# Tag Info

46

The Tor functors are the derived functors of the tensor product. The starting observation is that if $0 \to M' \to M \to M'' \to 0$ is a ses of modules and $N$ is any module (let's work over a fixed commutative ring $R$), then $M' \otimes N \to M \otimes N \to M'' \otimes N \to 0$ is exact, but you don't necessarily have exactness at the first step. (This is ...

23

You will be a lot more motivated to learn about Tor once you observe closely how horribly tensor product behaves. Let us look at the simplest example possible. Consider the ring $R=\mathbb C[x,y]$ and the ideal $I=(x,y)$. These are about the most well-understood objects, right? What is the tensor product $I\otimes_RI$? This is quite nasty, it has torsions: ...

20

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 $f^{*}: \mathcal{C}oh(Y) \rightarrow ... 19 I can sympathize with this question because I am about to teach a first (graduate level) course in commutative algebra. No homological algebra of any sort is a prerequisite: I'll be happy if all of my students are comfortable with exact sequences. On the other hand, just a little bit of Tor is extremely helpful when studying commutative algebra (and ... 17 First of all: French effacer means "to erase", and it is of course a composite of the prefix e- (latin ex-) [away from] face. I'm not aware of a different meaning than erase (maybe wipe up or annihilate are also viable translations in some circumstances), and what you're describing is probably more or less what Grothendieck had in mind. As far as I know ... 15 You may wish to read about local cohomology. The Wikipedia article is mostly about the sheaf theory, but over an affine scheme and for quasicoherent sheaves, you can think of it as follows: if$R$is a ring (say, noetherian),$I \subset R$an ideal, then for an$R$-module$M$,$H^i_I(M)$is the right-derived functor of the functor$M \mapsto \varinjlim ...

11

One motivation is deformation theory, see MO:111084. Basically, the cotangent complex turns out to be the left derived functor of the Kaehler differentials functor $\Omega^1$. In intersection theory, Serre's intersection formula is another example, see MO:12236. The Tor's in this formula are simply the homotopy groups of the derived tensor product, taken ...

9

Consider the one-to-one function of $\mathbb{Z}$-modules \begin{align*} f\colon\mathbb{Z}&\longrightarrow\mathbb{Z}\\ a&\longmapsto 2a \end{align*} If we tensor with $\mathbb{Z}/2\mathbb{Z}$, the map $f$ induces a map $$\mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}\ \longrightarrow\ ... 8 Theo's answer is excellent. Let me add a bit on how this formalism is used in practice. Say you want to show that \mathrm{Tor} commutes with flat base change of rings. That is, if M, N are A-modules, B a flat A-algebra, then \mathrm{Tor}_B(M_B, N_B) \simeq \mathrm{Tor}_A(M,N) \otimes_A B functorially -- it's not even obvious a priori how to get ... 8 The statement can be made more precise:$$ L_n L_m F = \begin{cases} L_nF, & \text{if } m = 0, \\0, & \text{if }m \gt 0. \end{cases}$$The point is that L_mF is (co-)effaceable for m \gt 0, that is L_mF(P) = 0 for all projective P. See my answer here for some background on that. First of all, it is easy to see that L_nL_0F \cong L_nF ... 7 With the proper definitions you have the following facts. I'll be using Keller's terminology in Derived Categories and their uses: For any functor F: \mathscr{A} \to \mathscr{B} (between abelian or even exact categories), the class of (right) F-acyclic objects are a subcategory Ac \subset \mathscr{A} closed under extensions. The restriction of F to ... 6 Total derived functor is an exact functor between triangulated categories (i.e. respects distinguished triangles and translation functor). Exactness of each individual R^i F perhaps is not the best way to think about the situation. From technical point of view when you try to calculate R^iF you need a resolution, but you can choose any resolution ... 5 For 3: This is almost cheating, I know... Mac Lane studied the functors \mathrm{Trip}_n which arise from derivating (?) the functor M\otimes N\otimes P of three variables. There are references in his book Homology. The interesting thing is, this cannot be expressed in terms of \mathrm{Tor}. For 1: If \mathcal I is the set of all non-zero ideals ... 5 Derived functors don't matter in this case. You just have an exact sequence:$$ \dots \longrightarrow A \stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C \stackrel{h}{\longrightarrow} D \stackrel{i}{\longrightarrow} E \longrightarrow \dots $$in which f and i are isomorphisms. But this means that morphisms g and h are zero:$$ B ...

5

Since $F,G$ are not related at all, the answer is of course: No. But the answer is Yes if $F=GP$. More generally, let $P : A \to B$ be any exact functor which preserves injectives. The latter property holds for example when $P$ has an exact left adjoint (nice exercise). In particular, this applies to the case that $P$ is an equivalence of categories. Then ...

