Burch’s theorem

Let A be an r x (r – 1) matrix. Set δ_i = (-1)^i times the determinant of the (r – 1) x (r – 1) matrix you get by deleting the ith row. Then we have the equality

(◊) (δ_1, …, δ_r) A = 0.

Let I be an ideal in a Noetherian regular local ring R with dim(R) = dim(R/I) + 2 = depth(R/I) + 2. Then I has a minimal resolution of the form

0 —> R^{r – 1} —> R^r —> I —> 0

where r is the minimal number of generators for I. Denote A the matrix defining the map R^{r – 1} —> R^r. In this situation Burch’s theorem tells us that I is generated by the δ_i, and in fact the map R^r —> I is (up to a unit) given by the row vector (δ_1, …, δ_r).

Why is this useful? Well, suppose you want to deform R/I (see this post). It is often easy to see that there are lots of deformations, but what isn’t so easy is to prove that there are any unobstructed deformations. But in the situation above we can just choose a family of matrices A(t) formally depending on an auxiliary parameter t. Then the minors δ_1(t), …, δ_r(t) of A(t) generate an ideal I(t) in R[[t]]. Then R[[t]]/I(t) is a flat deformation of R/I by the criterion from the post on deformation theory: all the relations lift to R[[t]] because the equation (◊) is universal and hence holds also for our matrix A(t).

As an example consider a fat point in C^2, for example given by the ideal I = (x^n, x^{n – 1}y, x^{n – 2}y^2, …, y^n). The matrix A is the matrix whose ith column look is (0, …, 0, x, -y, 0, …, 0) with x in the ith spot. We can deform this by picking (0, …, 0, x – ta_i, -y + tb_i, 0, …, 0) with a_i, b_i 2n pairwise distinct complex numbers. The deformed scheme for t = 1 has n(n + 1)/2 reduced points, namely the points (a_i, b_j) with j ≥ i.

In fact you can show that any deformation is given by deforming the matrix (by applying the Burch’s theorem which is more general than what I said above to the resolution of the deformed ideal), and hence all deformations are unobstructed and the deformation space of the singularity defined by I is smooth. This in particular shows that the Hilbert scheme of points of a smooth surface is smooth.


The key to the solution to (***) which we formulated here is the construction of a derivation on the Tate resolution of R/I. I am not going to explain this in detail, partly because I do not have a good intuition for why this derivation should exist. So I am only going to give you the flavor of the thing: although the following is correct in spirit it is likely not completely correct in all details (in particular, some of the arguments below should not be done on the level of co/homology but rather on the level of complexes).

Let R = k[[x_1, …, x_n]] and I = (f_1, …, f_r) ⊂ m_R with r minimal for I. Consider as before the sequence

0 — > Rel —> R^{⊕ r} —> I —> 0

Note that if (***) is false, then we obtain a surjective map ξ : Rel —> R/I which annihilates the submodule TrivRel of trivial relations. This in particular implies that Rel/TrivRel = R/I ⊕ (Other part). Because R^{⊕ r} —> R —> R/I —> 0 is the beginning of a minimal resolution we see that we obtain an element e(ξ) ∈ Ext_R^2(R/I, R/I). Cupping with e(ξ) determines an operation

e(ξ) : Tor_n(R/I, M) —> Tor_{n – 2}(R/I, M)

for any module M (in particular M = k or M = R/I). Use Tate’s method to find a free dga with divided powers A which is a resolution of R/I as in the post here. It turns out that the fact that ξ is zero on trivial relations implies that we can represent e(ξ) by a derivation j : A —> A compatible with divided power structure which is homogeneous of degree -2 (this is where Gulliksen whose name I mentioned before comes in). On the other hand, the decomposition Rel/TrivRel = R/I ⊕ (Other part) implies that there exists a nonzero element x in Tor_2(R/I, k) such that δ(x) is nonzero (as an element of k). But this is a contradiction because some divided power of x is zero (as R is regular so finite projective dimension) and on the other hand δ^n(γ_n(x)) = (δ(x))^n is nonzero.

