---
abstract: 'We give some examples of abelian categories $\ca$ for which the derived category ${{\mathbf D}}(\ca)$ is not left-complete. Perhaps the most natural of these is where $\ca$ is the category of representations of the additive group $\mathbb{G}_a$ over a field $k$ of characteristic $p>0$.'
address: |
    Centre for Mathematics and its Applications\
    Mathematical Sciences Institute\
    John Dedman Building\
    The Australian National University\
    Canberra, ACT 0200\
    AUSTRALIA
author:
- Amnon Neeman
bibliography:
- 'stan.bib'
title: 'Non-left-complete derived categories'
---

[^1]

Assumed background {#S0}
==================

In this article we assume the reader is familiar with derived categories and with [*t*]{}–structures on them. See Verdier [@Verdier96] for the theory of derived categories, and Beilinson, Bernstein and Deligne [@BeiBerDel82 Chapter 1] for an introduction to [*t*]{}–structures.

The counterexample {#S1}
==================

Suppose $\ca$ is an abelian category and ${{\mathbf D}}(\ca)$ is its derived category. For any object $x\in{{\mathbf D}}(\ca)$, we write $x^{\geq n}$ for the truncation of $x$ with respect to the standard . We have canonical maps $x^{\geq n}\la x^{\geq n+1}$, and a (non-canonical) map $$\CD
\ph_x\,\,:\,\,x @>>> \holim x^{\geq n}\ .
\endCD$$ The category ${{\mathbf D}}(\ca)$ is said to be *left-complete* if, for every object $x\in{{\mathbf D}}(\ca)$, any map $\ph_x$ as above is an isomorphism. Even though the map $\ph_x$ is not canonical, it can be shown that, for given $x$, if one $\ph_x$ is an isomorphism then they all are.

The reader can find much more about left-complete categories in Lurie [@Lurie06 Section 7] or [@Lurie11 Subsection 1.2.1, more precisely starting from Proposition 1.2.1.17]. See also Drinfeld and Gaitsgory [@DrinfeldGaitsgory11].

In this note we will see how to produce many $\ca$ for which ${{\mathbf D}}(\ca)$ is not left-complete. Our counterexamples will be of a very special form, which allows us to easily compute the homotopy inverse limit $\holim x^{\geq n}$. Let us now sketch what we will do.

We will suppose that the abelian category $\ca$ satisfies the axiom \[AB4\], that is coproducts are exact; this makes it easy to compute coproducts in the derived category ${{\mathbf D}}(\ca)$, just form the coproduct as complexes. Suppose $A$ is an object in our \[AB4\] abelian category $\ca$, and let $$x\eq \coprod_{i=0}^\infty A[i]\ .$$ It is clear that, for $n>0$, we have $$x^{\geq-n}\eq \coprod_{i=0}^n A[i]\eq \prod_{i=0}^n A[i]\ ,$$ where the last equality is because finite coproducts agree with finite products. Now the homotopy inverse limit of the products is a genuine inverse limit, and we have $$\holim x^{\geq n}\eq  \prod_{i=0}^\infty A[i]\ .$$ Thus our problem becomes to decide whether the map $$\CD
\ds\coprod_{i=0}^\infty A[i]@>\ph>>   \ds\prod_{i=0}^\infty A[i]
\endCD$$ is an isomorphism. Note that in this case the map is canonical; our homotopy inverse limit happens to be a genuine inverse limit, removing the arbitrariness. The left hand side is easy to work with; its cohomology is $A$ in each degree $n\leq0$. What we will show is how to produce examples where the right hand side has lots more cohomology. More precisely, we have $$\prod_{i=0}^\infty A[i]\eq A[0]\oplus\left(
\prod_{i=1}^\infty A[i]
\right)$$ and the expectation would be for the second term to have a vanishing $H^0$; what we will show is how to produce non-zero classes in $$H^0\left(
\prod_{i=1}^\infty A[i]
\right)\ .$$

It is time to disclose what will be our choice for the category $\ca$ and for the object $A\in\ca$.

