Regular local rings and complete intersections

Written by Claude Opus 5.5 (Anthropic) on 22 October 2026. Self-checked by the writing AI. Public domain (CC0).

For a noetherian local ring $A$ with residue field $k$, the cotangent complex $L_{k/A}$ measures how far $A$ is from being regular. Its first homology is always $\mathfrak m/\mathfrak m^2$; its second homology vanishes exactly when $A$ is regular; and its third homology vanishes exactly when $A$ is a complete intersection, that is, when the completion of $A$ is a regular local ring modulo an ideal generated by a regular sequence. A presentation of the completion by a regular local ring of the smallest possible dimension reads off the number of generators of $\mathfrak m$ and the number of relations from $D_1(k/A)$ and $D_2(k/A)$. The lesson proves:

$A$ is regular if and only if $D_2(k/A)=0$, if and only if $L_{k/A}\cong(\mathfrak m/\mathfrak m^2)[1]$;
$L_{k/A}$ does not change under completion, and power series rings over fields and Cohen rings are regular;
for a surjection from a regular local ring $P$ onto $A$ or its completion, with kernel $I$, the shift isomorphisms $D_q(k/A)\cong D_{q-1}(-/P,k)$ for $q\geq3$, and an exact sequence relating $D_2(k/A)$, $I/\mathfrak m_PI$, $\mathfrak m_P/\mathfrak m_P^2$ and $\mathfrak m/\mathfrak m^2$;
$A$ is a complete intersection if and only if $D_3(k/A)=0$, if and only if $D_q(k/A)=0$ for every $q\geq3$; the kernel of every presentation by a regular local ring is then generated by a regular sequence, and of none otherwise;
presentations with $I\subseteq\mathfrak m_P^2$ exist, and for them $\dim_kD_2(k/A)$ is the minimal number of generators of $I$.

Prerequisites. Weakly regular ideals of noetherian rings (Proposition 3.4, Theorem 3.5); The fundamental spectral sequence (Proposition 1.1 (3), Theorem 4.2, Exercise 5.1); Weak complete intersections (its conventions on the transitivity triangle and on $\partial_L$); The cotangent complex (Proposition 3.1, Theorem 4.1). From the AI-integrated Stacks project: regular local rings [Stacks, Tag 00KU], minimal generators of the maximal ideal of a regular local ring [Tag 00NQ], regularity modulo a regular sequence [Tag 00NU]; discrete valuation rings [Tag 00PD] and Cohen rings [Tag 0327]; power series rings over noetherian rings [Tag 0306]; the Cohen structure theorem [Tag 032A]; completion of noetherian rings: noetherian [Tag 0316], flat [Tag 00MB], complete with the same quotients [Tag 031C]; complete intersections [Tag 09PZ, Tag 09Q3]; Nakayama's lemma [Tag 00DV].

Conventions

Rings are commutative, in the usual sense. $(A,\mathfrak m,k)$ denotes a noetherian local ring with maximal ideal $\mathfrak m$ and residue field $k=A/\mathfrak m$; $\operatorname{edim}A=\dim_k\mathfrak m/\mathfrak m^2$ is its embedding dimension, the minimal number of generators of $\mathfrak m$ [Tag 00DV]. A regular local ring is a noetherian local ring whose maximal ideal is generated by $\dim A$ elements [Tag 00KU]. For a surjection $A\to B$ and a $B$-module $M$, $D_q(B/A,M)=H_q(L_{B/A}\otimes_BM)$ as in the lesson on the fundamental spectral sequence, and $D_q(B/A)=D_q(B/A,B)$. Since $L_{k/A}\otimes_kk=L_{k/A}$, $D_q(k/A,k)=D_q(k/A)=H_q(L_{k/A})$. The isomorphism $\partial_L\colon H_1(L_{B/A})\to I/I^2$ for $B=A/I$ is that of Proposition 1.1 (3) of the lesson on the fundamental spectral sequence; it is natural in the pair $(A,I)$. A homomorphism of local rings is local if it maps the maximal ideal into the maximal ideal; a surjective homomorphism of local rings is local.