It is really a beautiful trick (apparently due to Gulliksen) to play off against each other the derivation pushing things down in degree and the divided power structure to go back up into the area where the Tors are zero.


We continue the discussion. Let R be a regular complete local ring with residue field k. Let S = R[[x_1, …, x_n]]. Let I ⊂ R and J ⊂ S be ideals such that IS &sub J and such that A = R/I —> B = S/J is a flat ring homomorphism. Consider the map

(**) I/m_RI —> J/m_SJ

In the previous post we claimed that the cokernel of this map is (J + m_RS)/(m_SJ + m_RS). To see this choose f’_1, …, f’_r ∈ J whose images f_1, …, f_r in S/m_RS form a minimal system of generators for the ideal (J + m_RS)/m_RS which is the ideal cutting out B/m_AB in S/m_RS. Think of B as a flat deformation of B/m_AB over A. Then by the discussion in this post, the flatness assumption implies that any relation between f_1, …, f_r in S/m_RS lifts to a relation in S/IS. Hence any element h of J ∩ m_RS is an element of IS + m_RJ as desired.

But more is true. Namely, recall from this post that with these choices we obtain an obstruction map

Ob : Rel/TrivRel —> S/(JS + m_RS) \otimes_k I/m_RI

(unfortunately the notation between these two posts isn’t compatible) where Rel is the module of relations between the f_1, …, f_r in S/m_RS. Now because J ⊂ m_S we can compose Ob with the canonical map S/(JS + m_R S) —> k to get a reduced map

Ob_reduced : Rel/TrivRel —> I/m_RI.

At this point an argument along the lines of the argument in the first paragraph shows that (**) is injective if this reduced obstruction map is zero (in fact I think it is equivalent, but I didn’t check this).

Having arrived at this point we see that it suffices to prove the following (changing back to the notation in the post on deformation theory):

(***) Given a proper ideal I in R = k[[x_1, …, x_n]], a minimal set of generators f_1, …, f_r for I with module of relations Rel and submodule of trivial relations TrivRel ⊂ Rel. Then any R-module homomorphism Rel/TrivRel —> R/I has image contained in m_R/I.

It turns out that this result is contained in a paper by Vasconcelos and with a little bit more detail on the proof it is Lemma 2 in this paper by Rodicio. The key appears to be the technique of Tate to find divided power dga resolutions of R/I combined with a technique for constructing derivations on dgas which is due to Gulliksen. We will return to this in a future post.

Avramov’s theorem

Let A —> B be a flat local homomorphism of Noetherian local rings. By the local criterion for flatness this also implies that the map on completions A* —> B* is flat. Hence, in order to prove Avramov’s theorem that I mentioned here it suffices to prove it for a flat map of Noetherian complete local rings.

Let A be a complete local Noetherian ring. Write, using the Cohen structure theorem, A = S/I where S is a regular complete local ring. The complete intersection defect of A is the nonnegative integer

cid(A) = dim_k(I/mI) – dim(S) + dim(A)

where k = S/m is the residue field of S. Note that dim_k(I/mI) is the minimal number of generators of the ideal I. Since S is regular we see that A is a complete intersection if and only if cid(A) = 0.

Next, let A —> B be a flat local homomorphism of complete local Noetherian rings. Avramov proved, among other things, that

(*) cid(B) = cid(A) + cid(B/m_AB)

in this situation. It is clear that this proves the result we mentioned in the previous post.

What does (*) mean in more elementary terms? For simplicity, let us assume that the residue fields of A and B are identified by the map A —> B. Write A = R/I. Set S = R[[x_1, …, x_n]] for some large n and choose a surjection S –> B (here we use that the residue fields are equal). Set J ⊂ S equal to the kernel of S —> B so that B = S/J. Consider the induced map

(**) I/m_RI —> J/m_SJ

The equality (*) is equivalent to  the injectivity of (**). Namely, flatness of A –> B gives dim(B) = dim(A) + dim(B/m_AB). By construction dim(S) = dim(R) + dim(k[[x_1, …, x_n]]). Finally, the map J/m_SJ —> (J + m_RS)/(m_SJ + m_RS) is surjective with kernel equal to the image of (**). (Proof omitted, but see next post.)

