# Tag Info

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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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To my very regret, old-fashioned terminologies are all over the place in mathematics and still prevent us from using the universal benefit of (the deep idea of) category theory, for example the unification of various scattered notions in mathematics. So let me answer what $A \times B$ and $A \oplus B$ should denote (although most books have not adopted this ...

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(ADDITION) Late remark: you say in your question that the only application you know is Rotman's proof of the fundamental theorem of arithmetic via Jordan-Hölder, indeed that is Corollary 4.56 of his book "Advanced Modern Algebra". Just for completeness, I believe it is worth mentioning including here his discussion on the next two pages about the ...

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The equality $RI\otimes_R N=R\otimes_R IN$ is very subtly false: the point is that it does not hold in $I\otimes_RN$, which is the only place where it could hold. But, since tensor product is $R-$bilinear, can't we write (for example) $1\cdot i\otimes n=1\otimes i\cdot n \:$? No, we can't! Because $1\otimes i\cdot n$ does not make sense in ...

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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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Here's what always makes sense to me. Yoneda's lemma tells us that if we really want to understand a ring $R$ we should study the sets $\text{Hom}_{\text{Ring}}(R,S)$ for all the other rings $S$ we could possibly imagine. That said, studying ALL the rings $S$ seems a little naive--is there no way to lighten our load? Well, intuitively if we have some class ...

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This is an excellent question, and your intuition is not too far off. Note that the counterexamples given in the other answers involve a non-finitely generated module ($\mathbb Q$ thought of as a $\mathbb Z$-module), and constructions like annihilators behave much better for finitely generated modules than for non-finitely generated ones (as a general ...

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The containment $\mathrm{Ann}_A(M)+\mathrm{Ann}_A(N)\subseteq\mathrm{Ann}_A(M\otimes_AN)$ can be proper. For an example, take $A=\mathbf{Z}$, $M=\mathbf{Z}/2\mathbf{Z}$, and $N=\mathbf{Q}$. Then $\mathrm{Ann}_A(M)=2\mathbf{Z}$, and $\mathrm{Ann}_A(N)=0$, but $M\otimes_AN=\mathbf{Z}/2\mathbf{Z}\otimes_{\mathbf{Z}}\mathbf{Q}=0$ because the left tensor factor ...

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Consider the simplest possible nontrivial (left) $R$-module: $R$ itself. It's certainly finitely generated, by $\{ 1 \}$. The submodules are exactly the (left) ideals of $R$. So you want a ring which has (left) ideals which are not finitely generated. For example, you could use a non-Noetherian commutative ring, such as $\mathbb{Z}[X_1, X_2, X_3, \ldots ]$.

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This answer is similar to the others; perhaps it will help to see the same points made by yet another person. First of all, it might help to note that $\mathbb C[x,y,z]/(xz-y)$ is isomorphic to $\mathbb C[x,z]$. So you are looking at the map $\mathbb C[x,y] \to \mathbb C[x,z]$ defined by $x \mapsto z, y \mapsto x z$, and asking why it is not flat. ...

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This is easy enough to prove directly: For the direction finitely presented $\implies$ compact, one can easily prove it by hand, and there is also the following somewhat slicker argument: For any module $M$ and any filtered direct limit $N = \varinjlim_i N_i$, there is a natural transformation $\varinjlim_i Hom(M,N_i) \to Hom(M,N).$ This is certainly an ...

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$\def\id{\operatorname{id}}$Suppose $M\otimes N$ is isomorphic to $R^n$. Pick a basis $\{x_1,\dots,x_n\}$ of $M\otimes N$, with $x_i=\sum_{j=1}^{r_i}m_{i,j}\otimes n_{i,j}$ for each $i\in\{1,\dots,n\}$. Let $r=r_1+\cdots+r_n$, let $\{e_{i,j}:1\leq i\leq n, 1\leq j\leq r_i\}$ be a basis of $R^r$, and consider the map $f:R^r\to M$ which maps $e_{i,j}$ to ...