Let $k$ be a field, let $R_1$ be a finitely generated $k$ algebra, and let ${\mathfrak{m}}$ be a $k$–point of $\spec {R_1}$. In other words, ${\mathfrak{m}}\subset R$ is a maximal ideal with $R_1/{\mathfrak{m}}\cong k$. We make a string of definitions:

$
R_n=\otimes_{i=1}^n R_1
$, where the tensor is over the field $k$.

The inclusion $R_n\la R_{n+1}$ is the inclusion of the tensor product of the first $n$ terms.

$R=\colim R_n$.

The map $\Phi_i:R_1\la R$ is the inclusion of the $i\mth$ factor.

The category $\ca$ will be the category of all those $R$–modules, on which $\Phi_i({\mathfrak{m}})$ acts trivially for all but finitely many $i$. The object $A\in\ca$ will be the colimit over $n$ of the $R_n$–modules $k= \otimes_{i=1}^n [R_1/{\mathfrak{m}}]$.

The main result is

Assume that $k=R_1/{\mathfrak{m}}$ is not projective over the localization ${(R_1)}_{\mathfrak{m}}$ of the ring $R_1$ at the maximal ideal ${\mathfrak{m}}$. With the category $\ca$ and the object $A\in\ca$ as in Construction \[C0.1\], there is a non-zero element in $$H^0\left(
\prod_{i=1}^\infty A
\right)\ .$$

The case where $R_1=k[x]/(x^p)$ is of particular interest. If the field $k$ is of characteristic $p$ then the category $\ca$ happens to be the category of representations of the additive group $\mathbb{G}_a$, and we learn that its derived category is not left-complete.

We trivially have $$\prod_{i=1}^\infty A[i]\eq \left(\prod_{i=1}^n A[i]\right)\oplus\left(
\prod_{i=n+1}^\infty A[i]\right)\ ,$$ and hence $$H^0\left(
\prod_{i=1}^\infty A[i]
\right)\eq
H^0\left(
\prod_{i=1}^n A[i]
\right)\oplus
H^0\left(
\prod_{i=n+1}^\infty A[i]
\right)\ .$$ On the other hand, with the finite product we have no problem computing $$H^0\left(
\prod_{i=1}^n A[i]
\right)\eq
H^0\left(
\coprod_{i=1}^n A[i]
\right)\eq0\ ,$$ and Theorem \[T0.2\] now allows us to deduce that $$H^0\left(
\prod_{i=n+1}^\infty A[i]
\right)\quad\neq\quad
0\ .$$ Translating we have that $$H^n\left(
\prod_{i=1}^\infty A[i]
\right)\quad\neq\quad
0$$ for all $n\geq 0$. The complexes $A[i],\, i>0$ all belong to ${{{\mathbf D}}(\ca)}^{<0}$, but the product $\prod_{i=1}^\infty A[i]$ is not bounded above.

   The author would like to thank Drinfeld and Gaitsgory for asking the question that led to these counterexamples.

The proof {#S2}
=========

We begin with a little lemma.

Let $k$ be a field, and let $R$ and $S$ be finitely generated $k$–algebras. Suppose further that we are given $k$–points of $\spec R$ and $\spec S$; that is ${\mathfrak{m}}\subset R$ and ${\mathfrak{n}}\subset S$ are maximal ideals, with $$R/{\mathfrak{m}}\quad\cong\quad k\quad\cong\quad S/{\mathfrak{n}}\ .$$ Let $E$ be an injective envelope of $k=R/{\mathfrak{m}}$ over the ring $R$, and $F$ an injective envelope of $k=S/{\mathfrak{n}}$ over the ring $S$. Then $E\otimes_k F$ is an injective envelope of $k$ over the ring $R\otimes_k S$.

We will first prove the case where $R$ and $S$ are polynomial rings.

