In his really nice thesis, Tobias Dyckerhoff proved the following theorems about matrix factorizations(of possibly infinite rank) over a regular local k-algebra R with a function w and residue field k such that the Tyurina algebra, T= $R/(w,dw)$ is finite dimensional. This last condition says that w has an isolated singularity. For further reference, let S denote the ring R/(w).

1) The homotopy category of matrix factorizations has a compact generator as a triangulated category, which he denotes as $k^{stab}$.

2) As a consequence of 1), he derives that there is a natural complex which represents the identity functor thought of as an element of MF($R\otimes R,1\otimes w-w\otimes 1)$ which he denotes as the stabilization of the diagonal $R^{stab}$.

3) MF(R,w) is a Calabi Yau dg-category.

Now my question is how much of the above remains true for when the singularity is non-isolated? In some writings, Kontsevich, while not explicitly saying so, writes as if the homotopy category always has a compact generator and that the category is there by "dg-affine", e.g. equivalent to D(A), the derived category of modules over a dg-algebra. Is this indeed known to be true or false? If not, is there a way to prove 2) without making reference to 1)? I'm asking because I haven't found anything about this stuff in the literature, but a lot of things in this field are not written or written in physics literature that I'm not familiar with.

I've checked a few examples with non-isolated singularities and it appears that for example in the category of factorizations $(k[[x,y]], xy^2)$, that while $k^{stab}$ doesn't generate as Dyckerhoff proves, one has $(k\oplus k[[x]])^{stab}$, which I think does generate. The way I want to argue this is Dyckerhoff's theorem 3.6 that it is enough to show that $Tor_S(k\oplus k[[x]], M)$ implies that $Tor_S(N,M)=0$, where N is a finitely generated T module and M is any S module.Then one does an analysis of finitely generated modules over T(I didn't think about the characteristic 2 case) and does some devissage with short exact sequences. Please let me know if this sounds off. I also think that with a bit more calculation one can prove similarly that in (k[[x,y,z]], xyz) the module $(k[[x]]\oplus k[[y]]\oplus k[[z]]\oplus k)^{stab}$ is a compact generator.

Added: I think the right generalization of the above two examples is the following, in http://websupport1.citytech.cuny.edu/faculty/hschoutens/pdf/finiteprojdim.pdf, the author introduces the notion of a "net". The above method should give a compact generator, whenever the net of finitely generated modules over T is generated as a net by finitely many modules. This happens for example when T has finitely prime ideals. The modules A/p, where p is a prime, generate the net of finitely generated projective modules over T, which is enough to prove the vanishing above. In particular, this should take care of the case when T has dimension 1. A question is what are some conditions on (R,w) which lead to the net of finitely generated modules over T being generated by finitely many objects?

Assuming that this right, I think that to derive 2 and 3 for these examples becomes a formality in view of Dyckerhoff's section 5. One just replaces his compact generator with the new one.

  • 1
    $\begingroup$ In Schouten's work he shows that the net generated by the cylic modules of the form A/p, for p singular, is the entire category of finitely generated modules. A net lives in the abelian category of modules, but it's easy to see using Schouten's work that the objects A/p, for p singular, generate the bounded finite derived category. Moreover they're compact by work of Krause. However the set of singular primes set is not finite usually. Roquier have shown that a compact generator exists for the finite bounded derived category of most commutative rings; I believe all k-algebras of finite type. $\endgroup$ Aug 27, 2010 at 19:50
  • $\begingroup$ It's unclear, to me at least, how the two constructions are related. $\endgroup$ Aug 27, 2010 at 19:50
  • $\begingroup$ Thanks! Maybe arxiv.org/PS_cache/math/pdf/0310/0310134v3.pdf is precisely what I need to understand.... $\endgroup$ Aug 27, 2010 at 21:23

2 Answers 2


The answer to (1) is yes for any local abstract hypersurface $S$ whose singular locus is closed (which is barely a hypothesis, and free in the case of interest). Let us write $\mathrm{Sing} \;S$ for the singular locus, which we can write as a union finitely many irreducible components corresponding to primes $\mathfrak{p}_i$ for $i=1,\ldots,n$. Then the image of the object

$$ \bigoplus_{i=1}^n S/\mathfrak{p}_i $$

in the category of matrix factorizations is a compact generator. The point is that there is a notion of support for objects of the category of all matrix factorizations. For a hypersurface this support gives a classification of the localizing subcategories in terms of subsets of the singular locus. The image of the given object is supported everywhere so must generate.

