Math-453 · mathematics · Algebra

Math-453

Small Cohen-Macaulay Conjecture

Status OpenLLM status Open unverified

Top Open Problems openaiunknowngpt 6 astra ultraunknown Not reported cataloguedOpenOpen

Statement

Definitions and mathematical statement

Let $(R,\mathfrak m)$ be a complete Noetherian local ring of Krull dimension $d$. A system of parameters is a $d$-tuple $x_1,\ldots,x_d$ generating an $\mathfrak m$-primary ideal. A nonzero finitely generated $R$-module $M$ is maximal Cohen--Macaulay if $\operatorname{depth}_R M=d$: equivalently a system of parameters is an $M$-regular sequence, meaning each $x_i$ acts injectively on $M/(x_1,\ldots,x_{i-1})M$ and $M/(x_1,\ldots,x_d)M\ne0$.

The conjecture asks whether every such $R$ admits a maximal Cohen--Macaulay module. The word small means finite generation; an infinitely generated big Cohen--Macaulay module does not answer this question. The requested object is a module, not necessarily a finite algebra or a faithful module.

Formulation and concept references: [S1].

Short English statement

Does every complete Noetherian local ring have a nonzero finitely generated module of full possible depth?

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Math-453-A2 · Literature report · LLM-generated · unknown ·

OpenAI claim summary for Math-453

The proposed complex example is normal and complete, with Krull dimension three. Every nonzero finite module over it has smaller depth, directly opposing universal existence of a small Cohen--Macaulay module. This is presented as a claimed negative answer.

Small Cohen-Macaulay Conjecture

OpenAI's claim

OpenAI presents a counterexample in the selected manuscript. A normal complete local domain of dimension three over $\mathbb C$ is claimed to admit no nonzero finite module of depth three.

Mathematical statement of the claimed result

Result 1.
The source claims a complete Noetherian normal local domain $R\supset\mathbb C$, with residue field $\mathbb C$, satisfying \[ \dim R=3,\qquad \forall M\ne0\text{ finitely generated over }R:\ \operatorname{depth}_RM<3. \] Its surface obstruction takes a smooth integral projective complex surface $X$, an ample globally generated divisor $H$, and the completed vertex ring $R(X,H)=\widehat{T(X,H)_{T_{>0}}}$, $T(X,H)=\bigoplus_{j\ge0}H^0(X,\mathcal O_X(jH))$. Existence of a small Cohen--Macaulay module would imply \[ K_X^2-2c_2(X)\le6H^2. \] For the constructed polarization at parameter $n=7$, its uncompleted three-dimensional vertex localization $T_{T_{>0}}$ also admits only the zero module among finitely generated modules of maximal Cohen--Macaulay depth.

Outline of the argument

The construction completes a polarized surface cone. Extending a hypothetical maximal Cohen--Macaulay module to a regular model, balancing slopes by modifications, and comparing Chern characters yields a numerical surface inequality; an explicit branched surface violates that inequality.

Formal verification

Family 195 has no Lean scope page in this release. Its selected result is not presented as formalized. This site has not executed the Lean code. The scope descriptions do not certify the remaining manuscript arguments.

Status caveat

OpenAI attributes these manuscripts to work by an internal model. Its release cautions that arguments without formalization can contain mistakes. This catalogue has not independently checked the proofs; the labels report the mathematical claims attributed to OpenAI.

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Math-453-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-453

Small Cohen-Macaulay Conjecture

Definitions and mathematical statement

Let $(R,\mathfrak m)$ be a complete Noetherian local ring of Krull dimension $d$. A system of parameters is a $d$-tuple $x_1,\ldots,x_d$ generating an $\mathfrak m$-primary ideal. A nonzero finitely generated $R$-module $M$ is maximal Cohen--Macaulay if $\operatorname{depth}_R M=d$: equivalently a system of parameters is an $M$-regular sequence, meaning each $x_i$ acts injectively on $M/(x_1,\ldots,x_{i-1})M$ and $M/(x_1,\ldots,x_d)M\ne0$.

The conjecture asks whether every such $R$ admits a maximal Cohen--Macaulay module. The word small means finite generation; an infinitely generated big Cohen--Macaulay module does not answer this question. The requested object is a module, not necessarily a finite algebra or a faithful module.

Formulation and concept references: [S1].

Short English statement

Does every complete Noetherian local ring have a nonzero finitely generated module of full possible depth?

Research attempt

GPT-6 Astra Ultra; 27 September 2026. Status: UNRESOLVED.

The useful starting point is that faithfulness is not required. This makes reduction to one component legitimate, but it does not make passage from a big module to a small one legitimate. I work out that distinction, including the low-dimensional construction and a specific failure of the proposed finite-submodule shortcut. The results below are reductions and standard subcases, not a new solution of the general conjecture.

Reduction to a highest-dimensional domain.
Choose a minimal prime $P$ of $R$ with $\dim(R/P)=d$, and put $S=R/P$. Suppose that $N$ is a nonzero finite maximal Cohen--Macaulay $S$-module. It is finite over $R$. If $x_1,\ldots,x_d$ is a system of parameters of $R$, its images generate an ideal primary to the maximal ideal of $S$, since $S/(x_1,\ldots,x_d)S$ is a quotient of the Artinian ring $R/(x_1,\ldots,x_d)$. These images are a system of parameters of $S$. Their regularity on $N$ is exactly their regularity when $N$ is regarded as an $R$-module. Thus a positive answer for complete local domains of dimension $d$ gives a positive answer in that dimension for arbitrary complete local rings. Conversely domains are among the original inputs. This reduction preserves the dimension; an arbitrary minimal prime would not suffice if the ring were not equidimensional.