Let $R'=k[x_1^{},x_2^{},\ldots,x_m^{}]$ be a polynomial ring, and let ${\mathfrak{m}}$ be the maximal ideal generated by $\{x_1^{},x_2^{},\ldots,x_m^{}\}$. Then we know the injective envelope $E'$ of $k=R'/{\mathfrak{m}}$ explicitly: it is the quotient of $S=k[x_1^{},x_1^{-1},x_2^{},x_2^{-1},\ldots,x_m^{},x_m^{-1}]$ by the $R'$–submodule generated by all monomials $x_1^{i_1^{}}x_2^{i_2^{}}\cdots x_m^{i_m^{}}$ with at least one of the $i_j^{}>0$. As a $k$–vector space $E'=k[x_1^{-1},x_2^{-1},\ldots,x_m^{-1}]$, and the $R'$–module structure is obvious when we declare $x_1^{i_1^{}}x_2^{i_2^{}}\cdots x_m^{i_m^{}}=0$ if some $i_j^{}>0$. If $S'=k[y_1^{},y_2^{},\ldots,y_n^{}]$ and ${\mathfrak{n}}\subset S'$ is the ideal generated by $\{y_1^{},y_2^{},\ldots,y_n^{}\}$, then the fact that $$E'\otimes_k F'\eq k[x_1^{-1},x_2^{-1},\ldots,x_m^{-1}]\otimes_k
k[y_1^{-1},y_2^{-1},\ldots,y_n^{-1}]$$ is the injective hull of $k$ over $R'\otimes S'$ is by inspection.

Now for the general case: assume $R=R'/I$ and $S=S'/J$ where $R'$ and $S'$ are polynomial rings, and $I\subset R'$ and $J\subset S'$ are ideals contained in the ${\mathfrak{m}}$ and ${\mathfrak{n}}$ above. Then the injective hull $E$ of $k=R/{\mathfrak{m}}$ over the ring $R$ is the largest $R$–submodule of the $R'$–module $E'$, that is the $R'$–submodule $E\subset E'$ of all elements annihilated by the ideal $I$. The lemma therefore comes down to the fact that the submodule of $E'\otimes_k F'$ annihilated by the ideal $I\otimes_k S'+R'\otimes_k J$ is precisely $E\otimes_k F$.

   Let $\ov R$ be the localization of $R_1$ at the maximal ideal ${\mathfrak{m}}$. We are assuming that $k$ is not projective over $\ov R$, that is the projective dimension of $k$ is at least one. Choose and fix a minimal free resolution of $k=\ov R/{\mathfrak{m}}\ov R$ as an $\ov R$–module. Let us write this resolution as $$\CD
@>>>P_2@>>> P_1 @>>> P_0 @>>> k @>>> 0\ .
\endCD$$ Then the modules $P_i$ are all finite and free over the ring $\ov R$, the differentials are all matrices over $\ov R$, and the minimality guarantees that the entries in these matrices all belong to the ideal $\ov{\mathfrak{m}}={\mathfrak{m}}\ov R
\subset \ov R$. Now let $E$ be the $\ov R$–injective envelope of the module $k$; applying the functor $\Hom_{\ov R}(-,E)$ to the projective resolution above, we produce an injective resolution $I^*$ of $k$, which we write out as $$\CD
0@>>> k@>>>I^0@>>> I^1 @>>> I^2 @>>>
\endCD$$ We know that each $I^j=\Hom(P_j,E)$ is a finite coproduct of copies of $E$, and that the differentials $I^j\la I^{j+1}$ are matrices whose entries belong to the ideal $\ov{\mathfrak{m}}$. The fact that the projective dimension of $k$ is at least one tells us that $P_1\neq 0$, and therefore $I^1\neq0$. Note that an injective envelope $E$ of $k$ over the localized ring $\ov R={(R_1)}_{\mathfrak{m}}$ is also an injective envelope of $k$ over the ring $R_1$, hence we have produced an injective resolution of $k$ over $R_1$. Next we

Choose a non-zero element $a$ in the image of the map $k\la I^0$.

Choose a non-zero element $b\in I^1$, with ${\mathfrak{m}}b=0$.

If we view $k$ as a module over the ring $R_n=\otimes_{i=1}^nR_1$, then the tensor product $J_n^*= \otimes_{i=1}^nI^*$ is certainly a resolution of $k$ as an $R_n$ module, and Lemma \[L2.1\] guarantees further that

Each $J_n^i$ is injective as a module over $R_n$.

