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Recently, I have been writing up some notes on derived functors and I came across the Eilenberg-Watts theorems [1], which essentially explain why it is hard to find derived functors besides Ext and Tor. Before asking my question, allow me to briefly state these theorems.

Let $R$ and $S$ be rings and $M$ and $S-R$ bimodule. A basic property of the functor $M\otimes_R-:R-\mathrm{Mod}\to S-\mathrm{Mod}$ is that it is an additive covariant right-exact functor. In fact, it also commutes with direct sums. Curiously these properties are enough to characterize it: any covariant additive $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ that is right-exact and commutes with direct sums is in fact naturally equivalent to some $M\otimes_R -$ for some $S-R$ bimodule $M$. This is the statement of the Eilenberg-Watts theorem for tensor functors.

For completeness, I should state the other version for Hom. If $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ is an additive left-exact contravariant functor which converts direct sums into direct products (i.e. $T(\oplus M_i) \cong \prod T(M_i)$) then there is an $R-S$ bimodule $M$ such that $T$ is naturally equivalent to $\\mathrm{Hom}_{R}(-,M)$.

Finally, if $T:R-\mathrm{Mod}\to \mathbb{Z}-\mathrm{Mod}$ is left-exact covariant that commutes with inverse limits then there is a left $R$-module $M$ such that $T$ is naturally equivalent to $\mathrm{Hom}_{R}(M,-)$.

The Prompting for the Question

From the above, any functor satisfying the hypotheses of these Eilenberg-Watts type theorems are going to be naturally equivalent to either a tensor or a Hom, and thus its derived functors will just be Tor or Ext respectively. For instance, if $G$ is a group then $G-\mathrm{Mod}\to \mathbb{Z}-\mathrm{Mod}$ given by $A\mapsto A_G$, where $A_G$ is the quotient of $A$ by the submodule generated by $ga - a$ for all $g\in G$ and $a\in A$ is just the usual coinvariant functor, whose left derived functors are the homology groups $H^i(G,A)$. By Eilenberg-Watts, $-_G$ must be equivalent to some tensor functor, and in fact it is easy to prove that $A\mapsto A_G$ is naturally equivalent to $A\mapsto \mathbb{Z}\otimes_{\mathbb{Z}G}A$

Incidentally, the proof given by C.E. Watts is explicit enough so that the above natural equivalence is apparent.

The Question

Notice that in each of these the hypothesis of playing nice with limits is required. I am actually interested in functors which do not play nicely with limits. For instance,

What are some examples of covariant right-exact functors $T:R-\mathrm{Mod}\to S-\mathrm{Mod}$ that do not commute with all direct sums? [Edit: $T$ also should not be left exact in this case.]

Such a $T$ of course cannot be a left-adjoint for otherwise it would commute with direct sums. Such a $T$ could be interesting because its left-derived functors may not be "like" the Tor functor. The question also goes for dropping the playing-nice-with-limit hypotheses in the other forms of the theorem. I tried a Google search but could not seem to find anything relevant.

Since I am asking for a list of examples, I have made this a community wiki. Thanks!

[1] Watts, "Intrinsic Characterizations of Some Additive Functors". Proceedings of the American Mathematical Society, Vol. 11, No. 1 (Feb., 1960), pp. 5-8

Addendum (edit)

Thanks everyone for their answers; I think I should have been more precise and asked a question more along the lines of:

What are some derived functors that are not Ext or Tor?

Which I believe some of the existing answers are. In essence I wanted examples that were neither tensors nor Homs in disguise...

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up vote 13 down vote accepted

Here is a specific example, though it admits obvious generalizations. Let $R=S=\mathbb{Z}$, and consider the functor from abelian groups to abelian groups defined by $$T(X) = \mathrm{Ext}^1_{\mathbb{Z}}(\mathbb{Q}_p/\mathbb{Z}_p,X),$$ where $p$ is a prime. This is right-exact since $\mathrm{Ext}^2_{\mathbb{Z}}\equiv0$. It does not commute with direct sums, a fact which is clear when you observe that $T$ is also the $0$th left derived functor of $p$-completion, and $p$-completion does not preserve infinite sums, even when applied to free abelian groups.

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Nice example. Reading between the lines, by $p$-completion you mean inverse limit of tensor product with $\mathbb Z/p^k$, and by $n$th left derived functor of $F$ you mean $n$th homology of $F$ applied to a projective resolution. But I don't believe I've ever heard that term used this way. I mean, in my experience that construction is only applied when $F$ is right exact (so that the $0$ derived is $F$ again); or if it is applied it doesn't necessarily called derived functors. What also strikes me ... – Tom Goodwillie Apr 11 '12 at 21:39
... is that the really "good" homotopy-invariant version of $p$-completion is a right derived of a left derived: homotopy inverse limit of derived tensor with $\mathbb Z/p^k$. – Tom Goodwillie Apr 11 '12 at 21:40
@TomGoodwillie I realize that this response is about 3 years out of date, but referring to those as the 0'th left derived functors (even when not right exact) goes back at least to Cartan-Eilenberg. – Tyler Lawson Sep 28 '15 at 15:09

Double dual (when $R=S$ is semisimple).

Or $X\mapsto \Pi_j (M_j\otimes_R X)$ where $\lbrace M_j\rbrace $ is an infinite collection of nontrivial bimodules.

