Regular Local Rings and Homological Dimension — Examples

Finite injectivity makes the extension integral and gives $d=\dim A=\dim B$. Choose a regular system $x_1,\dots ,x_d$ of $A$. The quotient $C=B\mathrm{/}{\mathfrak{m}}_{A}B$ is a finite-dimensional algebra over $k_A$ and a nonzero local ring. Its descending powers of the maximal ideal stabilize as vector subspaces; at stabilization Nakayama makes that power zero. Thus $\sqrt{{\mathfrak{m}}_{A}B}={\mathfrak{m}}_B$, and the images of the $x_i$ are parameters of $B$.

If $B$ is Cohen–Macaulay, those parameters are $B$-regular. Regarded as an $A$-module, $B$ consequently has depth at least $d$. The support-dimension bound gives depth at most $d$ (its annihilator is zero by injectivity). Homological regularity makes its projective dimension over $A$ finite; Auslander–Buchsbaum gives projective dimension zero and freeness.

Conversely, if $B$ is free over $A$, the regular parameter sequence of $A$ remains injective successively on the finite direct sums describing $B$ and its successive quotients. The terminal quotient is nonzero. Since this is a system of parameters of $B$, the regular-parameter criterion makes $B$ Cohen–Macaulay. If $d=0$, the sequence is empty and $A$ is a field; all steps remain valid.

A field has the single prime and maximal ideal $(0)$, so its dimension and cotangent dimension are both zero. It is regular by definition. The empty tuple generates its maximal ideal and $k\stackrel{1}{\to }k$ is its length-zero augmented resolution.

For a zero-dimensional regular local ring, $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2=0$. Nakayama applied to the finite ideal $\mathfrak{m}$ gives $\mathfrak{m}=0$. Every element outside the maximal ideal is a unit, so the ring is a field.

Every nonzero $a\in R$ is $t^{v(a)}u$ with $v(u)=0$, so $u$ is a unit. Thus the nonunits form $(t)$, the unique maximal ideal. In each nonzero ideal, choose an element of least valuation; it divides every element of that ideal and hence generates it. The zero ideal is principal too, so $R$ is Noetherian. A nonzero proper prime contains some $t^{n}u$ with $n\ge 1$, hence contains $t$ and equals $(t)$. Since $R\subseteq K$ is a domain and $(t)\mathrm{\ne }0$, its only primes are $(0)$ and $(t)$, giving dimension one. These choices are of one element for a fixed ideal, not of a family.

Since $t\mathrm{\notin }(t^2)$ by valuation, the map $k=R\mathrm{/}(t)\to (t)\mathrm{/}(t^2)$ sending $a+(t)$ to $at+(t^2)$ is bijective: surjectivity is immediate, and cancellation proves injectivity. Thus the cotangent dimension is one, $R$ is regular local, and its minimal generating tuple $(t)$ is a regular system of parameters.

For each $n\ge 0$, the map $a+(t)\mapsto a{t}^n+(t^{n+1})$ is well-defined and identifies $k$ with $(t^n)\mathrm{/}(t^{n+1})$, again by cancellation and principality. Multiplication of these classes agrees with polynomial multiplication. Hence the graded map $k[T]\to {\mathrm{gr}}_{(t)}R$ sending $T$ to $t+(t^2)$ is an isomorphism in every degree, and therefore an isomorphism of graded rings. No choice principle is used.

A field is regular; polynomial extension and localization make $R$ regular. The quotient by the indicated prime inverts every nonzero polynomial in the remaining variables, hence is $k(x_{r+1},\dots ,x_n)$. The maximal ideal is generated by the first $r$ variables.

The strict coordinate chain $(0)⊊(x_1)⊊\cdots ⊊(x_1,\dots ,x_r)$ survives in the localization and has length $r$. The $r$ generators bound cotangent dimension by $r$, and the embedding bound gives $r\le \dim R\le \mathrm{edim}R\le r$. Thus those generators are minimal and form regular parameters. If $r=0$, the ring is the fraction field; if $r=n$, the residue field is $k$, including $n=r=0$.

