# Tag Info

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 ...

18

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 ...

11

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 ...

11

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 ...

9

If $I$ is an ideal of $R$, the dual of $R/I$ is isomorphic to $\mathrm{Ann}(I) = \{r \in R : rI = 0\}$, and this doesn't have to be finitely generated. Take for instance $R = k[y,x_1,x_2,\dotsc]/(y x_i : i \geq 1)$ and $I=(y)$. A more natural question would be: How can we characterize commutative rings with the property that duals of f.g. modules over that ...

9

Note that $K\otimes_{R} M$ is isomorphic as an $R$-module to the localization of $M$ at the multiplicative subset of $R$ consisting of all non-zero elements. (I think this point of view would be helpful if you're familiar with the notion of localization of modules.) The kernel of the map $M\to K\otimes_{R} M$ is equal to the $R$-submodule of $M$ consisting ...

9

It is not true that $RI\otimes N = R\otimes IN$. You can't "factor out" elements of the ideal linearly because the ideal is being thought of as a module over $R$, and $1$ is (generally) not in the ideal. Try to write down what you think the isomorphism should be and you'll see it. For concreteness, take $I=(a)$ where $a$ is some non-invertible element in ...

9

You just have to work with the definitions (i.e. universal properties) in order to answer this question. It is not an extra convention or something like that (unfortunately, many mathematicians believe this). If $(E_i)_{i \in I}$ is a family of $R$-modules with underlying sets $|E_i|$, and $F$ is some $R$-module, a map $\prod_i |E_i| \to |F|$ is called ...