The statement about the classification of localizing subcategories is not actually published anywhere (Iyengar has announced this as part of a more general result on complete intersections and I have an (independent and different) proof, also for complete intersections, which also works for certain singular schemes - but Iyengar has not yet released a preprint and the results of mine are in my thesis which isn't publicly available yet). Although there is this paper by Takahashi, the main result of which is sufficient to give the statement about the generator above.

The generator above actually works for any noetherian ring which is locally a hypersurface if one works in one's favourite infinite completion of $D_{\mathrm{Sg}}(S)$.

Edit: As promised in the comments (well sort of, it didn't turn out to be soon) here is the link to the preprint concerning subcategories of singularity categories from which one can deduce the existence of compact generators in certain cases as above.

  • $\begingroup$ This is awesome! Very geometric! $\endgroup$ Aug 28, 2010 at 0:51
  • $\begingroup$ Thanks Greg for sharing your work! I take it based upon your discussion that if X is a Noetherian scheme of finite type with enough locally frees(maybe seperated) and is a hypersurface, then assuming for simplicity that Z the singular locus is irreducible and let $i_*(O_Z)$ be the pushforward of the structure sheaf on Z with the reduced induced subscheme structure. This would be a compact generator for the some completion of $D_sg(X)$ too? $\endgroup$ Aug 28, 2010 at 7:55
  • $\begingroup$ @Daniel: The infinite completion I have in mind, for a noetherian separated scheme $X$, is $K_{\mathrm{ac}}(\mathrm{Inj}\;X)$ the homotopy category of acyclic complexes of quasi-coherent injective $O_X$-modules. What you say above is correct if $X$ is defined by a section of an ample line bundle on a noetherian regular separated scheme. For any noetherian separated scheme $X$ with hypersurface singularities there is a weaker sense in which $i_*(O_Z)$ generates $K_{\mathrm{ac}}(\mathrm{Inj}\;X)$, but I am not sure if it generates in the usual sense and kind of suspect it doesn't in general. $\endgroup$ Aug 29, 2010 at 7:56
  • $\begingroup$ cont'd: - $i_*(O_Z)$ is not actually an object of $K_{\mathrm{ac}}(\mathrm{Inj}\;X)$ but there is a functor from the derived category making this homotopy category an infinite completion of the singularity category which one can apply. The reference for this is Krause's paper "The Stable Derived Category of a Noetherian Scheme". Also, thanks Daniel and Kevin - I'll try to remember to edit in a link when this stuff is ready (which will be soon). $\endgroup$ Aug 29, 2010 at 8:03

This won't answer your question completely, but I thought I would expand on my comment above. This is very much an algebraic approach to your question.

The homotopy category of finitely generated matrix factorizations over a hypersurface $S$ is equivalent to the stable category of maximal Cohen-Macaulay $S$-modules, which is essentially due to Eisenbud. By a theorem of Buchweitz the later category is equivalent to the quotient $D^f_b(S)/ Thick(S)$, where $D^f_b(S)$ is the full subcategory of the derived category whose objects are the complexes with finitely generated cohomology. A reference for this is Buchweitz's manuscript which is available at https://tspace.library.utoronto.ca/handle/1807/16682. So to answer your question it is enough to find a generator for $D^f_b(S)$.

In the paper of Roquier you cited above he shows that for a commutative algebra of finite type over a field, such a compact generator exists (he works much more generally). I haven't spent much time with the proof, but from my understanding it is not at all constructive.

On the other hand Schoutens has shown that for a commutative noetherian ring $R$, modules of the form $R/\mathfrak{p}$, such that $R_{\mathfrak{p}}$ is not regular, can be used to construct the category of finitely generated $R$-modules by taking extensions, direct summands and cosyzygies. There are some details I'm skipping but Schoutens does this carefully and his work can be used to show that the objects $R/\mathfrak{p}$, such that $R_{\mathfrak{p}}$ is not regular, and $R$, generate $D^f_b(R)$.

Usually there are an infinite number of primes in the singular locus...I've thought about how to use Schoutens theorem to come up with a finite set that generate but I haven't made much progress. One thing to note is that if your local hypersurface has a 1-dimensional singular locus, and is the quotient of a power series ring, then there will be a finite number of primes in the singular locus. This is the case in the examples you listed.

  • $\begingroup$ Thanks Jesse! You summarize a lot of the main ideas/issues really well! I didn't know this theorem about the bounded category of coherent modules. I'm used to thinking that perfect modules are better than D^b(CohX) when X isn't smooth so this is really surprising to me. I think I maybe able to use this observation to prove what I want, I'll have to check. One final subtlety, which I added in an edit and maybe you didn't catch is that I want to consider matrix factorizations of possibly infinite rank... because otherwise the notion of compact doesn't make sense at all. I noted the 1-d case too. $\endgroup$ Aug 27, 2010 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.