Re: restriction of the derived tensor product

On Oct 11, 12:58 pm, Axel Vogt <&nore...@xxxxxxxxxxx> wrote:
John H Palmieri wrote:
Suppose that k is a field, S is a commutative k-algebra and
R=S[x]/(x^2). Let X and Y be two objects in the (unbounded) derived
category of R, and write X|S and Y|S for X and Y restricted to S -- that
is, for X and Y viewed as objects in the derived category of S. Is it
true that

(*) if X tensor_R Y = 0, then X|S tensor_S Y|S = 0,

where "tensor" means the derived tensor product in the derived category?

I have a proof, and while I hope the result is true, I don't believe my
proof. The proof goes like this: first, the restriction X|S can be
written as X|S = X tensor_R (R|S). Then use associativity and
commutativity of the various derived tensor products (throughout, tensor
means derived tensor product, and = means isomorphic):

X|S tensor_S Y|S = (X tensor_R R|S) tensor_S (Y tensor_R R|S)
= X tensor_R (R|S tensor_S (Y tensor_R R|S))
= X tensor_R (R|S tensor_S (R|S tensor_R Y))
= X tensor_R ((R|S tensor_S R|S) tensor_R Y)
= X tensor_R (Y tensor_R (R|S tensor_S R|S))
= (X tensor_R Y) tensor_R (R|S tensor_S R|S)

Then certainly X tensor_R Y = 0 implies that the right-hand side is
zero. Now, I don't believe this because it doesn't use anything
specific about the rings involved, and it seems to apply, for example,
if S=k and R=any k-algebra; but there are certainly objects X and Y in
the derived category of some k-algebra R so that X tensor_R Y is zero,
while (X|k) tensor_k (Y|k) is non-zero.

The question I care about most is the one marked with a (*), but I'm
also curious as to what's wrong with this proof...

With some technical additions I think (*) is Proposition 5.9 in
Hartshorne "Residues and Duality" (where his f is through your
R ---> S, usually the conditions in R&D can not be simply dropped).

(I've responded to this once, but a bit too hastily.)

Hartshorne requires some boundedness hypotheses which I don't want,
but a similar result is in Spaltenstein's paper "Resolutions of
unbounded complexes". However, neither result is what I want, I
think: after discussing it with a colleague, I am convinced that the
map f^* in Hartshorne (and in Spaltenstein) is an induction functor,
not a restriction functor. So I think this doesn't actually help me.

As I said in the original message, my (*) should not hold in general;
in particular, it should not hold when S=k is a field and R is an
arbitrary commutative k-algebra -- I think it fails if R=k[x], for
instance. I still hope that it might hold in the particular situation
I set up, where R=S[x]/(x^2). Somehow, the fact that the inclusion
map S --> R induces a bijection on Spec ought to be relevant...

.