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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 ...

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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: ...

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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 ...

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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 ...

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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 $M \mapsto \varinjlim ... 9 Consider the one-to-one function of$\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 $$\mathbb{Z}/2\mathbb{Z}\cong \mathbb{Z}/2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}\ \longrightarrow\ ... 8 The statement can be made more precise:$$ 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 ... 8 In your calculation of the group \textrm{Tor}^1(M, N), it seems that the calculation of the kernel is incorrect. The kernel must be the whole \mathbb{Z} \otimes N since an elementary tensor k \otimes [n] is mapped to 284k \otimes [n] = 142k \otimes [2n] = 142k \otimes [0] = 0. By the way, i is generally written as a subscript in ... 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 Partial answer. 1) Yes. Maybe the simplest way is to view an element of the Picard group as an equivalence class of Weil divisors. Fix an affine space U_n (resp. V_m) in \mathbb P^n (resp. \mathbb P^m) with complement H_n, L_m isomorphic to a projective space of dimension one lower. Let D be a Weil divisor on \mathbb P^n\times \mathbb P^m. ... 6 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 ... 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 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 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 ... 3 I guess the point here is that$N \otimes_R -$and$- \otimes_R N$are naturally isomorphic functors. Therefore, you get an isomorphism of chain complexes$N \otimes_R P^{\bullet} \cong P^{\bullet} \otimes_R N$, which implies that the two complexes have isomorphic homology. So, it doesn't matter if you apply$N \otimes_R -$or$- \otimes_R N$. 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 ...

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Maybe it would help to work with algebras rather than coalgebras: since everything is finite-dimensional, we can dualize. The dual to the coalgebra $E(x)$ is again the exterior algebra $E(x)$, and we can work with $E(x)$-modules rather than comodules. In this case, the goal is to show that $\mathrm{Ext}^{\bullet}_{E(x)}(k, k)$ is a polynomial algebra, ...

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Q1: Yes, take projective resolutions $P^*,Q^*$ of $A,B$ resp., build the tensor bicomplex $R^{*,*}$ and take the total homology of that bicomplex. This is how you can define Tor as a bifunctor. Q2: Yes, and the choices of the resolutions don't matter. You can prove that the total homology of the bicomplex in Q1 is the same as the homology of $B \otimes ... 2 I am a little bit confused by this question, but let me share my (limited) perspective. Almost all I know is included in Hartshorne's Algebraic Geometry. I will later write why this is a little bit limited. The importance of flasque sheaves in computation of sheaf cohomology stems from the following fact that can be proved by hand Let$0 \rightarrow ...

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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 ...

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I like "Sheaves in Topology" by Dimca a lot. I also second the suggestion of Gelfand/Manin. They actually have two books for some reason, "Homological Algebra" and "Methods of Homological Algebra", which are quite similar but have slightly different focus/applications. Both of them are worthwhile, and I think either one of them could be useful to you.

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Let $R=\mathbb{Z}$ for simplicity and $n=1$. The $\mathrm{Tor}^1(-,-)$ is characterized by the following properties: $\mathrm{Tor}^1(F,M)$ for all free $\mathbb{Z}$-module $F$. $\mathrm{Tor}^1(\mathbb{Z}/n\mathbb{Z},M)=Ker(M\rightarrow M, m\mapsto nm)$ $\mathrm{Tor}^1(L\oplus M, N)=\mathrm{Tor}^1(L,N)\oplus \mathrm{Tor}^1(M,N)$ ...

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Let $R,S$ be rings, let $\phi:R \to S$ be a ring map. Then we have functors $\uparrow = \uparrow_R^S : \operatorname{mod}_R \to \operatorname{mod}_S$ ("induction") and $\downarrow = \downarrow^S_R : \operatorname{mod}_S \to \operatorname{mod}_ r$ ("restriction") defined as follows: $M \downarrow$ is the $R$-module with the same underlying set as $M$ and ...

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Compare with a projective resolution $P_\bullet\to M\to 0$. By projectivity, we obtain (from the identiy $M\to M$) a complex morphism $P_\bullet\to A_\bullet$, which induces $F(P_\bullet)\to F(A_\bullet)$. With a bit of diagram chasing you shold find that $H_\bullet(F(P_\bullet))$ is the same as $H_\bullet(F(A_\bullet))$. A bit more explict: We can build a ...

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