Define a series by coefficients $a_{\alpha }\in k$ for $\alpha \in {\mathbb{N}}^n$, with coefficientwise addition and convolution multiplication. For each fixed multi-index only finitely many pairs sum to it, so multiplication is defined and associative by finite reindexing. Compatible truncations in total degrees below $q$, for all $q\ge 1$, identify this ring with the inverse limit of $k[x_1,\dots ,x_n]\mathrm{/}(x_1,\dots ,x_n{)}^q$. In these quotients every polynomial with nonzero constant term has a finite geometric-series inverse, so the same inverse limit is the completion of the coordinate local polynomial ring.

That local polynomial ring is regular: it is a localization of a polynomial ring over a field, and the coordinate prime chain and $n$ maximal-ideal generators give dimension $n$. Completion is Noetherian local, preserves dimension and embedding dimension, and preserves regularity. Its maximal ideal is generated by the variable images and its residue field is $k$. For $n=0$, the index set ${\mathbb{N}}^0$ has one element and the ring is just $k$.

Every element has a unique form $a+b\epsilon$. It is a unit precisely when $a\mathrm{\ne }0$, with inverse $a^{-1}-b{a}^{-2}\epsilon$. Hence the unique maximal ideal is $(\epsilon )$. Every prime contains the nilpotent $\epsilon$, so this is the only prime and the dimension is zero.

The square of the maximal ideal is zero and the class of $\epsilon$ is a nonzero $k$-basis of it. Therefore the cotangent dimension is one, strictly larger than Krull dimension. The ring is finite-dimensional over $k$ and hence Noetherian, so the regularity definition applies and fails.

The quotient $A=k[x,y]\mathrm{/}(y^2-x^3)$ has unique representatives $a(x)+yb(x)$ by division by the monic polynomial in $y$. Under $x\mapsto t^2$, $y\mapsto t^3$, the two summands have even and odd powers of $t$, respectively; their vanishing forces both to be zero. Hence $A$ embeds in $k[t]$ and is a domain in every characteristic. The origin ideal remains a proper nonzero maximal ideal after localization.

The ambient local domain $S=k[x,y{]}_{(x,y)}$ has cotangent basis $x,y$: modulo $(x,y{)}^2$, every localized polynomial has a unique constant and linear part because its denominator has nonzero constant term. Hence $\mathrm{edim}S=2$. The chain $(0)⊊(x)S⊊(x,y)S$ gives $\dim S\ge 2$, while [F1] gives $\dim S\le 2$.

The kernel $(f)S$ of $S\to R$ is a nonzero prime by step 1.1. Since $R$ is a nonfield local domain, $(0)⊊(x,y)R$ gives $\dim R\ge 1$. A chain of length two in $R$ would lift to a chain of primes $(0)⊊(f)S⊊\mathfrak{p}⊊(x,y)S$ of length three in $S$, contradicting step 1.2. Thus $\dim R=1$. As $f\in (x,y{)}^{2}S$, the quotient introduces no relation in the cotangent space, so $\mathrm{edim}R=2$. By [F2], $R$ is not regular. The argument works in every characteristic.

The coordinate local ring is regular of dimension two: the coordinate chain and two generators give the dimension. Its variables are regular parameters. The displayed maps are exactly its two-variable Koszul maps; their composition is $-xyc+yxc=0$ and the Koszul theorem gives exactness.

Every entry is in $(x,y)$, so the resolution is minimal. Its ranks in degrees zero, one, two are $1,2,1$ and it is zero above two. The rank formula gives the asserted Betti numbers, independently of the characteristic.

The ring is local with maximal ideal $(\epsilon )$. For multiplication by $\epsilon$, the image and kernel both equal $(\epsilon )$: $\epsilon (a+b\epsilon )=a\epsilon$. The augmentation $R\to k$ has that same kernel. Thus the infinite augmented complex is exact in every degree and all positive matrix entries are in the maximal ideal.

Under the stated AC premise, the rank formula gives ${\beta }_i(k)=1$ in every degree. These nonzero Betti numbers are unbounded in degree, so the projective-dimension criterion gives infinity. A finite initial truncation would have a nonzero left kernel and is not a finite resolution.

