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.

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**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).

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**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!