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What about representable functors? $\hom(M,-) : \mathrm{Mod}(R) \to \mathrm{Ab}$ is right exact iff $M$ is projective, and it preserves direct sums iff $M$ is even finitely generated projective. Of course this is almost a special case of the excellent answer by Tom Goodwillie, but I wanted to add this simple observation.

Remark: If $M$ is not projective, it is harder to characterize those modules $M$ such that $\hom(M,-)$ preserves direct sums, see this MO discussion and the associated nlab article.

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Thanks Martin. I think I should have been more precise by asking for functors which are neither Hom nor tensor in disguise, so I could get derived functors than are neither Ext nor Tor. However, using this I guess we get examples of exact functors that do not preserve arbitrary sums... – Jason Polak Apr 11 '12 at 18:53
Sure, I've read your question. But since covariant Hom-functors are always left exact, one computes their right derived functors, namely Ext groups. But Eilenberg Watts (2nd form, which is equivalent the 1st form by GAFT) talks about contravariant Hom-functors. So my example isn't really the Hom-functor in disguise in your question. – Martin Brandenburg Apr 11 '12 at 19:25

Given associative rings $R$ and $S$, any covariant additive functor $R{-}proj\to S{-}mod$ from the category of projective $R$-modules to the category of $S$-modules can be extended to a right exact functor $R{-}mod\to S{-}mod$ in a unique way. The new functor preserves the infinite direct sums if and only if the original functor did.

So any additive, but not infinite direct sum-preserving functor $R{-}proj\to S{-}mod$ leads to a kind of example you are looking for. One can replace the projective modules with the free ones here if one wishes. Similarly, a covariant additive functor $R{-}inj\to S{-}mod$ from the category of injective $R$-modules to the category of $S$-modules can be extended to a left exact functor $R{-}mod\to S{-}mod$ in a unique way.

The following specific examples come from the theory of comodules and contramodules.

Let $A$ and $B$ be finite-dimensional associative algebras over a field $k$, and let $M$ be an $A$-$B$-bimodule. Then there is a naturally defined covariant right exact functor $F_M\colon A{-}mod\to B{-}mod$ constructed as follows. Given a left $A$-module $P$, consider the two maps $Hom_k(Hom_k(A,M),P)\simeq Hom_k(M,\:A\otimes_kP)\rightrightarrows Hom_k(M,P)$, one of which is induced by the map $M\to Hom_k(A,M)$ defined by the rule $m\mapsto (a\mapsto am)$, while the other one is induced by the action map $A\otimes P\to P$. These are two morphisms of left $B$-modules, the $B$-module structures being induced by the right action of $B$ in $M$. The $B$-module $F_M(P)$ is defined as the cokernel of the difference of of the above two morphisms.

Being defined as the cokernel of a morphism of exact functors, the functor $F_M$ is clearly right exact. One can compute that $F_M(A\otimes V) \simeq Hom_k(M,V)$ for any $k$-vector space $V$, so $F_M$ does not preserve infinite direct sums if $M$ is infinite-dimensional.

The functor $F_M(P)$ is called $Cohom_{A^*}(M,P)$, where $A^*$ and $B^*$ are viewed as (finite-dimensional) $k$-coalgebras, $M$ is an $A^*$-$B^*$-bicomodule, $P$ is considered as an $A^*$-contramodule, and the functor takes values in $B^*$-contramodules. Viewed in this way, the functor is defined for any (not necessarily finite-dimensional) $k$-coalgebras in place of $A^*$ and $B^*$.

Similarly, given a $B$-$A$-bimodule $N$, one defines a covariant left exact functor $G_N\colon A{-}mod\to B{-}mod$ as follows. For any left $A$-module $L$, consider the pair of maps $N\otimes_k L\rightrightarrows N\otimes_k A^*\otimes_k L$ induced by the maps $N\to N\otimes_k A^*$ and $M\to A^*\otimes M$ dual to the action maps $N\otimes_k A\to N$ and $A\otimes_k L\to L$. This is a pair of morphisms of left $B$-modules with the $B$-module structures coming from that of $N$. Set $G_N(L)$ to be the kernel of the difference of this pair of maps.

Being defined the kernel of a morphism of exact functors, the functor $G_N$ is obviously left exact. One can compute that $G_N(Hom_k(A,V)) \simeq N\otimes_k V$, so $G_N$ does not preserve infinite products if $N$ is infinite dimensional. The functor $G_N(L)$ is otherwise known as the cotensor product $N\square_{A^*}L$.

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A left-exact covariant example is local cohomology: Let $(A, \mathfrak{m})$ be a commutative local ring. The functor $H^0_{\mathfrak{m}}$ from $A$-mod to $A$-mod is $$H^0_{\mathfrak{m}}(M) = \bigcup_{n \geq 1} \{ x \in M : \mathfrak{m}^n x=0 \}$$

This is left exact covariant. For each finite $n$, the corresponding term of the union is $\mathrm{Hom}_A(A/\mathfrak{m}^n, M)$, but there is no $A$-module $N$ such that $H^0_{\mathfrak{m}}(M) = \mathrm{Hom}_A(N,M)$. And, indeed, it does not commute with inverse limits: If $A$ is the $p$-adics, then $H^0_{\mathfrak{m}}(\mathbb{Z}/p^k) = \mathbb{Z}/p^k$ for every $k$, but $H^0_{\mathfrak{m}}(\mathbb{Z}_p)=0$.

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