In $k[x,y]$, $(xy)=(x)\cap (y)$, as divisibility of monomials shows. It is therefore radical. Every prime of the quotient contains $x$ or $y$; in the origin localization either branch is the local line $k[t{]}_{(t)}$, with prime chain of length one. Thus $R$ is reduced of dimension one. Both $x$ and $y$ survive and their product is zero, so it is not a domain.

The relation $xy$ is quadratic, so $\mathfrak{m}\mathrm{/}{\mathfrak{m}}^2$ has independent basis $\bar{x},\bar{y}$. Its dimension two strictly exceeds the dimension one just computed, and the definition makes $R$ nonregular. Localization does not change these cotangent classes since denominators have nonzero constant term. This works also in characteristic two.

In the ambient coordinate local ring $S=k[x,y{]}_{(x,y)}$, the associated graded ring is $k[X,Y]$: a rational function with denominator of nonzero constant term has initial form equal to its numerator initial form divided by that constant. This identifies each graded piece and respects products. Therefore orders add on products of nonzero elements of $S$. In particular for $f=y^2-x^3$, $\mathrm{in}(f)=Y^2$, and $\mathrm{in}(hf)=\mathrm{in}(h)Y^2$ for every nonzero $h\in S$.

The quotient map $S\to R$ sends ${\mathfrak{n}}^n$ onto ${\mathfrak{m}}^n$ for each $n$, so by [F1] it induces a surjection ${\mathrm{gr}}_{\mathfrak{n}}S\twoheadrightarrow {\mathrm{gr}}_{\mathfrak{m}}R$. Its degree-$n$ kernel consists of classes of $a\in {\mathfrak{n}}^n$ with $a\in (f)+{\mathfrak{n}}^{n+1}$. Write $a=hf+b$ with $b\in {\mathfrak{n}}^{n+1}$. If its degree-$n$ class is nonzero, it is exactly the initial form of $hf$, hence a multiple of $Y^2$. Conversely every homogeneous multiple of $Y^2$ is the initial form of a polynomial multiple of $f$. Thus the graded kernel is exactly $(Y^2)$, proving the displayed quotient directly. Since $f\in {\mathfrak{n}}^2$, the cotangent classes of $x,y$ form its degree-one basis, and the induced map is the claimed polynomial map.

The displayed diagonal map is injective since $R$ is a domain. Its image consists exactly of pairs whose first coordinate lies in $(x)$, which is the augmentation kernel. Hence the complex is exact.

The second coordinates form the two-term identity summand, whose identity homotopy contracts it. Removing it leaves multiplication by $x$ on the first coordinates. Since $x$ belongs to the maximal ideal, this remaining resolution is minimal by the unit-cancellation criterion.

The coordinate chain of prime ideals gives dimension at least three, while the three generators of the maximal ideal give embedding dimension at most three and hence dimension at most three. Thus $R$ is regular of dimension three with parameters $x,y,z$. Their Koszul complex is a minimal resolution.

The exterior bases have ranks $1,3,3,1$ in degrees zero through three and zero above. The rank and projective-dimension formulas give these Betti numbers and projective dimension three, since the top rank is one.

The ring is regular local of dimension two, by polynomial regularity and the coordinate chain and generator count. The residue-field computation gives projective dimension two. Its depth is zero because every maximal-ideal element kills the nonzero module $k$. The presentation $0\to \mathfrak{m}\to R\to k\to 0$ is minimal.

The syzygy theorem gives $\mathrm{pd}\mathfrak{m}=1$. The ring depth is two, since $x,y$ is a regular sequence and depth is bounded by dimension. Auslander–Buchsbaum gives $\mathrm{depth}\mathfrak{m}=2-1=1$. Concretely its minimal resolution is $0\to R\stackrel{c\mapsto (-yc,xc)}{\to }{R}^2\to \mathfrak{m}\to 0$: reducing a relation modulo $x$ shows $b=xc$, and then cancellation gives $a=-yc$.