1. Regular local rings

Lemma 1.1. $H_0(L_{k/A})=0$, and $\partial_L\colon H_1(L_{k/A})\to\mathfrak m/\mathfrak m^2$ is an isomorphism. So $\dim_kD_1(k/A)=\operatorname{edim}A$.

Proof. Proposition 1.1 (3) of the lesson on the fundamental spectral sequence, for $k=A/\mathfrak m$. $\square$

Theorem 1.2. For a noetherian local ring $(A,\mathfrak m,k)$ the following are equivalent:

$A$ is regular;
$\mathfrak m$ is generated by a regular sequence;
$D_2(k/A)=0$;
$D_q(k/A)=0$ for every $q\geq2$;
$\partial_L$ induces an isomorphism $L_{k/A}\cong(\mathfrak m/\mathfrak m^2)[1]$ in $D(k)$.

Proof. (1) implies (2): a minimal set of generators of the maximal ideal of a regular local ring is a regular sequence [Tag 00NQ].

(2) implies (1): if $x_1,\dots,x_r$ is a regular sequence generating $\mathfrak m$, then $A/(x_1,\dots,x_r)=k$ is a regular local ring of dimension $0$, so $A$ is regular [Tag 00NU].

(2) and (3) are equivalent by Proposition 3.4 of the lesson on weakly regular ideals of noetherian rings, applied to $I=\mathfrak m$: there $D_2(k/A,\kappa)$ is $D_2(k/A)$.

(2) implies (5) and (4): by the same proposition $\mathfrak m$ is weakly regular, and Theorem 4.2 of the lesson on the fundamental spectral sequence gives (5), which implies (4). Finally (4) implies (3). $\square$

2. Completion and presentations

Lemma 2.1. Let $\hat A$ be the $\mathfrak m$-adic completion of $A$. Then $\hat A$ is a noetherian local ring with maximal ideal $\mathfrak m\hat A$ and residue field $k$, and $A\to\hat A$ is flat. The functoriality map $L_{k/A}\to L_{k/\hat A}$ is an isomorphism; so $D_q(k/A)\cong D_q(k/\hat A)$ for every $q$, and $\operatorname{edim}\hat A=\operatorname{edim}A$.

Proof. $\hat A$ is noetherian [Tag 0316] and flat over $A$ [Tag 00MB]. It is $\mathfrak m\hat A$-adically complete, and $\hat A/\mathfrak m^n\hat A=A/\mathfrak m^n$ [Tag 031C]; in particular $\hat A/\mathfrak m\hat A=k$. An element $u\in\hat A\smallsetminus\mathfrak m\hat A$ maps to a unit of $k$; if $v\in\hat A$ lifts its inverse, then $uv=1-w$ with $w\in\mathfrak m\hat A$, and $\sum_jw^j$ converges, so $u$ is a unit. Hence $\hat A$ is local with maximal ideal $\mathfrak m\hat A$. As $\hat A$ is flat over $A$, $\operatorname{Tor}^A_i(k,\hat A)=0$ for $i>0$, and $k\otimes_A\hat A=\hat A/\mathfrak m\hat A=k$. By Proposition 3.1 of The cotangent complex, the functoriality map $L_{k/A}=L_{k/A}\otimes^{\mathbf L}_kk\to L_{k/\hat A}$ is an isomorphism. The last assertion follows from Lemma 1.1. $\square$

Lemma 2.2. Let $\Lambda$ be a field or a Cohen ring, and $n\geq0$. Then $R=\Lambda[[x_1,\dots,x_n]]$ is a regular local ring, whose maximal ideal is generated by $x_1,\dots,x_n$ and the maximal ideal of $\Lambda$.