4

I see nothing in the definition of derived functors that wouldn't be preserved by equivalence of categories. So $R^nGP(X)=PR^nG(X)$. (Derived functors are, as far as I know, defined only up to isomorphism, so "$=$" here means "canonically isomorphic".) EDIT: Martin Brandenburg is right. I assumed $F=GP$, but this is not stated in the question.

4

I'll try to do the non-derived category case. Let $X$ be a projective variety over $k$. We will do an induction on $i$. The result is clear for $i=0$, since in this case $Ext^{0}(\mathcal{F}, \mathcal{G}) = Hom(\mathcal{F}, \mathcal{G}) \simeq \Gamma(X, \mathcal{H}om(\mathcal{F},\mathcal{G}))$ which is finite-dimensional by Hartshorne, II.5.19, ...

4

Let $\newcommand\ZZ{\mathbb Z}R=\ZZ/p^n$, with $n>1$, be your ring and consider the module $M=\ZZ/p$. The obvious map $R\to M$ has kernel the ideal generated by $p$, so the sequence $R\xrightarrow{p}R\to M$ is exact. The kernel of the map $p:R\to R$ is generated by $p^{n-1}\in R$. We thus have an exact sequence of the form R\xrightarrow{\quad ...

4

You'll find Bernhard Keller's notes for a short course on the subject in his web page. His exposition is characteristically lucid and clear. His focus is representation theory, so they may not match your interests, though. I also like a lots the to-the-pointness approach taken by Dieter Happel in his book about triangulated categories.

4

The modern way about thinking about derived functors stems from work of Quillen and Verdier in the 1960s and has a very pleasing formulation, albeit at the cost of introducing some fairly sophisticated machinery. I will try to explain this briefly. Let $\mathcal{A}$ be an abelian category and let $\mathbf{Ch}^{\ge 0}(\mathcal{A})$ to be the category of ...

4

Similar to $\rm{Ext}^n$, $\rm{Tor}^R_n(A,B)$ for any $n$ can be viewed as an abelian group described explicitly as a so-called torsion product, which as a set consists of equivalence classes of triples $(f,L,g)$ where $L$ is a length $n$-complex of finitely generated projective modules, $f:L\to A$ and $g:\rm{Hom}(L,R)\to B$ are chain maps with $A,B$ ...

3

My answers to your four questions: (1) Yes, "Apply Ext$(-,G)$ to" is a standard way of saying it. (2) Your proof for the case $C=\mathbb Z/p$ looks OK. It might become a bit shorter and easier if you started with the exact sequence $0\to\mathbb Z\to\mathbb Z\to\mathbb Z/p$, where the map $\mathbb Z\to\mathbb Z$ is multiplication by $p$. That way, you ...

3

In the context of model categories, the left (reps. right) derived functors are a generalization of the traditional notion of a derived functor. The left (resp. right) derived functor $LF$ of a functor $F\colon \mathbf C\to \mathbf D$ where $\mathbf C$ is a model category is exactly a right (resp. left) Kan extension along the localization $\gamma\colon ... 3 As Darij Grinberg points out, we only have to prove that$\mathrm{Ext}^n$is$R$-linear in both variables. For this we have to recall how the action on morphisms is defined: If$f : A' \to A$is a homomorphism, and$P^* \to A$and$P'^* \to A'$are projective resolutions, there is a (unique up to homotopy) homomorphism of complexes$\overline{f} : P'^* \to ...

3

Gelfand and Manin's "Methods of homological algebra" explains the subject quite nicely, though it takes them some pages to develop the theory (and there are many typos, at least in the first edition). Personally, I also use the first three chapters of Huybrecht's "Fourier-Mukai transforms in algebraic geometry", which is a bit more condensed but has a nice ...

3

No, the extended sequence needn't be exact. Take first $0\to A\to B\to D\to 0$ exact so that $0\to \hom(M,A)\to \hom(M,B)\to\hom(M,C)\to\operatorname{Ext}^1(M,A)$ is exact. Take any $E$ such that $\hom(M,E)\neq 0$. Then set $C=D\oplus E$. $0\to A\to B\to C$ is exact, but then the sequence $0\to \hom(M,A)\to\hom(M,B)\to\hom(M,C)\to \operatorname{Ext}(M,A)$ ...

3

Here's a guess. From the bibliography of Schiffmann Gérard's Théorie élémentaire des catégories (1962) which can be found on page 33 of Séminaire Banach (1972): (It may be hard to find or lost as Chevalley had a habit of only keeping a single copy of his papers.)

3

You can compute Ext via projective resolutions of the first argument. In this case, we have the periodic projective resolution $\dotsc \xrightarrow{p^{n-1}} \mathbb{Z}/p^n \xrightarrow{\cdot p} \mathbb{Z}/p^n \xrightarrow{\cdot p^{n-1} } \mathbb{Z}/p^n \xrightarrow{\cdot p}\mathbb{Z}/p^n \xrightarrow{\text{pr}} \mathbb{Z}/p \to 0.$ Let us apply ...

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