Wikipedia says that in certain situations, $Ext^\ast_A(R,R)$ becomes a ring, such as when $A$ is an augmented $R$-algebra, but the outline is too sketchy for me to understand.

I can't find this in Weibel. Isn't $Ext^\ast_A(M,M)$ a ring (even an $\mathbb{N}$-graded algebra over center $Z(A)$) for any algebra $A$, via the splicing (concatenation) of extensions (exact sequences) $$0\to M\to E_k\to\ldots\to E_1\to M\to0\text{ and }0\to M\to E'_l\to\ldots\to E'_1\to M\to0,$$ $$\text{producing }0\to M\to E_k\to\ldots\to E_1\to E'_l\to\ldots\to E'_1\to M\to0\in Ext^{k+l}_R(M,M)?$$

A more detailed sketch or a reference would be welcome. How can this multiplication be defined when we interpret $Ext^n_A(M,M)$ as $H_n(Hom_A(P_\ast,M))$ for some projective resolution $P_\ast\to M$?

If I understand correctly, this makes the group and algebra and Lie algebra cohomology $\bigoplus_{n\geq0}Ext^n_{R[G]}(R,R)$ and $\bigoplus_{n\geq0}Ext^n_{A\otimes A^{op}}(R,R)$ and $\bigoplus_{n\geq0}Ext^n_{U(\mathfrak{g})}(R,R)$ into rings, correct?

Can some similar construction be made for $Tor$?

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    $\begingroup$ You should check these notes by May: math.uchicago.edu/~may/MISC/TorExt.pdf $\endgroup$ Mar 29, 2015 at 9:25
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    $\begingroup$ Found another reference: Mac Lane's Homology, section VIII (Products). There's also Cartan & Eilenberg which is really comprehensive and immensely general. And I'm going to leave a link to a question I just asked here, because it is somehow related and I think it would be useful to have them formally linked: math.stackexchange.com/questions/1222604/… . $\endgroup$ Apr 6, 2015 at 16:54
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    $\begingroup$ In general, note that Tor is a coalgebra, not an algebra! $\endgroup$
    – Pedro
    Jan 6, 2018 at 15:56

2 Answers 2


You should think of $\text{Ext}^{\bullet}_A(R, R)$ as a shadow of a more fundamental operation, namely taking derived endomorphisms $\text{RHom}_A(R, R)$ of $R$, regarded as an object in the derived category of $A$-modules. Here the ring structure is obvious; the multiplication comes from composition. More generally, there is a composition map $\text{RHom}(A, B) \times \text{RHom}(B, C) \to \text{RHom}(A, C)$, which you can write down by replacing $A, B, C$ with projective resolutions and then composing chain maps.

The situation for $\text{Tor}$ is different; you should also think of $\text{Tor}$ as a shadow of a more fundamental operation, namely taking derived tensor products $M \otimes_A^L N$, but even in the underived setting the tensor product of two modules has no reason to acquire a ring structure. One should only expect a ring structure if $M$ and $N$ are themselves already $A$-algebras, which happens e.g. if $A$ has an augmentation $A \to R$ and $M = N = R$.

  • $\begingroup$ The way you get from an operation on RHom to an operation on Ext is by taking homology, and this will only have nice properties if you're in a setting where you can apply the Kunneth theorem. $\endgroup$ Apr 23, 2014 at 4:37
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    $\begingroup$ Never worked with derived categories. Is there an easy answer that just deals with modules and chain complexes? I'm looking for more explicit formula for the product. Are the claims in my post correct? Any recommended references? $\endgroup$
    – Leo
    Apr 23, 2014 at 15:37

Let $k$ be a commutative ring and let $A$ be an augmented $k$-algebra, so that in particular $k$ is an $A$-module. Fix a projective resolution $(P,d_P)$ of $k$. Then $EP=\text{End}_A(P)$ is a complex with differential given by $df = d_P f -(-1)^{|f|} f d_P$, and it has a product given by composition of maps: for endomorphisms $f$ and $g$, set $f\smile g = fg$ (the composition). If $d$ is the differential of $EP$ you can check that $$d(f\smile g) = df\smile g +(-1)^{|f|} f\smile dg $$ so $EP$ is a differential graded algebra. This means its homology, which is $\text{Ext}_A(k,k)$, is also a differential graded algebra with the product induced by $\smile$.

You can also consider a free resoluion $P= V\otimes A$ so that $\text{Hom}_A(V\otimes A,k)= \text{Hom}_k(V,k)=V^*$, and you can construct a section $i$ of the map $p : EP=\text{Hom}_A(P,P) \to \text{Hom}_A(P,k)$ given by postcomposition with the augmentation of $P\to k$ so that $ip$ is homotopic to the identity. Then you can define a product on $V^*$ by the rule $f\smile' g =p( if\smile ig)$ which induces the same product on $H(V^*) = \text{Ext}_A(k,k)$.

In both cases one can in fact produce on $\text{Ext}$ a minimal $A_\infty$-algebra structure, with $\smile$ being $m_2$.


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