Proof. A field is a regular local ring of dimension $0$. A Cohen ring is a complete discrete valuation ring [Tag 0327], hence a regular local ring of dimension $1$ [Tag 00PD]. The ring $R$ is noetherian [Tag 0306]. Let $\mathfrak m_\Lambda$ be the maximal ideal of $\Lambda$ and $\mathfrak n=\mathfrak m_\Lambda R+(x_1,\dots,x_n)$. Elements of $\mathfrak n$ are not units, as their images in $\Lambda=R/(x_1,\dots,x_n)$ lie in $\mathfrak m_\Lambda$. A series $f\notin\mathfrak n$ has a constant term $c\in\Lambda^\times$, and $f=c(1-g)$ with $g\in(x_1,\dots,x_n)$; since $g^j$ has no terms of degree $<j$, the series $\sum_jg^j$ is defined and inverts $1-g$. So $R$ is local with maximal ideal $\mathfrak n$. The quotient $R/(x_1,\dots,x_{i-1})$ is $\Lambda[[x_i,\dots,x_n]]$, on which multiplication by $x_i$ is injective, and $R/(x_1,\dots,x_n)=\Lambda\neq0$; so $x_1,\dots,x_n$ is a regular sequence. As $\Lambda$ is a regular local ring, so is $R$ [Tag 00NU]. $\square$

2.3. Complete intersections. A noetherian complete local ring $C$ is a complete intersection if there is a surjective homomorphism of local rings $P\to C$, with $P$ a regular local ring, whose kernel is generated by a regular sequence [Tag 09PZ]; a noetherian local ring is a complete intersection if its completion is one [Tag 09Q3]. Every noetherian complete local ring $C$ has surjections from regular local rings: by the Cohen structure theorem [Tag 032A], $C\cong\Lambda[[x_1,\dots,x_n]]/J$ with $\Lambda$ a field or a Cohen ring, and $\Lambda[[x_1,\dots,x_n]]$ is regular by Lemma 2.2. Theorem 2.6 shows that the condition does not depend on the surjection.

Proposition 2.4. Let $C$ be a noetherian local ring with residue field $k$, $P$ a regular local ring with maximal ideal $\mathfrak m_P$, and $P\to C$ a surjective homomorphism with kernel $I\subseteq\mathfrak m_P$. Then $D_q(k/C)\cong D_{q-1}(C/P,k)\quad(q\geq3),$ and there is an exact sequence $\begin{aligned}&0\to D_2(k/C)\to I/\mathfrak m_PI\\&\quad\to\mathfrak m_P/\mathfrak m_P^2\to\mathfrak m_C/\mathfrak m_C^2\to0,\end{aligned}$ in which $I/\mathfrak m_PI\to\mathfrak m_P/\mathfrak m_P^2$ is induced by the inclusion $I\subseteq\mathfrak m_P$ and $\mathfrak m_P/\mathfrak m_P^2\to\mathfrak m_C/\mathfrak m_C^2$ by $P\to C$.

Proof. The transitivity triangle of $P\to C\to k$ (Theorem 4.1 of The cotangent complex) gives the long exact sequence $\begin{aligned}&\cdots\to D_q(C/P,k)\to D_q(k/P)\\&\quad\to D_q(k/C)\to D_{q-1}(C/P,k)\\&\quad\to D_{q-1}(k/P)\to\cdots,\end{aligned}$ in which $D_q(k/P)=0$ for $q\geq2$, by Theorem 1.2. For $q\geq3$ the outer terms around $D_q(k/C)\to D_{q-1}(C/P,k)$ vanish, which gives the isomorphisms. In low degrees the sequence is $\begin{aligned}&0\to D_2(k/C)\to D_1(C/P,k)\\&\quad\to D_1(k/P)\to D_1(k/C)\\&\quad\to D_0(C/P,k).\end{aligned}$ Since $H_0(L_{C/P})=0$, $D_0(C/P,k)=0$, and right exactness of $\otimes_Ck$ gives $D_1(C/P,k)=H_1(L_{C/P})\otimes_Ck$, which $\partial_L$ identifies with $(I/I^2)\otimes_Ck=I/\mathfrak m_PI$, because $I^2\subseteq\mathfrak m_PI$. By Lemma 1.1, $D_1(k/P)\cong\mathfrak m_P/\mathfrak m_P^2$ and $D_1(k/C)\cong\mathfrak m_C/\mathfrak m_C^2$. The maps are induced by the functoriality of the cotangent complex for the squares from $P\to C$ to $P\to k$ and from $P\to k$ to $C\to k$; by naturality of $\partial_L$ in the pair, they are the maps induced by $I\subseteq\mathfrak m_P$ and by $P\to C$. $\square$

