Koszul complex is important for homological theory of commutative rings. However, it's hard to guess where it came from. What was the motivation for Koszul complex?

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    $\begingroup$ In addition to the answer below. If you know about the Koszul dual of a Koszul algebra $A$, then it is isomorphic to the Ext-algebra $Ext_A(k,k)$ where $k$ is the (direct sum of distinct isoclasses of) simple module(s) of $A$. The natural computation of Ext-groups come from looking at complex, the Koszul complex is exactly (quasi-isomprhic to) the projective resolution of the simple module(s). Beilinson-Ginzburg-Soergel's paper is the place to look at this approach. $\endgroup$
    – Aaron
    Commented Jun 28, 2012 at 17:06
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    $\begingroup$ There is a related question here mathoverflow.net/questions/146353/history-of-koszul-complex, along with some interesting answers. $\endgroup$
    – user143746
    Commented Apr 17, 2014 at 6:20

2 Answers 2


I don't know the historical origins, but it is not so hard to make up a story:

Consider the basic example $$0 \to k[x] \to k[x] \to k \to 0,$$ where the middle arrow is mult. by $x$. This is a resolution of $k = k[x]/(x)$ as a $k[x]$-module.

Now suppose you want to generalize this to obtain a resolution of $k = k[x_1,...,x_n]/(x_1,...,x_n)$ as a $k[x_1,...,x_n]$-module. It is not hard to see that you need "one copy" of the above sequence for each variable; tensoring these all together over $k$ gives you the usual Koszul resolution of $k$ over $k[x_1,...,x_n]$.

It is not hard to pass now to the more general context of elements $a_1,\ldots,a_n$ in a ring $A$, and to imagine the the Koszul complex of $a_1,\ldots,a_n$ will related to the module $A/(a_1,\ldots,a_n)$.

  • $\begingroup$ Thanks. That makes sense. I was thinking about the defintion of Koszul complex by Cartan-Eilenberg(p.151). Let $K$ be a field. Let $A = K[x_1, ..., x_n]$ be the ring of polynomials. Let $E(y_1, ..., y_n)$ be the exterior algebra on $n$ letters over $K$. Let $M$ be a $A$-module. Cartan-Eilenberg defined Koszul complex as $M \otimes E(y_1, ..., y_n)$. $\endgroup$ Commented Jun 17, 2012 at 6:26

In this answer I would rather focus on why is the Koszul complex so widely used. In abstract terms, the Koszul complex arises as the easiest way to combine an algebra with a coalgebra in presence of quadratic data. You can find the modern generalization of the Koszul duality described in Aaron's comment by reading Loday, Valette Algebraic Operads (mostly chapters 2-3).

To my knowledge the Koszul complex is extremely useful because you can use it even with certain $A_\infty$-structures arising from deformation quantization of Poisson structures and you relate it to the other "most used resolution in homological algebra", i.e. the bar resolution.

For a quick review of this fact, please check my answer in Homotopy equivalent chain complexes

As you can see it is a flexibe object which has the property of being extremely "explicit". This helped alot its diffusion in the mathematical literature.


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