# When is every “solid” perfect complex faithful?

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).

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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

OK, I think you are right that there are non-faithful solid guys even over a two dimensional regular local ring. For example, say $R = k[[x, y]]$ and we consider the complex $E$ given by $$A = \left( \begin{matrix} t_1 & t_2 \\ t_3 & t_4 \end{matrix} \right) : R^{\oplus 2} \longrightarrow R^{\oplus 2}$$ where $t_1, t_2, t_3, t_4$ are pairwise distinct monomials in $x$ and $y$ all of the same degree (for example $17$) such that $t_1 t_4 - t_3 t_2 = 0$. Such a quadruple of monomials exists and then $E$ is solid as follows from the discussion in the question. Then we consider $E \otimes E^\wedge$ and we get the complex $$M_2(R) \longrightarrow M_2(R) \oplus M_2(R) \longrightarrow M_2(R)$$ where the first arrow maps $X$ to $(XA, AX)$ and the second arrow maps $(Y, Z)$ to $AY - ZA$. There is the map $\eta$ from $R$ placed in degree $0$ to the displayed complex, sending $1$ to $(\mathbf{1}_2, \mathbf{1}_2)$. But, instead of using the characterization with $\eta$ you give, we can use the dual characterization with a trace map which sends $(Y, Z)$ to $\text{Tr}(Y) - \text{Tr}(Z)$. The characterization is that the dual complex $E^\wedge$ is faithful if and only if there exists an element $(Y, Z)$ of the kernel of the boundary with $\text{Tr}(Y) - \text{Tr}(Z) = 1$. If you are confused about the signs, dear reader, then I suggest you let $k$ have characteristic $2$. OK, but now let's look at the leading terms... say $$Y = \left( \begin{matrix} y_1 & y_2 \\ y_3 & y_4 \end{matrix} \right) + h.o.t \quad\text{and}\quad Z = \left( \begin{matrix} z_1 & z_2 \\ z_3 & z_4 \end{matrix} \right) + h.o.t$$ with $y_i, z_i \in k$. Multiplying out we get $$AY - ZA = \left( \begin{matrix} t_1(y_1 - z_1) + t_2 y_3 - t_3 z_2 & ... \\ ... & ... \end{matrix} \right) + \text{terms of degree} > 17$$ OK, so for this to be zero we see that we have $y_3 = z_2 = 0$ and $y_1 = z_1$ because the monomials $t_1, t_2, t_3, t_4$ are $k$-linearly independent. My millennia long experience with these types of things now suggests that working out the other dots we find that similarly $y_4 = z_4$. However, this contradicts the assumption that $\text{Tr}(Y) - \text{Tr}(Z) = 1$.