OK, now why is (**) injective? I claim that this is a question about the obstruction space for the deformation theory of the algebra B/m_AB over k. I will discuss this in the next post.

Complete intersections

Let us say that a Noetherian local ring is a strong complete intersection if it is of the form S/(f_1, …, f_r) where S is a regular local ring and f_1, …, f_r is a regular sequence. It turns out that if R = S/I = S’/I’ where S, S’ are regular local rings, then I is generated by a regular sequence if and only if I’ is (this is not a triviality!). But, as there exist Noetherian local rings which are not the quotient of any regular local rings, this definition of a strong complete intersection does not make a whole lot of sense. Note that it is clear that if R is a strong complete intersection, then so is R_p for any prime ideal p of R.

The Cohen structure theorem tells us we can write the completion of any Noetherian local ring as the quotient of a regular local ring. Thus we say a Noetherian local ring is a complete intersection if its completion is a strong complete intersection. By the Cohen structure theorem we can write the completion as a quotient of a regular local ring, so this definition makes sense.

The problem: Why is the localization of a complete intersection at a prime a complete intersection?

The solution to this conundrum comes from a theorem of Avramov: If R —> R’ is a flat local homomorphism of Noetherian local rings, and if R’ is a complete intersection, then R is a complete intersection. How do we use this? Suppose that R is a complete intersection and p a prime ideal of R. Let R’ be the completion of R. As R —> R’ is faithfully flat, we can find a prime p’ of R’ lying over p. By assumption R’ is a strong complete intersection, hence R’_{p’} is a strong complete intersection. Since the local ring map R_p —> R’_{p’} is flat, Avramov’s theorem kicks in and we see that R_p is a complete intersection!

Cool, no?

Deformation theory

This post is a brief review of deformation theory. I personally learned how to compute deformation spaces of singularities from Jozef Steenbrink, Theo de Jong and Duco van Straten (it is very enjoyable to compute deformation spaces of singularities late at night provided one has a large supply of either coffee or beer).

Consider a field k. Set R = k[x_1, …, x_n]. Let I be an ideal of R. Choose a set of generators f_1, …, f_r of I. Denote Rel the module of relations, i.e., such that we have a short exact sequence

0 —> Rel —> R^{oplus r} —> I —> 0

Suppose that A is an Artinian local ring with residue field k. Set R_A = A[x_1, …, x_n]. An embedded deformation of R/I over A is an ideal I_A ⊂ R_A such that I_A \otimes_A k = I and such that R_A/I_A is flat over A. It turns out that this is equivalent to the following:

  1. I_A can be generated by elements f’_1, …, f’_r whose reductions modulo m_A are equal to f_1, …, f_r, and
  2. for any (g_1, …, g_r) ∈ Rel there exist g’_1, …, g’_r in R_A such that ∑ g’_if’_i = 0.

If you’ve ever tried to compute the deformation space of a singularity then you’ve seen this. In particular, if A = k[ε] is the ring of dual numbers, then f_i’ = f_i + ε h_i and condition 2 implies that f_i → h_i defines a map from I to R/I. Thus the tangent space to this deformation functor is

T^1 = Hom_R(I, R/I) = Hom_R(I/I^2, R/I)

Note that this is typically an infinite dimensional space (except if R/I is Artinian), which make sense because we are doing embedded deformations. To get the first order deformation space of R/I as a singularity we can divide out by the module of derivations of R over k, but for this blog post we prefer not to do so.