Corollary 2.5. Let $P$ be a regular local ring and $I\subseteq\mathfrak m_P$ an ideal, $C=P/I$ with residue field $k$. Then $I$ is generated by a regular sequence if and only if $D_3(k/C)=0$, and then $D_q(k/C)=0$ for every $q\geq3$.

Proof. By Proposition 2.4, $D_q(k/C)\cong D_{q-1}(C/P,k)$ for $q\geq3$. By Proposition 3.4 of the lesson on weakly regular ideals of noetherian rings, applied to $I\subseteq\mathfrak m_P$, $I$ is generated by a regular sequence if and only if $D_2(C/P,k)=0$; and then $I$ is weakly regular, so $D_q(C/P,k)=0$ for every $q\geq2$ (Theorem 3.5 there). $\square$

Theorem 2.6. For a noetherian local ring $(A,\mathfrak m,k)$ the following are equivalent:

$A$ is a complete intersection;
$D_q(k/A)=0$ for every $q\geq3$;
$D_3(k/A)=0$.

If they hold, the kernel of every surjective homomorphism from a regular local ring onto $\hat A$ is generated by a regular sequence; if they fail, no such kernel is.

Proof. Let $P\to\hat A$ be a surjection from a regular local ring, with kernel $I$; it exists by 2.3, and $I$ lies in the maximal ideal of $P$. By Lemma 2.1, $D_q(k/A)\cong D_q(k/\hat A)$ for every $q$. By Corollary 2.5 for $C=\hat A$, $I$ is generated by a regular sequence if and only if $D_3(k/A)=0$, and then $D_q(k/A)=0$ for every $q\geq3$. This holds for every such surjection. So (1), which asks it for one surjection, implies (2); (2) implies (3); and (3) gives it for every surjection, hence (1). $\square$

If $A=P/I$ with $P$ a regular local ring, Corollary 2.5 and Theorem 2.6 show directly, without completing, that $A$ is a complete intersection if and only if $I$ is generated by a regular sequence.

Proposition 2.7. Let $(A,\mathfrak m,k)$ be a noetherian local ring.

For every surjection $P\to\hat A$ from a regular local ring with kernel $I$, $\begin{aligned}&\dim_kD_2(k/A)\\&\quad=\mu(I)-\operatorname{edim}P+\operatorname{edim}A,\end{aligned}$ where $\mu(I)=\dim_kI/\mathfrak m_PI$ is the minimal number of generators of $I$. The following are equivalent: $I\subseteq\mathfrak m_P^2$; $\dim P=\operatorname{edim}A$; $D_2(k/A)\cong I/\mathfrak m_PI$.
There is a surjection $P\to\hat A$ from a regular local ring with kernel contained in $\mathfrak m_P^2$. For it, $\dim P=\operatorname{edim}A$ and $\dim_kD_2(k/A)=\mu(I)$.