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Here is a New version of the answer I'll leave the old version below so that the comments remain understandable. Let $K$ be a commutative ring and $x,y,z$ be indeterminates. Put $$M:=\frac{K[x,y,z]}{(xz-y)}\quad.$$ In particular, $M$ is an $K[x,y]$-module. We claim that $M$ is not $K[x,y]$-flat. Set $$t:=1\otimes y-z\otimes x\in ... 15 This is an application of the third isomorphism theorem, although the theorem does not play a crucial role in it. Let a, b be positive, say, integers. Then$$ a \mathbf{Z} + b \mathbf{Z} = \gcd(a, b) \mathbf{Z}, $$and$$ a \mathbf{Z} \cap b \mathbf{Z} = \operatorname{lcm}(a, b) \mathbf{Z}. $$Now the third isomorphism theorem gives you the isomorphism ... 15 My list of pathologies: A submodule of a finitely generated module does not have to be finitely generated. Example: Let K be a field and R = K[x_1, x_2, \ldots] be the ring of polynomials in infinitely many variables over K. R considered as an R-module is generated by f(x_1, x_2, \ldots) = 1, i.e., it is finitely generated. Now, let S = ... 14 We consider \mathbb{Z}-modules (i.e., abelian groups). Since \mathbb{Q} is divisible, if A is a torsion abelian group, then A\otimes\mathbb{Q} is trivial. Let G be the direct product of cyclic group of order p^n, with p a prime, and n increasing; that is:$$G = \prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}.$$Then ... 14 The ideal I=\langle X_1,X_2,...,X_n,... \rangle \subset \mathbb R[X_1,X_2,...,X_n,...]=A can be seen as a submodule of the free A-module of dimension one A=A^1, and that module is not finitely generated. Do you see why? (Hint: even in a polynomial ring with infinitely many indeterminates, each polynomial involves only finitely many variables. In other ... 13 Lemma One way to see why free submodules of free modules over a commutative ring have to have lesser or equal rank uses this lemma. The lemma is shown here, that surjections of finitely generated modules over commutative rings are necessarily isomorphisms. If R^m were isomorphic to a submodule of R^n and m>n, you would easily be able to construct ... 13 There are no ways to consider \mathbb{Q} as an \mathbb{R}-module (i.e. \mathbb{R}-vector space), if you follow the usual convention of requiring that modules be unital, i.e. 1\cdot q=q for any q\in\mathbb{Q}. This is because a vector space over \mathbb{R} (or indeed, any field) is free; because \mathbb{R} is uncountable, any \mathbb{R}-vector ... 13 Sure. Let A be the integers localized at (2); that is,$$A = \left\{ \frac{a}{b}\in\mathbb{Q}\;\Bigm|\; a,b\in\mathbb{Z}, b\gt 0, \gcd(a,b)=\gcd(2,b)=1\right\}.$$The field of quotients of A is \mathbb{Q}, and is equal to A[\frac{1}{2}], so it is generated as an A-algebra by 1 and \frac{1}{2}. More generally, any UFD R with only finitely ... 13 Pick a maximal linearly independent subset \{c_{\beta}\} of C. Now push the a_{\alpha} to B using f, and for each c_{\beta} pick c'_{\beta}\in B such that g(c'_{\beta}) = c_{\beta}. Now suppose that you have a finite linear combination of the a_{\alpha} and the c'_{\beta} that is equal to 0,$$n_{\alpha_1}f(a_{\alpha_1}) + ...

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There is indeed some nice intuition behind these definitions, and the good news is that not even all that deep. Remember two things: First, that this cohomology all comes by the "fixed by" functor $M\to M^G$, and second, that these crossed homomorphisms come from the definition of cochains, and more directly, the coboundary operator from $n$-chains to ...

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The part that is missing is pretty much the essence of the following (incomplete) alternative proof. Determine the kernel of $$\begin{array}{rlrl} g: & \mathbb{Z} & \rightarrow & (\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \\ & z & \to & z (1 \otimes 1) \end{array}$$ ...

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Let's follow up on the comments. A first attempt at dualizing the condition would dualize "direct summand" (it's actually self-dual, since a direct summand is either an onto homomorphism $F\to P$ that has a section, or a one-to-one homomorphism $P\to F$ that has a retraction) and "free module". The "free module on $X$" is $\mathbf{F}(X)$, where ...

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Yep! There's only a difference in the direct product and the direct sum in the infinite case. For example, if $\mathbb N$ is our infinite indexing set and $\{A_n\}_{n\in \mathbb N}$ are $K$-algebras, then $\prod_{n\in \mathbb N}A_n$ is the set of tuples $(x_1,x_2,\ldots,x_i,\ldots)$ such that $x_n\in A_n$. However, we define $\bigoplus_{n\in N}A_n$ to be ...

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Yes, such an injective morphism has non-zero determinant. Actually, if $M$ is a finitely generated free module over the commutative ring $R$ and $u:M\to M$ is an endomorphism, one has the precise equivalence: $$u \;\text {is injective}\iff \det(u) \; \text {is not a zero divisor in}\; R.$$ The proof is based on the fact that elements $m_1,m_2, ... 11 Consider a finite generating set and find a common denominator for its elements. Now, look at what you've got... Alternatively, if$M$is such a submodule, it has no torsion and we know from the structure theorem for f.g. modules over a PID that it must be free. Its rank can be computed by first tensoring with$K$over$R\$ and computing the dimension over ...

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