This question will probably seem quite silly to those well-versed in algebraic geometry (about which I admittedly hardly know anything); in the preface of Atiyah-Macdonald's book on commutative algebra they mention that one should take a course in homological algebra if one "wishes to pursue algebraic geometry in any depth". I am wondering which parts of homological algebra are mainly used in algebraic geometry; I actually know something about homological algebra and I would appreciate it if somebody could point out the methods used in algebraic geometry, and the role which they play in the development of the theory. I am particularly interested in derived categories and the derived category approach to derived functors (in contrast to the classical definition); apparently these concepts were somehow motivated by developments in AG.

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    $\begingroup$ Derived functors are extensively used in algebraic geometry. The right derived functors of the global sections of a sheaf are what are known as the sheaf cohomology. These are useful, for example, in understanding invariants of line bundles on projective varieties. $\endgroup$
    – jmracek
    Dec 17, 2013 at 21:32
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    $\begingroup$ Algebro-geometric spaces are locally modeled on commutative rings, and homological algebra plays just as important a role in algebraic geometry as in commutative algebra. The most basic principle is to study the abelian category of (quasi-coherent, resp. coherent) sheaves on a space, in analogy with studying the abelian category of modules over a ring. $\endgroup$
    – user314
    Dec 18, 2013 at 11:56
  • $\begingroup$ I have made an edit so as to make the question seem less broad. $\endgroup$ Dec 21, 2013 at 19:18

1 Answer 1


I imagine one could write books on the subject, but probably the first moment one encounters homological algebra in algebraic geometry is the following.

If $f: X\to Y$ is a proper map of varieties, then given a coherent sheaf $\mathcal{F}$ on $Y$ one can form the pull-back $f^{*} \mathcal{F}$. Now, the pull-back functor $f^{*}: \mathcal{C}oh(Y) \rightarrow \mathcal{C}oh(X)$ is often exact (under a technical assumption of flatness) and so one has some control over it even without homological algebra.

However, given a coherent sheaf $\mathcal{G}$ on $X$, one can also push it forward to obtain $f_{*} \mathcal{G}$. Notably, the functor $f_{*}$ fails to be exact (it is only left exact) and so its right derived functors are of crucial importance.

Already very interesting case is given by taking $Y$ to be a point. Then $f_{*} = \Gamma$, the global sections functor and its derived functors are called sheaf cohomology, denoted by $H^{i}$. These vector spaces are important in their own right and turn out to be finite-dimensional if $X$ is projective. Thus, their dimension is a powerful numerical invariant of the coherent sheaf $\mathcal{G}$ and, by extension, of $X$.

However, the usual recipe to compute $H^{i}$ (resolve your sheaf by injective sheaves) is almost useless in practice, as injective sheaves are very complicated beasts. There turns out to be another, very direct way to compute $H^{i}$ using so called $\check{C}$ech cohomology. One needs a substantial amount of (admittedly, not very deep, but that depends on who you are talking to...) homological algebra to prove that these very computable $\check{C}$ech cohomology groups agree with the derived functors of $\Gamma$, which have excellent formal properties. This is all nicely explained in the standard textbook by R. Hartshorne.

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    $\begingroup$ $f^*$ is not exact ... $\endgroup$ Dec 18, 2013 at 9:40
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    $\begingroup$ @MartinBrandenburg You are completely right. I must have been thinking in terms of $f^{-1}$ which was of course wrong, thanks for pointing that out! However, one might argue that $f^{*}$ is "as exact as tensor product" and so in many cases in algebraic geometry this functor is indeed exact. On the other hand, non-exactness of $f_{*}$ is a fundamental phenomena related to the global geometry of the space. $\endgroup$ Dec 21, 2013 at 13:19
  • $\begingroup$ Great answer. As someone with only a cursory background in algebraic geometry, this certainly helped me understand a few of the motivations. $\endgroup$
    – Dan Rust
    Dec 24, 2013 at 15:48

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