Example please!

Can you, dear reader, send me (or post in comments) an example of a Noetherian ring A and finite A-modules M, N such that the canonical map

RHom_A(M, A) ⊗^L_A N —–> RHom_A(M, N)

is not an isomorphism in D(A)? Much obliged.


Update Jan 27, 2013. Bhargav emailed me the following observation. Any mistakes are mine.

Lemma If A has a dualizing complex ω and the map is an isomorphism for every M, N then A has to be Gorenstein.

Proof. Denote the map above f_{M, N}. Then f_{M, N} is also an isomorphism for any M, N \in D^b_{Coh}(A) by triangles. Now choose M = N = ω. Then RHom(ω, ω) = A, so ω is invertible for the derived tensor product. This forces ω to be (locally) the shift of an invertible module by the following lemma. End.

Lemma If A is a Noetherian local ring and K is in D^b_{Coh}(A) and K ⊗^L_A M = A for some object M of D(A), then K is the shift of an invertible A-module.

Proof. Observe that K can be represented by a bounded above complex K^* of finite free A-modules all of whose differentials are zero modulo the maximal ideal m_A of A. Let k be the residue field of A. We have

k = A ⊗^L_A k = (K ⊗^L_A M) ⊗^L_A k = (K ⊗^L_A k) ⊗^L_k (M ⊗^L_A k)

The lemma is clear for D(k) as this is the category of graded k-vector spaces. We conclude that K ⊗^L_A k which is represented by K^* ⊗_A k is isomorphic to k[a] for some integer a. Thus we conclude that K^{-a} = A and K^n = 0 for other n as desired. End.

The conclusion from the comments below is that the ring of dual numbers k[\epsilon] does satisfy the property that f_{M, N} is an isomorphism for all M, N finite. [This is wrong! See update below.] This is a Gorenstein ring so there is no contradiction. On the other hand the ring R = k[x, y, w]/(x^2, y^2, xw – yw, w^2) does not satisfy the property, which now also follows from Bhargav’s observation as this ring isn’t Gorenstein (the socle has dimension 2).


Update Jan 29, 2013. Actually, Bhargav send me the following update. As usual any and all mistakes are mine.

Lemma If the map is an isomorphism for all finite M,N, then A is regular.

Proof. A is Gorenstein as before, so A has finite injective dimension. Hence, RHom(M,A) is a finite A-complex for any M. Then RHom(M,A) (x) N is bounded above (being the derived tensor product of two bounded above A-complexes). On the other hand, the right hand side is not bounded above if A is not regular. For example if A is local with residue field k, we could take M = N = k, in which case Ext^i_A(k,k) is non-zero for arbitrarily large i (as the minimal free resolution does not terminate). End.

In particular, this shows that the map is not an isomorphism for A = k[x]/(x^2), and M = N = k. This contradict the discussion above in the comments. The mistake I (and I think also Ben) made is that in computing the LHS for A = k[ε] = k[x]/(x^2), and M = N = k I took a free resolution of k over k[ε], then I took the dual of this complex and used it to compute the left hand side. But a complex of free A-modules isn’t K-flat so can’t be used to compute the derived tensor product… Argh!

Apologies for all the confusion!

21 thoughts on “Example please!

  1. At the very least, $M$ and $N$ have to have infinite projective dimension over $A$. But, the obvious example of $A=k[x]/(x^2)$ and $M=N=k$ doesn’t work.

    • OK, how about a slight modification of your idea. Suppose we have a Noetherian ring R with two elements x and y such that Ker(x) = Im(x) and Ker(y) = Im(y) in R. Take M = R/xR and N = R/yR. Then I think the RHS is represented by the complex
      N — x –> N — x –> N — x –> ….
      whereas the LHS is the total complex associated to the double complex which has R in every spot of the lower right quadrant (4th quadrant) and has multiplication by x going horizontally and multiplication by y going vertically. I think that in this situation the LHS can have nonzero cohomology in degree -1 for example for the ring R = k[x, y, w]/(x^2, y^2, xw – yw, w^2). Namely, the element z = xw = yw in the spot (0, -1) I think gives a nonzero cohomology class. Hmm…?