Next, I want to consider the obstruction space to our given deformation problem. Suppose that we have a small extension A —> B of Artinian local rings with kernel K. Suppose that f’_1, …, f’_r is a bunch of elements of R_A whose images in R_B do define an embedded deformation. To see if f’_1, …, f’_r is also a deformation we need to check if relations G = (g_1, …, g_r) ∈ Rel lift to relations among the f’_1, …, f’_r. By the assumption that we have a deformation over B we know that we can pick (g’_1, …, g’_r) such that ∑ g’_i f’_i is an element Ob(G) of KR_A = R \otimes_k K. Picking different choices of g’_i changes Ob(G) by an element of KI. Hence the obstruction is a well defined map

Rel —> R/I \otimes_k K

Note that the trivial relations f_if_j = f_jf_i of course do lift to relations among the f’_i. Hence we see that the obstruction map is an element of

Ob ∈ Hom_R(Rel/TrivRel, R/I) \otimes_k K

where TrivRel ⊂ Rel is the module of trivial relations. However, the obstruction map above depends on the initial choice of lifts f’_i to elements of R_A (with more or less given images in R_B as we’re given the deformation over B). Altering the choice of these f’_i modifies Ob by an element of Hom_R(R^{oplus r}, R/I) \otimes_k K. Combining all of the above we see that

T^2 = Coker(Hom_R(R^{oplus r}, R/I) —> Hom_R(Rel/TrivRel, R/I))

Again this is usually an infinite dimensional vector space over k.

I’m going to try and say something intelligent about this obstruction space in a future blog post. But for the moment I just make the comment that Rel/TrivRel is equal to the first Koszul homology group for the ring R and the sequence of elements f_1, …, f_r.

Computing Tor

Let R be a Noetherian ring and let I, J be ideals of R. Then Tor^R_*(R/I, R/J) is a differential graded algebra (with zero differential). How does one get this algebra structure?

In a paper published in 1957, John Tate came up with the following strategy: Try to find a strictly commutative differential graded R-algebra A endowed with divided powers (as in this post) together with a given augmentation ε : A —> R/I such that

  1. H_i(A) = 0 for i > 0 and H_0(A) = R/I, and
  2. A is obtained from R by successively adjoining divided power variables.

The first condition means that A is quasi-isomorphic to R/I as a dga (with divided powers) and the second implies that A is a free resolution of R/I as an R-module. Hence we see that Tor_*(R/I, R/J) is the homology of A \otimes_R R/J = A/JA which is a dga with divided powers.

Tate shows that you can construct such dga resolutions of R/I by successively adjoining variables to kill cycles; starting with the Koszul complex for a set of generators of I. In the book by Gulliksen and Levin it is checked that the dga which Tate gets is endowed with divided powers.

I’d just like to make here the observation that this also determines divided powers on Tor^R_*(R/I, R/J). This despite the problem that in general the homology of a dga with divided powers isn’t endowed with divided powers as I mentioned here.

Namely, let B be a dga with divided powes. It turns out that the only obstruction to defining γ_n on H_*(B) is that it may happen that y ∈ B of even degree is a coboundary but γ_n(y) isn’t.  For example if B is the divided power algebra over F_2 on x in degree 1 and y in degree 2 and d(x) = y, then γ_2(y) isn’t a coboundary! But, if there exists a surjection φ : A —> B of dgas with divided powers where A is such that H_i(A) = 0 for i > 0, then this disaster doesn’t happen. The reason is that writing y = d(x) and x = φ(x’) for some x’  ∈ A, then y’ = d(x’) is a coboundary in A, hence γ_n(y’) is a cocycle in A by the compatibility of divided powers with d, hence γ_n(y’) = d(x”) as H_i(A) = 0, hence γ_n(y) = d(φ(x”)).

And of course, in the situation of Tate’s construction above, we have that A/JA is the quotient of a dga acyclic in positive degrees!

Divided powers

Consider a differential graded algebra (A, d) sitting in homological degrees 0, 1, 2, … and with d : A_n —> A_{n – 1}. Then the cohomology H(A) is also a differential graded algebra (with zero differential of course).

We say that (A, d) is strictly commutative if xy = (-1)^e yx with e = deg(x)deg(y) and x^2 = 0 when x has odd degree. In this case H(A) is a strictly commutative differential graded algebra.

