Hot answers tagged

61

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 ...


31

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 ...


28

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 ...


26

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


25

One way is to use the adjunction property of the tensor product. For any $R$-module $P$, we have the following sequence of canonical $R$-module isomorphisms: $$\begin{eqnarray}\hom(\varinjlim(M_i \otimes N), P) \cong&& \varprojlim \hom(M_i \otimes N, P) \\ \cong && \varprojlim \hom(N, \hom(M_i, P)) \\ \cong && \hom(N, \varprojlim \...


25

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 = \...


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

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 ...


20

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 ...


20

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 ...


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 $I\...


19

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 ...


19

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 $$\prod_{n=1}^{\infty}\...


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 ...


18

Let $F$ be a free $R$-module, where $R$ is a PID, and $U$ be a submodule. Then $U$ is also free (and the rank is at most the rank of $F$). Here is a hint for the proof. Let $\{e_i\}_{i \in I}$ be a basis of $F$. Choose a well-ordering $\leq$ on $I$ (this requires the Axiom of Choice). Let $p_i : F \to R$ be the projection on the $i$th coordinate. Let $F_i$ ...


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 rule)....


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

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 K[x,y,z]\...


16

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


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. ...


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 ...


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 ...


15

It suffices to show that if $K_i\to M_i\to N_i$ is exact at $M_i$ for each $i$ (and the appropriate squares commute), then $\varinjlim K_i\to \varinjlim M_i\to \varinjlim N_i$ is exact at $\varinjlim M_i$. (To get that $\varinjlim$ sends short exact sequences to short exact sequences from this, simply apply the argument to $0\to K_i\to M_i$, exact at $K_i$;...


15

A module can have a nonzero annihilator. This overlaps a little with the torsion module example you gave above. If $R/I$ is a non-trivial quotient of a commutative ring, $R$, then it is an $R$ module with annihlator $I\neq 0$. A module can be Artinian but not Noetherian, and it can be Noetherian but not Artinian. Any non-Artinian but Noetherian ring ...


15

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 ...


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 $m_{i,...


14

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 ...


14

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}) + ...


14

The issue of when projective modules are free is discussed in $\S 3.5.4$ of my commutative algebra notes. In particular one gets very easy (but not very satisfying) examples by looking at disconnected rings: e.g. $\mathbb{C} \times \{0\}$ is quite clearly projective but not free over $\mathbb{C} \times \mathbb{C}$. It is more interesting to ask for ...



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