      • I think your example does indeed work. I think of it a little differently. First of all, in the derived category RHom(M,A) is equivalent to H^0, which is ker(x). So, we can compute RHom(M,A)\otimes^L N by taking a projective resolution of ker(x). But, there is a surjection R->ker(x)=im(x) with kernel ker(x). So, this is the infinite sequence stretching to the left with differentials equal to x in all cases. Tensoring with y, we can consider the degree 0 term, which is the cokernel of R/yR->R/yR given by tensoring with x. In the case at hand the image of x in R/yR is x and xw. So, the cokernel is the residue field k. On the other hand, the degree 0 cohomology of RHom(M,N) is the kernel of R/yR->R/yR, which is a 2-dimensional vector space over the residue field generated by x and w. Hopefully there are no silly mistakes here…

        • I wanted to point out that this recipe does still work, despite the updates. In fact, had I followed this recipe originally for k[e]/(e^2) it would have resulted in the desired example immediately. The issue is that after dualizing and obtaining RHom(M,R), which is bounded below, one obtains the quasi-isomorphic complex M again. The original projective resolution is indeed a K-flat complex, since it is bounded above. Doing this for R=k[e]/(e^2) and M=N=k yields a LHS with cohomology k in all non-positive degrees and 0 otherwise, while the RHS is k in all non-negative degrees and 0 otherwise. They are as unequal as can be!

  2. I think I’ve confused myself somehow…

    Is it not true that an object M in the derived category satisfies this property with respect to every N if and only if M is a perfect complex? (Sketch proof: The collection of such M is a thick subcategory that contains A, and so all perfect complexes satisfy this. On the other hand, the condition itself says that RHom(M, -) is isomorphic to M^* \otimes (-) which preserves coproducts, and so M is a compact object of D(A) hence perfect).

    So is this the same as asking for an example of a finite A-module that is not a perfect complex in the derived category? Wouldn’t, like… Z/p as a Z/p^2-module work? It has infinite projective dimension so it can’t be a perfect complex (sketch proof: over an affine scheme, perfect complexes are globally quasi-isomorphic to strict perfect complexes.)

    Apologies if I’m missing something here.

    • Oh whoops! I see what happened… the condition is not for “arbitrary N” but only for finite “N”, and my example doesn’t work then, I think. Hm… Back to the drawing board.

      • Yes! Also the characterization of perfect objects of D(A) as the compact objects of D(A) uses a lot more than just Mod_A ⊂ D(A) doesn’t it? I still like your idea though.

  3. Ok, here’s another observation:

    As noted earlier, the collection of E. in D(A) satisfying the assertion when M = E. and N is an arbitrary finite A-module is closed under (i) quasi-isomorphism, (ii) triangles, and (iii) arbitrary direct sums. Therefore it is a localizing subcategory of D(A), and by a result of Neeman these all look like _{p \in P} where P \subset Spec(R) is a collection of primes. So if a counterexample exists, there must also be one of the form M=k(p)… but I don’t know how helpful this is, since k(p) might not be finite.

  4. Since homming out of a sum commutes with the sum, and I guess I was assuming that the tensor commutes with sums in the derived category- is that not true?

  5. I like the first lemma! Very pretty. The second one is known and has appeared in print in various places. It holds more generally for any commutative ring by Fausk, a result which appeared in 2003 in JPAA. It is even true for simplicial commutative rings, commutative dg-algebras, and connective E_infinity-ring spectra.

    @Dylan: it is true that tensor products preserves sums (in each variable separately). No condition is needed on the objects. The fact is reflective of the fact that the tensor product itself is built as a colimit.

    @Dylan: it should not be true that for a fixed complex E the collection of N satisfying the condition forms a localizing subcategory. If it did, since it contains the unit A, it would have to be the entire derived category D_qc(A). (Also, finite N cannot be closed under arbitrary coproducts for trivial reasons, but that is not the main point.)

    @Johan: the characterization of the perfect objects as the compact objects of D_qc(X) is a bit trickier. It is not hard to show that perfect objects are compact, but the converse is the hard part. The proof (for D_qc(X) where X is quasi-compact and quasi-separated) is due to Bondal and van den Bergh. They show that D_qc(X) is generated by a single perfect object. It follows that the compact objects are the idempotent completion of the perfect objects. Since the perfect objects are already idempotent complete, you’re done. I suppose in the affine case the argument is actually much easier, because A is already obviously a perfect generator of D_qc(A). So, I think Dylan’s intuition in this case is correct.

    @Johan: I don’t think your example can work, because R/xR and R/yR are both perfect in the derived category. I will check this using less fanciness in a moment to see where the computation actually goes awry (assuming it does).

  6. Well, I’m glad that is cleared up. To be honest, that accords much more closely with what I expected before the discussion began. Hence, my own surprise at the faulty calculation for k[e].

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