Proof. (1) By Lemma 2.1, $D_2(k/A)\cong D_2(k/\hat A)$ and $\operatorname{edim}\hat A=\operatorname{edim}A$. The exact sequence of Proposition 2.4 for $C=\hat A$ gives the dimension formula. Its second map $I/\mathfrak m_PI\to\mathfrak m_P/\mathfrak m_P^2$ is zero exactly when $I\subseteq\mathfrak m_P^2$, which by exactness holds exactly when $D_2(k/A)\to I/\mathfrak m_PI$ is bijective, and exactly when $\mathfrak m_P/\mathfrak m_P^2\to\mathfrak m/\mathfrak m^2$ is bijective, that is $\operatorname{edim}P=\operatorname{edim}A$. For the regular local ring $P$, a minimal set of generators of $\mathfrak m_P$ has $\dim P$ elements [Tag 00NQ], so $\operatorname{edim}P=\dim P$.

(2) Start with any surjection $P\to\hat A$ from a regular local ring (2.3). If its kernel $I$ is not contained in $\mathfrak m_P^2$, choose $f\in I\smallsetminus\mathfrak m_P^2$. The class of $f$ in $\mathfrak m_P/\mathfrak m_P^2$ is nonzero, so $f$ is part of a minimal set of generators of $\mathfrak m_P$, and $P/(f)$ is a regular local ring of dimension $\dim P-1$ [Tag 00NQ]. The surjection factors through $P/(f)\to\hat A$, with kernel $I/(f)$. Repeating this, the dimension decreases each time, so after finitely many steps the kernel lies in the square of the maximal ideal. The rest follows from (1). $\square$

So $\dim_kD_1(k/A)$ is the minimal number of generators of $\mathfrak m$, and $\dim_kD_2(k/A)$ the minimal number of relations among them, in a presentation of the completion by a regular local ring of the smallest possible dimension.

Remark 2.8. Quillen conjectured that a homomorphism of finite type $A\to B$ of noetherian rings is a local complete intersection when $B$ has finite flat dimension over $A$ and $L_{B/A}$ has finite projective dimension, and that a noetherian local ring with $D_q(k/A)=0$ for all large $q$ is a complete intersection. Avramov proved the first in a very general form, and the second follows from his theorem on rings of locally complete intersection type applied to $A\to k$: L. Avramov, Locally complete intersection homomorphisms and a conjecture of Quillen on the vanishing of cotangent homology, Ann. of Math. 150 (1999), Theorems 1.3 and 1.5, arXiv:math/9909192. These results are not used in the course.

3. Exercises

Exercise 3.1. Let $k$ be a field, $P=k[x,y]_{(x,y)}$ and $A=P/(xy)$. Show that $P$ is regular, that $A$ is a complete intersection but not regular, and that $\dim_kD_1(k/A)=2$, $\dim_kD_2(k/A)=1$ and $D_q(k/A)=0$ for $q\geq3$.

Exercise 3.2. With $P$ as in Exercise 3.1, let $A=P/(x^2,xy,y^2)$. Show that $\dim_kD_2(k/A)=3$, that $A$ is not a complete intersection, and that $D_3(k/A)\neq0$.

Exercise 3.3. Let $(A,\mathfrak m,k)$ be a noetherian local ring and $f\in\mathfrak m$ a nonzerodivisor. Show that $D_q(k/A)\cong D_q(k/(A/fA))$ for $q\geq3$, and conclude that $A$ is a complete intersection if and only if $A/fA$ is.

Exercise 3.4. Show that a noetherian local ring $A$ with $\operatorname{edim}A\leq1$ is a complete intersection.