For example, in $k[[x,y]]/(x^2,xy)$ the quotient by $(x)$ is $k[[y]]$. It is a finite module of depth one over the original one-dimensional ring, although its annihilator is nonzero. Requiring a faithful module would discard this simple construction and would change the stated task.

The constructions in dimensions at most two.
In dimension zero the residue field is a nonzero finite module of depth zero. In dimension one apply the reduction to a domain $S$. A parameter $x$ has nonzero image in $S$: if it vanished, the quotient by it would still have dimension one. Multiplication by $x$ is injective on the domain, and $S/xS\ne0$ by Nakayama's lemma. Hence $S$ itself is the required module over $R$.

In dimension two there is a further standard construction. A complete Noetherian local domain $S$ is excellent, so its normalization $T$ in its fraction field is finite. A finite algebra over a complete local ring is a product of complete local rings. Here $T$ is a domain, so that product has just one factor. Integral extension preserves dimension, and $T$ is therefore a normal local domain of dimension two. Serre's condition $(S_2)$ gives $\operatorname{depth}T\ge2$. If $x,y$ are parameters of $S$, then $T/(x,y)T$ is finite over the Artinian ring $S/(x,y)$, so $x,y$ are also parameters of $T$. They form a regular sequence on the Cohen--Macaulay ring $T$. Consequently $T$, as a finite $S$-module and then an $R$-module, supplies the answer. The excellence, finiteness, and normality facts used in this paragraph are the standard algebraic inputs recorded in [E1]; the module argument is explicit above.

The same normalization argument stops in higher dimension at a precise point. Normality supplies depth at least $\min(2,\dim T)$, not depth $\dim T$. Replacing the number two by $d$ would silently replace $(S_2)$ by the Cohen--Macaulay property. Thus finiteness of normalization is useful but does not provide the missing higher-depth assertion.

Why taking a finite piece of a big module fails.
Even a finite submodule of a Cohen--Macaulay module can lose regularity. Take $R=k[[x,y]]$, the already Cohen--Macaulay module $B=R$, and its finite submodule $M=(x,y)$. Multiplication by $x$ is injective on $M$. But the class of $x$ in $M/xM$ is nonzero. Indeed $x=xm$ with $m\in M$ would give $m=1$ in the domain, a contradiction. Nevertheless \[ y[x]=[xy]=0\quad\hbox{in }M/xM, \] because $y\in M$. Hence the next parameter fails to act injectively. The ambient regular sequence has not descended to the finite submodule. This is a countertest of an inference, not a counterexample to the conjecture: the module $R$ itself solves this particular input.

The example identifies the exact obstruction. Suppose $B$ has a regular sequence $x_1,\ldots,x_d$, and $M\subset B$ is finite. If \[ x_i m\in(x_1,\ldots,x_{i-1})M, \] regularity in $B$ shows only that $m\in(x_1,\ldots,x_{i-1})B$. To deduce regularity in $M$ one needs witnesses to this membership lying in $M$. In the example the missing witness is $1$, since $x=x\cdot1$ in $B$ but $1\notin M$.

One can try to adjoin witnesses. For a fixed finite $M$, each module \[ K_i=\{m\in M:x_i m\in(x_1,\ldots,x_{i-1})M\} \] is finite, so finitely many generators and their expressions in $B$ require only finitely many new elements. This produces a finite larger submodule in which the old obstructing relations have witnesses. However, that larger submodule has its own modules $K_i$ and can create new relations. Iteration gives an ascending chain of finite submodules of $B$, not a chain of submodules of one fixed Noetherian module. Noetherianity of $R$ does not force such a chain to stop: even the successive finite free summands of $\bigoplus_{j\ge1}R$ give a strictly ascending chain. The direct-sum example is only a logical test of the termination argument; it asserts nothing about a preferred big Cohen--Macaulay construction.

A concrete sufficient condition and the remaining gap.
For a finite nonzero $M\subset B$ it would suffice to have \[ M\cap(x_1,\ldots,x_j)B=(x_1,\ldots,x_j)M \qquad(1\le j<d). \] The preceding membership argument would then prove regularity one parameter at a time; the final quotient is nonzero by Nakayama. This formulates a genuine finite saturation problem. It is not solved by showing the equalities for one collection of relations before enlargement. In particular, a bound on the number of steps, or a reason a finite submodule simultaneously has all these equalities, is the first missing claim in this approach.
Verification and outcome.
The annihilator example and the relation $y[x]=0$ were checked symbolically; they do not require numerical experimentation. The domain reduction explicitly checks parameter ideals and the nonzero condition. [S1] asks the small-module question in its discussion of Frobenius methods; those methods' hypotheses cannot be removed by the foregoing argument. The attempt proves the component reduction and reconstructs the cases $d\le2$, while isolating a finite saturation obstruction in higher dimension. No termination theorem, general finite module, or counterexample has been obtained. The full conjecture remains UNRESOLVED.

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