1
$\begingroup$

This is a sequel to the question

[1] Example of an additive functor admitting no right derived functor,

the purpose being to state a particular case of [1] which would be as easy as possible (though this particular case is still much too hard for me).

See [1] for any unexplained notation and for some of the definitions involved.

Let $k$ be a field, let $T$ be an indeterminate, let $A$ be the $k$-algebra $k[[T]]$, let $\mathcal C$ be the category opposite to the category of $A$-modules, let $\mathcal C_0$ be the full subcategory of finite dimensional objects of $\mathcal C$, and let $\mathcal C'$ be the category opposite to the category of $k$-vector spaces. Denote again by $k$ the residue field of $A$, so that $k\in\mathcal C_0$. Let $F:\mathcal C\to\mathcal C'$ be the functor $\text{Hom}_A(k,\ )$, let $F_0:\mathcal C_0\to\mathcal C'$ be its restriction, let $RF:\text D(\mathcal C)\to\text D(\mathcal C')$ be the right derived functor of $F$ (it exists by Theorem 14.4.3 p. 359 in the book Categories and Sheaves by Kashiwara and Schapira), and let $(RF)_0:\text D(\mathcal C_0)\to\text D(\mathcal C')$ be its restriction.

Assume that $RF_0:\text D(\mathcal C_0)\to\text D(\mathcal C')$ exists. My hope is to derive a contradiction from this assumption, but I'm still far from this. A weaker goal would be to prove that the natural morphism $RF_0\to(RF)_0$ is not an isomorphism.

Here is a first (very modest) step in this direction. For $X$ in $\text K(\mathcal C)$ let $\text{Qis}^X$ be the category of quasi-isomorphisms $X\to Y$ with $Y$ in $\text K(\mathcal C)$, and, for $X$ in $\text K(\mathcal C_0)$ let $\text{Qis}_0^X$ be the category of quasi-isomorphisms $X\to Y$ with $Y$ in $\text K(\mathcal C_0)$. If $X$ is in $\text K(\mathcal C_0)$, then $\text{Qis}_0^X$ is not necessary cofinal to $\text{Qis}^X$.

To prove this let $X\in\text K(\mathcal C_0)$ be equal to $k$ in degree 0 and to 0 in the other degrees, let $Y\in\text K(\mathcal C)$ be equal to $A$ in degree 0, to the maximal ideal $\mathfrak m$ of $A$ in degree 1, and to 0 in the other degrees, the differential $A\to\mathfrak m$ being the obvious epimorphism, and let $X\to Y$ be the obvious quasi-isomorphism. It is easy to see that there is no quasi-isomorphism $Y\to Z$ in $\text K(\mathcal C)$ with $Z$ in $\text K(\mathcal C_0)$. This shows that $\text{Qis}_0^X$ is not cofinal to $\text{Qis}^X$.

Again, my question is:

Assuming that $RF_0:\text D(\mathcal C_0)\to\text D(\mathcal C')$ exists, is the natural morphism $RF_0\to(RF)_0$ an isomorphism?

(It would probably better to consider an additive functor $F$ of abelian categories such that there is no obvious candidate for $RF$, but I haven't found such an example.)

$\endgroup$
  • $\begingroup$ Here is another additive functor $F$ of abelian categories for which one can hope to prove that it admits no right derived functor: Let $k$ and $A$ be as above, and let $\text{Mod}_{\text{fd}}(A)$ and $\text{Mod}_{\text{fd}}(k)$ be the categories of finite dimensional $A$-modules and $k$-vector spaces respectively. Then $F:\text{Mod}_{\text{fd}}(A)\to\text{Mod}_{\text{fd}}(k)$ is the additive functor $\text{Hom}_A(k,\ )$. $\endgroup$ – Pierre-Yves Gaillard Mar 18 '17 at 23:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.