Hot answers tagged

58

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 ...


29

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 ...


27

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: ...


26

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 ...


24

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 $\mathbb{Z}[X_1, X_2, X_3, \ldots ]$.


23

(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 ...


21

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 = ...


20

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 ...


19

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 ...


19

This is an application of the second 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 second isomorphism theorem gives you the ...


19

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 ...


18

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 ...


18

Let $R=\mathbb{Z}/6\mathbb{Z}$. Obviously, $R$ is a free module over itself. Because $\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/3\mathbb{Z}\cong R$, we have that $\mathbb{Z}/2\mathbb{Z}$, considered as an $R$-module, is projective, but it cannot be free - any non-trivial direct sum of $R$'s would have at least 6 elements. The Baer-Specker group is an example ...


18

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 ...


17

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 ...


17

If you're not familiar with tensor products, then learning them for this problem is overkill. But if you are familiar with them, then I think the following solution is nice. $\newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$ For a finitely generated abelian group $A$, note that $\mathrm{rank}(A) = \dim_\Q(A\otimes_\Z \Q)$. One can show (basically ...


17

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 ...


16

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 ...


16

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. ...


15

$\def\id{\operatorname{id}}$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 ...


15

The proof mentioned by Frederik and Loronegro is great because it provides a first example of how it can be useful to know that two functors are adjoint: left adjoints are right exact. However, you can also argue as follows. Let $D$ be the image of $\alpha \otimes \operatorname{id}$. You get an induced map $(B \otimes M)/D \to C \otimes M$. Let's try to ...


14

As Martin Brandenburg mentioned, this holds in a much more general context: Let $C,D$ be categories and $F\colon C\rightarrow D$, $G\colon D\rightarrow C$ functors, such that $F$ is left adjoint to $G$. Then $F$ preserves all colimits and $G$ preserves all limits. Especially G preserves kernels and therefore is left-exact, whenever you can talk about ...


14

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


13

This failure of freeness is a non-trivial result. One way to prove it is to begin with a lemma: If $F$ is a free abelian group and $C$ is a countable subgroup, then the quotient $F/C$ is the direct sum of a countable group and a free group. (I'm omitting "abelian" because I'm lazy and all groups here will be abelian.) [Proof of lemma: Fix a basis $B$ for ...


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}) + ...


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

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 ...


12

If $A$ is an abelian group, then $\cap_{n \geq 0} 2^n A$ is a subgroup of $A$, whose elements may be called $2^{\infty}$-divisible. Note that $0$ is the only $2^{\infty}$-divisible element of $\mathbb{Z}$. Therefore, the same is true for $\mathbb{Z}^{\mathbb{N}}$. But the element represented by $(2^0,2^1,2^2,\dotsc)$ is $2^{\infty}$-divisible in ...


12

Let $M$ be a module over a commutative ring. Recall three properties: $M$ is projective if it is a direct summand of a free module. $M$ is weakly stably free if there exists a free module $F$ such that $M \oplus F$ is free. $M$ is stably free if there exists a finitely generated free module $F$ such that $M \oplus F$ is free. Then free implies stably free ...



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