Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R$ be a noetherian commutative ring. Consider $D^{perf}(R)=K^b(R-proj)$ the category of bounded complexes of finitely generated projective $R$-modules, with maps of complexes up to homotopy.

QUESTION: For which $R$ is it true that every complex $E \in D^{perf}(R)$ whose support is the whole spectrum, $Supph(E)=Spec(R)$, is also $\otimes$-faithful?

Explanations:

(1) For simplicity, let us call an object $E$ "solid" if it is supported everywhere: $Supph(E)=Spec(R)$. This means that for every $P\in Spec(R)$, the object $E_P\in K^b(R_P-proj)$ is non-zero (i.e. not null-homotopic). Equivalently, the image of $E$ in $D(\kappa(P))$ is non-zero for every residue field $\kappa(P)$. Equivalently again, the ordinary support of the total homology of $E$ is the whole $Spec(R)$, as a finitely generated $R$-module. ($Supph$ is for support of the $h$omology and is called the "homological support".)

(2) An object $E$ in a tensor-category can be called "$\otimes$-faithful" (simply "faithful") if $-\otimes E$ is a faithful functor, i.e., for every map $f:E'\to E''$, the map $f\otimes E: E'\otimes E\to E''\otimes E$ is zero only if $f$ is zero.

(3) It is an exercise on rigid tensor-triangulated categories to show that $E\in D^{perf}(R)$ is faithful if and only if the unit $\eta:R\to E\otimes E^\vee$ is split injective, where $E^\vee$ is the dual of $E$ and of course $R=R[0]=\cdots 0\to 0\to R\to 0\to 0\cdots$ is the unit of the tensor on $D^{perf}(R)$. (Use that $\eta$ is split injective after applying $-\otimes E$ by the unit-counit relation.) Split injectivity of $\eta:R\to E\otimes E^\vee$ forces $E$ to be solid. In short, we always have faithful$\implies$ solid and the question is: For which $R$ is the converse true?

(4) It is well-known (and non-trivial) that a map $f:E'\to E''$ in $D^{perf}(R)$ which goes to zero in every residue field must be $\otimes$-nilpotent: $\exists\ n>0$ such that $f^{\otimes n}=0$. So, if $f\otimes E=0$ with $E$ solid, one can conclude $f^{\otimes n}=0$ for some $n>0$ but not necessarily $f=0$ in general.

(5) Note that it is necessary for "solid$\implies$ faithful" that $R$ be reduced. Indeed, let $a\in R$ with $a^2=0$ and let $E=cone(a)$ be the complex $E=\cdots0\to 0\to R\overset{a}\to R\to 0\to 0\cdots$. Then $f=a:R\to R$ satisfies $f\otimes E=0$ and $E$ is solid.

(6) Reduced is not enough though. I've an example of a non-faithful solid complex over $R=k[T^2,T^3]\subset k[T]$ ($k$ a field).

(7) When $R$ is Dedekind, every solid complex has a non-zero projective summand, hence is faithful. So, that is a class of examples where solid does imply faithful.

(8) In particular, I would be very interested in a regular ring $R$ with a non-faithful solid object. (Call that a conjecture in disguise, if you want.)

References:

M. Hopkins, "Global methods in homotopy theory", Homotopy theory (Durham, 1985), 73--96.

A. Neeman, "The chromatic tower for $D(R)$", Topology 31 (1992), no. 3, 519--532. Article (requires access).

R. Thomason, "The classification of triangulated subcategories", Compositio Math. 105 (1997), no. 1, 1--27. Article (requires access).

share|improve this question
1  
Very nice question! –  Hailong Dao Apr 20 '10 at 17:35
    
Doesn't (4) give an equivalent condition? ie $E$ is solid if and only if for every $f$, $f\otimes E=0$ implies that for some $n$, $f^{\otimes n}=0$. (The converse direction follows since $E\otimes k(p)\not=0$.) –  Don Stanley Apr 22 '10 at 7:05
    
Correct: (4) is equivalent to solid. The question is: For which rings R does "E solid" imply "E faithful". It is a question about R –  Paul Balmer Apr 22 '10 at 15:06
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.