4. Complete solutions

Solution 3.1. In the domain $P$, $x$ is a nonzerodivisor, $P/xP$ is a localization of the domain $k[y]$ in which $y$ is nonzero, and $P/(x,y)=k\neq0$; so $x,y$ is a regular sequence generating the maximal ideal, and $P$ is regular by Theorem 1.2. The element $xy\neq0$ is a nonzerodivisor of $P$, so $I=(xy)$ is generated by a regular sequence, and $A$ is a complete intersection by the remark after Theorem 2.6, with $D_q(k/A)=0$ for $q\geq3$. As $I\subseteq\mathfrak m_P^2$, $\mathfrak m_A/\mathfrak m_A^2=\mathfrak m_P/(\mathfrak m_P^2+I)=\mathfrak m_P/\mathfrak m_P^2$, which has dimension $2$; so $\dim_kD_1(k/A)=2$ by Lemma 1.1. In the exact sequence of Proposition 2.4 for $C=A$, $I/\mathfrak m_PI$ is spanned by the class of $xy$, and its map to $\mathfrak m_P/\mathfrak m_P^2$ is zero, as $xy\in\mathfrak m_P^2$; so $D_2(k/A)\cong I/\mathfrak m_PI\cong k$. As $D_2(k/A)\neq0$, $A$ is not regular (Theorem 1.2).

Solution 3.2. Here $I=(x^2,xy,y^2)=\mathfrak m_P^2$, and $x^2$, $xy$, $y^2$ form a minimal set of generators of $I$: their classes span $I/\mathfrak m_PI=\mathfrak m_P^2/\mathfrak m_P^3$, which has dimension $3$. As $I\subseteq\mathfrak m_P^2$, the exact sequence of Proposition 2.4 for $C=A$ gives $D_2(k/A)\cong I/\mathfrak m_PI$, of dimension $3$. The sequence $x^2,xy$ is not regular: $x\cdot xy=y\cdot x^2$ lies in $x^2P$, while $x$ does not, since $x=x^2a/s$ with $s\notin\mathfrak m_P$ would give $s=xa\in\mathfrak m_P$. By Proposition 3.4 of the lesson on weakly regular ideals of noetherian rings, if $I$ were generated by a regular sequence, every minimal set of generators would be one; so $I$ is not, $A$ is not a complete intersection (remark after Theorem 2.6), and $D_3(k/A)\neq0$ by Corollary 2.5.

Solution 3.3. Put $B=A/fA$. The ideal $fA$ is generated by the regular sequence $f$ ($B\neq0$, as $f\in\mathfrak m$), so it is weakly regular, $L_{B/A}\cong(fA/f^2A)[1]$, and $fA/f^2A\cong B$, $f$ being a nonzerodivisor (Proposition 3.4 and Theorem 3.5 of the lesson on weakly regular ideals of noetherian rings). Hence $D_q(B/A,k)$ is $k$ for $q=1$ and $0$ otherwise. The transitivity triangle of $A\to B\to k$ gives the exact sequences $D_q(B/A,k)\to D_q(k/A)\to$ $D_q(k/B)\to D_{q-1}(B/A,k)$, whose outer terms vanish for $q\geq3$. So $D_q(k/A)\cong D_q(k/B)$ for $q\geq3$, and in particular $D_3(k/A)=0$ if and only if $D_3(k/B)=0$. Theorem 2.6 applies to $A$ and to $B$, whose residue field is $k$.

Solution 3.4. By Proposition 2.7 (2), there is a surjection $P\to\hat A$ from a regular local ring with $\dim P=\operatorname{edim}A\leq1$. If $\dim P=0$, $P$ is a field and the kernel is $0$, generated by the empty sequence. If $\dim P=1$, $P$ is a discrete valuation ring [Tag 00PD], so the kernel is $0$ or generated by a nonzero element, which is a nonzerodivisor of the domain $P$ and lies in the maximal ideal; in either case it is generated by a regular sequence. So $\hat A$ is a complete intersection, and so is $A$.

Sources

The characterizations of regular local rings by $D_2(k/A)$ and of complete intersections by $D_3(k/A)$, and the reading of $D_1$ and $D_2$ as numbers of generators and relations, are due to Quillen, in Section 10 of his MIT notes on the homology of commutative rings (1968), available from Haynes Miller's page. Avramov's paper cited in Remark 2.8 records these criteria, with references to André's work on the homology of commutative algebras (Remark 1.6 there).