Let the inclusion $J_n^*\la J_{n+1}^*$ be the map taking $x\in J_n^*$ to $$x\otimes a\quad\in\quad
J_n^*\otimes I^0\quad\subset\quad
J_n^*\otimes I^*\eq J_{n+1}^*\ ,$$ where $a\in I^0$ is as in (i) above. We define $J^*$ to be $$J^*\eq\colim J^*_n\ ;$$ then $J^*$ is an injective resolution of $k$ in the category $\ca$.

To prove the theorem we need to find a non-zero element in $H^0\left(\prod_{i>0} k[i]\right)$, and our next observation is that the product in the derived category $\prod_{i>0} k[i]$ is obtained as the the ordinary product of injective resolutions. The complex $J^*[i]$ is an injective resolution of $k[i]$, and hence the derived product $\prod_{i>0} k[i]$ is just the usual product $\prod_{i>0} J^*[i]$. Now for every $i\geq1$ let $$S_i=\{i^2+1,\ldots,i^2+i\},$$ and observe that the sets $S_i$ are disjoint. In the injective $R_{i^2+i}$–module $$J^i_{i^2+i}=\coprod_{\sum \ell_m=i}I^{\ell_1}\otimes  I^{\ell_2}\otimes \cdots
\otimes I^{\ell_{i^2+i}}$$ or more specifically in the summand $${(I^{0})}^{\otimes i^2}\otimes  {(I^{1})}^{\otimes i}$$ we take the term $$\lambda_i\eq a^{\otimes i^2}\otimes b^{\otimes i}\ ,$$ where $a\in I^0$ and $b\in I^1$ are as in (i) and (ii) above. The embedding $J^*_{i^2+i}\la J^*$ of (iv) gives us an element which we will denote $\lambda_i\in J^i$. The elements $\lambda_i$ have the properties

Each $\lambda_i$ is a cycle; the differential $J^i\la J^{i+1}$ kills $\lambda_i$.

$\Phi_j({\mathfrak{m}})\lambda_i=0$ for all $i$ and $j$. We are assuming $i>0$, so each $\lambda_i$ must be a boundary because $H^i(J^*)=0$. But if $\mu_i\in J^{i-1}$ maps to $\lambda_i$, then there must exist an integer $j\in S_i$ so that $\Phi_j({\mathfrak{m}})$ does not kill $\mu_i$. Now form the element $$\prod_{i=1}^\infty \lambda_i\quad\in\quad \prod_{i=1}^\infty J^i\ ,$$ where the product is in the category of all $R$–modules.

The reader is reminded that the category $\ca$ is a subcategory of the category of $R$–modules. Both categories have infinite products; the products in the category of $R$–modules are just the usual cartesian products, while the products in $\ca$ are subtler. To form the product in $\ca$ of a bunch of objects in $\ca$, one first forms the usual cartesian product, and then consider inside it the largest object belonging to $\ca$, that is the collection of all elements satisfying part (v) of Construction \[C0.1\].

The element $\prod_{i=1}^\infty \lambda_i$ is a degree 0 cycle in the complex $\prod_{i\geq1}J^*[i]$, and it is annihilated by $\Phi_j({\mathfrak{m}})$ for all $j$. By Caution \[C2.5\] we have that $\prod_{i=1}^\infty \lambda_i$ belongs to $\prod_{i=1}^\infty J^i$ even when the product is understood in $\ca$. However, it is not a boundary in $\ca$. If we try to express $\prod_{i=1}^\infty \lambda_i$ as the boundary of $$\prod_{i=1}^\infty \mu_i\quad\in\quad \prod_{i=1}^\infty J^{i-1}\ ,$$ then we discover that each $\mu_i$ fails to be annihilated by some $\Phi_j({\mathfrak{m}})$ with $j\in S_i$. As the $S_i$ are disjoint, this produces infinitely many $\Phi_j({\mathfrak{m}})$ not annihilating $\prod_{i=1}^\infty \mu_i$, meaning it does not belong to $\ca$.

[^1]: The research was partly supported by the Australian Research Council