8

The key thing to remember about the operation "mod" is that it behaves "well" with respect to product (hence powers), and of course addition. This means that if you can simplify your life a lot distributing the calculation into many steps and taking "mod" at each stage. To compute $20^{15}$, you can first notice that $20 \pmod{17} = 3$. Then $20^{15} ... 8 Suppose$A$is a free$\Bbb{Z}$-module, and$b$is an element of a basis. Then there is no element$a\in A$such that$2a=b$, since otherwise$b$would have two different representations in terms of basis elements. Thus in your$M$there is nothing that could possibly be a basis element, so there is no basis. 8 Intro a right R-module is equivalent to a left R-module only when R is commutative This statement has a pair of serious problems to resolve. Because "equivalent" is undefined, it's unclear what the statement means. (Discussed briefly below.) It uses "only when," but that is the wrong logical direction: it should use just when. There are in fact ... 8 While it's true that in$M\otimes_RN$you can do $$xr\otimes y=x\otimes ry$$ you can't "exchange ideals" across the tensor product. A simple example should make this clear: set$R=\mathbb{Z}$,$I=2\mathbb{Z}$and$N=\mathbb{Z}/2\mathbb{Z}$. Since, as$\mathbb{Z}$-modules we have$\mathbb{Z}\cong 2\mathbb{Z}$, we have $$\mathbb{Z}\otimes N\cong ... 8 \require{AMScd}The factorization you ask about is an instance of the slogan "distinguished triangles are just like exact sequences". To wit, the axiom on existence of morphisms yields that in a distinguished triangle \begin{CD} X @>u>> Y @>v>> Z @>w>> X[1] \end{CD} v is a weak cokernel of u. For if f \colon Y \to W is a ... 8 I'm assuming that \mathbb Z_n means \mathbb Z/n\mathbb Z. I'm also assuming that "subproduct" means "subgroup of the product". If this is the case then yes, the subgroup generated by (1, 1, \ldots) \in \prod_{n > 1}\mathbb Z_n is isomorphic to \mathbb Z. To see that this is the case note that we can always define a homomorphism out of \mathbb ... 7 This is true if R is commutative. Otherwise, say that you are dealing with left R-modules, for instance. If you attempt to define multiplication by r by (rh)(m) = rh(m) for any homomorphism h \colon M \to N, then you run into the problem that the mapping rh may not be R-linear. For example, let h \colon R \to R be the identity map. Then ... 7 If R is a ring, M is a right R-module and N is a left R-module, then the functor of balanced maps F : \mathsf{Ab} \to \mathsf{Set}, A \mapsto \{\beta: |M| \times |N| \to |A| \text{ balanced}\} satisfies the assumptions of Freyd's criterion for representability: It is easy to check that it preserves limits. For the solution set condition, let ... 7 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 ... 7 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 ... 7 It holds when R is a PID (here injective \Leftrightarrow divisible). It is not true in general. The following papers study and characterize this property of R that tensor products of injectives are injective. Ishikawa, Takeshi. "On injective modules and flat modules." Journal of the Mathematical Society of Japan 17.3 (1965): 291-296. Enochs, ... 6 Assume we are in the category of abelian groups (for intuition's sake). a. \mbox{Hom}(\bigoplus A_i,B)\simeq\prod\mbox{Hom}(A_i,B). Let \phi_i:A_i\hookrightarrow\bigoplus A_i be the natural inclusion (we put 0's in all the other coordinates), and if f\in\mbox{Hom}(\bigoplus A_i,B), let f_i be its restriction to \phi_i(A_i) (Edit: If ... 6 Here's one idea. Use the following consequence of Nakayama: If M is a finitely generated R-module and f\colon M\to M is a surjective module homomorphism, then f is an isomorphism. Proof: View M as an R[x]-module, where the action of x on M is given by f. By assumption xM = M. The proof of Nakayama's lemma then gives that there is ... 6 More generally, if S is any subset of R and M,N are R-modules, then S^{-1} (M \oplus N) \cong S^{-1} M \oplus S^{-1} N. This is simply because S^{-1} (-) is left adjoint to the forgetful functor from S^{-1} R-modules to R-modules, hence preserves all colimits, in particular direct sums. But you can also check this directly. Beware that it ... 6 Let R be a semisimple ring which isn't a division ring, and take an idempotent e\notin \{0,1\}. Then we have that R=eR\oplus(1-e)R is a nontrivial decomposition of R, and both pieces are cyclic and projective (since they are summands of R) hence flat. Actually neither piece is a free module, but we'll argue here that at least one of them isn't ... 6 Any natural definition of R-action works only, when R is commutative. For example, if you try to define$$ (rf)(m)=r(f(m)) $$for all r\in R, f\in Hom_R(M,N), m\in M, then the mapping rf fails to be homomorphism of R-modules in general. If s\in S is such that sr=rs, then$$ (rf)(sm)=r(f(sm))=r(sf(m))=(rs)(f(m))\neq s((rf)(m)) $$in general. ... 6 Presumably, you'll want to define \psi=r.\phi by$$\psi(m)=r.\phi(m)$$(where$r\in R, \phi\in\mathrm{Hom}_R(M,N)$and$m\in M$.) However, this map, which is a morphism of abelian groups, need not be$R$-linear when$R$isn't commutative :$R$-linearity would imply that for all$r'\in R$(and all$m\in M$)$\psi(r'.m)=r'.\psi(m)$, i.e., by$R$-linearity of ... 6 In general, a left module over a ring$R$is equivalent to a right module over the ring$R^{\operatorname{op}}$, whose multiplication is the same as that in$R$, but in the opposite direction (i.e.$a\cdot_{\operatorname{op}} b = b \cdot a$). This is because in a left module, say$M$, we multiply by elements on the left, so, if by$\lambda_a: M\mapsto M$we ... 6 This is not true, in general. The modules$M$which occur as direct summands of a free module are precisely the projective modules, which are not in general free. The simplest counter-example is probably the ideal generated by$2$and$1+\sqrt{-5}$in the ring$\mathbf Z[\sqrt{-5}]$, which is projective over$\mathbf{Z}[\sqrt{-5}]$but not free. If you ... 5 First define a map$A\times M\rightarrow M$such that$(a,m) \mapsto am$. This is clearly bilinear, so it induces a homomorphism$A\otimes M\rightarrow M$such that$a\otimes m \mapsto am$(by the universal property of tensor products). Surjectivity is straight forward, just take$a=1$,$m\$ arbitrary. To show that the homomorphism is injective, note that any ...

5

The simplest case is when the group is equipped with a topology with respect to which it is compact (Hausdorff), so the proof of Maschke's theorem still works and the Peter-Weyl theorem is available. In particular, the representation theory of compact Lie groups is very well understood. The representation theory of noncompact Lie groups is still a major area ...

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