We say that (A, d) is a strictly commutative differential graded algebra endowed with divided powers if for every homogeneous element x of A in even degree d we have divided powers γ_n(x) of degree nd satisfying the usual rules for divided powers, and satisfying the compatibility

d(γ_n(x)) = d(x) γ_{n – 1}(x), for all n > 1

with the differential. Then H(A) is a strictly commutative differential graded algebra endowed with divided powers, right?

Wrong! Can you spot the mistake?

Baby not included

Are you attending the conference Moduli Spaces and Moduli Stacks in May? Or will you come by Columbia University in the next few months? If so, then you can order a Stacks Project T-shirt and pick it up here at the math department.

The picture shows me wearing the T-shirt; on the back it has “The Stacks Project” in white letters. Since I am trying to avoid spam orders, I will not put a link to the order form here, but I suggest you follow the link to the moduli conference above to find the link over there.

Quasi quasi-coherent sheaves

On any ringed topos there is a notion of a quasi-coherent sheaf, see Definition Tag 03DL. The pullback of a quasi-coherent module via any morphism of ringed topoi is quasi-coherent, see Lemma Tag 03DO.

Let (X, O_X) be a scheme. Let tau = fppf, syntomic, etale, smooth, or Zariski. The site (Sch/X)_{tau} is a ringed site with sheaf of rings O. The category of quasi-coherent O_X-modules on X is equivalent to the category of quasi-coherent O-modules on(Sch/X)_{tau}, see Proposition Tag 03DX. This equivalence is compatible with pullback, but in general not with pushforward, see Proposition Tag 03LC.

Let me explain this last point a bit. Suppose  f : X —> Y is a quasi-compact and quasi-separated morphism of schemes. Denote f_{big} the morphism of big tau sites. Let F be a quasi-coherent O_X-module on X. The corresponding quasi-coherent O-module F^a on (Sch/X)_{tau} is given by the rule F^a(U) = Γ(U, φ^*F) if φ : U —> X is an object of (Sch/X)_{tau}. In general, for a sheaf G on (Sch/X)_{tau} we have f_{big, *}G(V) = G(V \times_Y X). Hence we see that the restriction of f_{big, *}F^a to V_{Zar} is given by the (usual) pushforward via the projection V \times_Y X —> V of the (usual) pullback of F to V \times_Y X via the other projection. It follows from the description of quasi-coherent sheaves on (Sch/Y)_{tau} as associated to usual quasi-coherent sheaves on Y that f_{big, *}F^a is quasi-coherent on (Sch/Y)_{tau} if and only if formation of f_*F commutes with arbitrary base change. This is simply not the case, even for morphisms of varieties, etc.

On the other hand, we know that f_*F commutes with any flat base change (still assuming f quasi-compact and quasi-separated). Hence f_{big, *}F^a is a sheaf H on (Sch/Y)_{tau} such that H|_{V_{Zar}} and H|_{V_{etale}} are quasi-coherent. Moreover, the same argument shows that if G is any sheaf of O-modules on (Sch/X)_{tau} such that G|_{U_{Zar}} or G|_{U_{etale}} is quasi-coherent for every U/X then H = f_{big, *}G is a sheaf such that H|_{V_{Zar}} or H|_{V_{etale}} are quasi-coherent for any object V of (Sch/Y)_{tau}. Moreover, this property is also preserved by f_{big}^* as this is just given by restriction.

Thus a convenient class of O-modules on (Sch/X)_{tau} appears to be the category of sheaves of O-modules F such that F|_{U_{etale}} is quasi-coherent for all U/X. These “quasi quasi-coherent sheaves” are preserved under any pullback and pushforward along quasi-compact and quasi-separated morphisms. Via the approach I sketched here they give a notion of quasi quasi-coherent sheaves on the tau site of any algebraic stack with arbitrary pullbacks and pushforward along quasi-compact and quasi-separated morphisms. An interesting example of a quasi quasi-coherent sheaf is the sheaf of differentials Ω on the etale site that I mentioned in here.

Can anybody suggest a better name for these sheaves?