mathrm{pd}}_R(M\otimes R_1)$ is at most 1.

Proof: If R is regular, we can write $k=R/(f_1,\dots ,f_n)$, $f_i$ a regular system of parameters. An exact sequence $0\to M\stackrel{f}{\to }M\to M_1\to 0$, some f in the maximal ideal, of finite modules, ${\mathrm{pd}}_{R}M<\mathrm{\infty }$, gives us:

$0={\mathrm{Tor}}_{i+1}^R(M,k)\to {\mathrm{Tor}}_{i+1}^R(M_1,k)\to {\mathrm{Tor}}_i^R(M,k)\stackrel{f}{\to }{\mathrm{Tor}}_i^R(M,k),i\ge {\mathrm{pd}}_{R}M.$

But f here is zero since it kills k. Thus, ${\mathrm{Tor}}_{i+1}^R(M_1,k)\simeq {\mathrm{Tor}}_i^R(M,k)$ and consequently ${\mathrm{pd}}_R{M}_1=1+{\mathrm{pd}}_{R}M$. Using this, we get:

${\mathrm{pd}}_{R}k=1+{\mathrm{pd}}_R(R/(f_1,\dots ,f_{n-1}))=\cdots =n.$

The proof of the converse is by induction on $\dim R$. We begin with the inductive step. Set $R_1=R/f_{1}R$, $f_1$ among a system of parameters. To show R is regular, it is enough to show $R_1$ is regular. But, since $\dim R_1<\dim R$, by inductive hypothesis and the preceding lemma with $M=k$,

${\mathrm{pd}}_{R}k=\mathrm{gl.dim}R<\mathrm{\infty }\Rightarrow {\mathrm{pd}}_{R_1}k=\mathrm{gl.dim}{R}_1<\mathrm{\infty }\Rightarrow R_1\text{ regular}.$

The basic step remains. Suppose $\dim R=0$. We claim $\mathrm{gl.dim}R=0$ if it is finite. (This would imply that R is a semisimple ring; i.e., a field.) If that is not the case, then there is some finite module $M$ with $0<{\mathrm{pd}}_{R}M<\mathrm{\infty }$ and thus in fact we can find M with ${\mathrm{pd}}_{R}M=1$. By Nakayama's lemma, there is a surjection $u:F\to M$ such that $u\otimes 1:F\otimes k\to M\otimes k$ is an isomorphism. Denoting by K the kernel we have:

$0\to K\to F\stackrel{u}{\to }M\to 0$.

Since ${\mathrm{pd}}_{R}K={\mathrm{pd}}_{R}M-1=0$, K is free. Since $\dim R=0$, the maximal ideal $\mathfrak{m}$ is an associated prime of R; i.e., $\mathfrak{m}=\mathrm{ann}(s)$ for some s in R. Since $K\subset \mathfrak{m}M$, $sK=0$. Since K is not zero, this implies $s=0$, which is absurd. The proof is complete.

Depths

Let R be a ring and M a module over it. A sequence of elements $x_1,\dots ,x_n$ in $R$ is called a regular sequence if $x_1$ is not a zero-divisor on $M$ and $x_i$ is not a zero divisor on $M/(x_1,\dots ,x_{i-1})M$ for each $i=2,\dots ,n$.

Assume R is local with maximal ideal m. Then the depth of M is the supremum of any maximal regular sequence $x_i$ in m. It is easy to show (by induction, for example) that $\mathrm{depth}M\le \dim R$. If the equality holds, R is called the Cohen–Macaulay ring.

The Auslander–Buchsbaum formula relates depth and projective dimension.

References

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• Kaplansky, Irving, Commutative rings, Allyn and Bacon, 1970.
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