Let $(X,\mathcal{O}_X)$ be a scheme or a general ringed space. First recall that a complex of $\mathcal{O}_X)$-modules $\mathcal{E}^{\bullet}$ is called strictly perfect if $\mathcal{E}^{\bullet}$ is a two-side bounded complex of finitely generated locally free $\mathcal{O}_X)$-modules.

Then we have the following definition of pseudo-coherent complex of $\mathcal{O}_X)$-modules:

Let $(X, \mathcal{O}_X)$ be a ringed space. Let $\mathcal{E}^{\bullet}$ be a complex of $\mathcal{O}_X$-modules. Let $m \in \mathbf{Z}$.

  1. We say $\mathcal{E}^\bullet$ is $m\textit{-pseudo-coherent}$ if there exists an open covering $X = \bigcup U_i$ and for each $i$ a morphism of complexes $\alpha_i : \mathcal{E}_i^\bullet \to \mathcal{E}^\bullet|_{U_i}$ where $\mathcal{E}_i$ is strictly perfect on $U_i$ and $H^j(\alpha_i)$ is an isomorphism for $j > m$ and $H^m(\alpha_i)$ is surjective.

  2. We say $\mathcal{E}^\bullet$ is $\textit{pseudo-coherent}$ if it is $m$-pseudo-coherent for all $m$.

See http://stacks.math.columbia.edu/tag/08CA

If $\mathcal{E}^\bullet$ is pseudo-coherent, then locally the cohomology $H^{\bullet}(\mathcal{E})$ is bounded above but not bounded below.

It seems that even if $\mathcal{E}^\bullet$ is pseudo-coherent, for different $m$ we may choose different cover $\{U_i\}$ and different $\mathcal{E}_i^\bullet$.

$\textbf{My question}$ is: is the definition of pseudo-coherent complex equivalent to the following:

We say $\mathcal{E}^\bullet$ is $\textit{pseudo-coherent}$ if there exists an open covering $X = \bigcup U_i$ and for each $i$ a morphism of complexes $\alpha_i : \mathcal{E}_i^\bullet \to \mathcal{E}^\bullet|_{U_i}$ where $\mathcal{E}_i$ is a $\textit{bounded above}$ complex of finitely generated locally free sheaves on $U_i$?

  • can I ask a question for my own sanity? If you're working over a local ring R, is $\bigoplus_{i \geq 0} R[i]$ pseudo-coherent? – bananastack Mar 20 '15 at 15:52
  • @bananastack Yes. Actually for each $m$ we can take $\bigoplus_{0\leq i\leq -m}$ to be our strictly perfect complex in part one of the definition. – Zhaoting Wei Mar 20 '15 at 16:24
up vote 3 down vote accepted

No, this is true under noetherian hypothesis. See the relevant exposé by Illusie in SGA 6, It is related to the phenomenon that not every finitely presented module is coherent whenever the ring is not coherent itself.

  • Does my definition itself stand for other types of complexes? – Zhaoting Wei Mar 20 '15 at 22:56
  • Your definition is the standard one. The idea is that on non-noetherian schemes $D^-_c$ does not behave well, so the notion of pseudo-coherent is a good substitute. – Leo Alonso Mar 20 '15 at 23:35
  • The Noetherian condition can be loosened to be locally Noetherian, isn't it? – Zhaoting Wei Mar 20 '15 at 23:53
  • You're right. Beyond that, you need the full definition of pseudo-cohernce. – Leo Alonso Mar 22 '15 at 14:29
  • What Leo is saying is that by a happy accident, in the noetherian case, "Oka's theorem" is true -- the structure sheaf is coherent. In fact, as Leo indicates, if the structure sheaf is coherent,then the phenomenon you described is true.But unfortunately, it is not always true that structure sheaves are coherent. – Pramathanath Sastry May 7 '17 at 16:17

Yes, provided that $X$ is a scheme (or, for instance, an algebraic stack). In particular, we need not assume that $X$ is locally noetherian.

Let $(X, \mathcal{O}_X)$ be a locally ringed space (or a locally ringed topos). Let $\mathcal{E}^\bullet$ be a complex of $\mathcal{O}_X$-modules and let $U_\alpha$ be a covering of $X$. Then it is clear that $\mathcal{E}^\bullet$ is pseudo-coherent if and only if $\mathcal{E}^\bullet|_{U_\alpha}$ is pseudo-coherent for all $\alpha$. Moreover, it is clear that a bounded above complex of finite locally free $\mathcal{O}_X$-modules is pseudo-coherent (the map from the brutal truncation at $m$ gives a perfect complex witnessing that it is $m$-pseudo-coherent). In particular, your alternative definition (stated at the end of the question) is stronger than the usual definition.

Now assume that $X$ is an affine scheme. Then a complex of $\mathcal{O}_X$-modules is pseudo-coherent if and only if its is quasi-isomorphic to a bounded above complex of finite free $\mathcal{O}_X$-modules. Indeed, if $\mathcal{E}^\bullet$ is pseudo-coherent, then it has quasi-coherent cohomology, so it is quasi-isomorphic to a complex of quasi-coherent modules. This reduces the problem to the corresponding statement for modules over a ring, which is Stacks Project, Tag: 064U.

This shows that your alternative definition is equivalent to the usual definition when $X$ is a scheme (or an algebraic stack).

I don't believe that your alternative definition is equivalent to the usual one for arbitrary locally ringed spaces, but I don't have a counterexample.

Note: Your definition of strictly perfect complex is not the correct one for arbitrary ringed spaces. A complex on a ringed space is strictly perfect if it is a bounded complex of direct summands of locally free complexes.

  • I might be a little confused, but are you sure that the map from the "brutal" truncation goes in the correct direction? – Denis Nardin Nov 3 '17 at 10:24
  • Yes, cohomological grading is used (arrows in the complex go from smaller to higher degrees). – Daniel Bergh Nov 3 '17 at 10:37

Your Answer

 
discard

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.