Degree truncation identifies series modulo $(x,y{)}^q$ with polynomials modulo that ideal. In the truncated polynomial ring, every denominator allowed in $R$ is a unit, by a finite geometric-series expansion of its nonconstant part. Thus both quotients have the stated monomial basis, and their inverse limit is $k[\text{\hspace{0.17em}⁣}[x,y]\text{\hspace{0.17em}⁣}]$, identifying it as the maximal-adic completion.

The coordinate chain and two maximal-ideal generators give $\dim R=\mathrm{edim}R=2$. The cotangent-completion theorem preserves embedding dimension and the completion dimension theorem preserves dimension independently. The completion is Noetherian local and regular. At $q=1$ the quotient is $k$; at $q=2$ the basis $1,x,y$ exhibits the two cotangent classes.

Translate coordinates $u_i=x_i-a_i$. The ambient local ring $S=k[u_1,\dots ,u_n{]}_{(u_1,\dots ,u_n)}$ is regular, with cotangent basis the $u_i$. It is a domain, so the nonzero polynomial $f$ is a nonzerodivisor. The quotient regularity criterion says $S\mathrm{/}(f)$ is regular exactly when $f\mathrm{\notin }{\mathfrak{m}}_S^2$. The hypotheses cannot hold for n=0, because then f is a nonzero constant.

Monomial expansion after translation gives $f\equiv {\sum }_i(\mathrm{\partial }f\mathrm{/}\mathrm{\partial }{x}_i)(a)u_i(\mathrm{m}\mathrm{o}\mathrm{d}{\mathfrak{m}}_S^2)$; the constant term vanishes. Independence of the cotangent basis makes this class nonzero precisely when at least one coefficient is nonzero. This proves both implications over every characteristic. It is a rational-point hypersurface statement and makes no assertion about smoothness over arbitrary residue-field extensions.

For a surjection $R\to A=R\mathrm{/}I$ of local rings the maximal ideal of $A$ is ${\mathfrak{m}}_R\mathrm{/}I$ and the residue fields agree. Thus its cotangent space is the quotient ${\mathfrak{m}}_R\mathrm{/}(I+{\mathfrak{m}}_R^2)$. If $R$ is regular of dimension $d$, [F2] gives $\mathrm{edim}A\le d$.

For a supplied regular cover put $c={\dim }_k((I+{\mathfrak{m}}_R^2)\mathrm{/}{\mathfrak{m}}_R^2)$. Lift a basis of this subspace to $x_1,\dots ,x_c\in I$ and extend their cotangent classes to a basis of ${\mathfrak{m}}_R\mathrm{/}{\mathfrak{m}}_R^2$. Under AC, apply [F1] successively to these $c$ basis members; after each quotient the remaining classes are a cotangent basis. The resulting ring $R\mathrm{/}(x_1,\dots ,x_c)$ is regular of dimension $d-c=\mathrm{edim}A$ by the cotangent quotient calculation of step 1.1. It still surjects onto $A$, so the lower bound is attained. If $c=0$ retain the original cover; if $c=d$ the new cover is the residue field.

Before localization every polynomial in $t$ has a unique expression $a(t^2)+tb(t^2)$, so $k[t]$ is free on $1,t$ over $k[s]$. After localizing the base at $(s)$, call the resulting rank-two free algebra $C$. If $h(t)$ has nonzero constant term, write $h=a(s)+tb(s)$. Then $(a+tb)(a-tb)=a(s{)}^2-sb(s{)}^2$ is a unit of the base, since its constant term is $a(0{)}^2\mathrm{\ne }0$. Thus every such $h$ is a unit of $C$, proving $C=k[t{]}_{(t)}$, also in characteristic two.

The source and target are regular one-dimensional coordinate local rings (their elements are units times powers of their variable, giving DVRs). The maximal ideal contracts correctly, and freeness makes the map flat. Modulo the source maximal ideal the fibre is $k[t]\mathrm{/}(t^2)$, whose only prime is $(t)$, with zero square and one-dimensional cotangent space. Its Krull dimension is zero, so it is singular by the regularity definition.