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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

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 $\mathbb ... 5 Hint:$R=\mathbb{Z}$,$C=\mathbb{Z}/2\mathbb{Z}$. There is no problem in$\phi$being well defined; there's some problem in showing it's injective. ;-) The map is well defined, because such is the map$\tilde\phi\colon C\to C\otimes_RK$defined by $$x\mapsto x\otimes1$$ and composing with the natural isomorphism$C\otimes_RR\to C$gives your$\phi$. 5 For ease of notation, let$A = \operatorname{im}\varphi_1 \subset N$,$T = \operatorname{im}\varphi_2 \subset S$, and$Q = T/\operatorname{im} (\varphi_2\circ\varphi_1) = T/\varphi_2(A)$. Consider the map$\psi = \pi \circ \varphi_2 \colon N \to Q$, where$\pi \colon T \to Q$is the canonical projection. The kernel of$\psi$is $$\ker \psi = \psi^{-1}(0) = ... 5 The answer is yes. Hint: You can easily see that considering the kernel, image resp. of an R-homomorphism. Conversely, if assumptions ii) or iii) hold for such R-homomorphisms, then M has to be simple, where you can consider the possible submodules of M to check. 5 Note first that for any ring R and any R-module M, if the cardinality \lvert M\rvert is infinite and greater than \lvert R \rvert, then any generating set of M (in particular, any basis) must have cardinality equal to \lvert M\rvert (this is simple combinatorics). In particular, any basis of \bf C over \bf Q must have cardinality of the ... 4 As P is f. g. we have an exact sequence 0\rightarrow Q\rightarrow A^n\rightarrow P\rightarrow 0, Q denoting the kernel of the map A^n\rightarrow P. As P is projective, this exact sequence splits, A^n\cong Q\oplus P. The exact sequence A^n\cong Q\oplus P\rightarrow A^n\rightarrow P\rightarrow 0 shows P to be f. p. (where Q\oplus P\rightarrow ... 4 A counterexample Let R be your favorite commutative ring which isn't Noetherian, and let I be any maximal ideal of it. Then M=R/I is a simple (and so certainly Noetherian) module for this not-Noetherian ring. Why am I wrong? The wrong link is the idea that the chain of I_iM stabilizing should imply that the chain of I_i should stabilize. It ... 4 Hint: You could use the universal property of the tensor product. Let's look at \ell_s:M\times N\to M\otimes_R N where \ell_s(m,n):=(sm)\otimes n. It's definitely well-defined since the module operation on M is well-defined. What other properties does this map have? Of course after all is said and done we're going to relabel \ell_s(m,n) as ... 4 Nope. Consider two 4 \times 4 matrices:$$\begin{pmatrix}0 & & & \\ & 0 & & \\ & & 0 & 1\\&&0&0\end{pmatrix}\begin{pmatrix}0 & 1 & & \\ 0 & 0 & & \\ & & 0 & 1\\&&0&0\end{pmatrix}$$(where I've written the matrices in Jordan block form; ... 3 Don't use elements, because then you will miss the obvious and elegant proof: M is an (S,R)-bimodule, this means that we have an right R-linear map S \otimes_R M \to M satisfying the usual axioms. The tensor product is associative (this is what happens in rschwieb's answer, but why proving this once again in a special case?) and functorial, hence we ... 3 As you remarked, injective modules are divisible, that is, rM=M for all r\in R, r\ne 0. The key step is to show that every non-zero homomorphism f:M\to R is surjective. Let x\in M such that f(x)\ne 0. Set r=f(x). Since rM=M there exists y\in M such that ry=x. Then rf(y)=r, so f(y)=1, and this is enough. Now use that M is ... 3 The straightforward answer for "what is a basis for \mathbb{C}/\mathbb{Q}" is that we don't know. The sneaky answer is that we do know there is one, because any maximal linearly independent set is a basis, and exists by Zorn's Lemma. This shows that \mathbb{C} is a free \mathbb{Q}-module, since that concept is literally equivalent to the existence of ... 3 Well, the key words are onto endomorphisms are isomorphisms. (Actually, the paper proves more than the title says!) 2) The answer is negative and as you remarked this also solves 1). Let F be the submodule of M generated by x_1,\dots, x_{n+1}. We have an isomorphism F\to A^{n+1}. Since M is generated by n elements there exists K\le A^n such ... 3 Note that \operatorname{im}(f) \supseteq \operatorname{im}(f^2) \supseteq \operatorname{im}(f^3) \supseteq \cdots so by the DCC, there is an integer n such that \operatorname{im}(f^n)=\operatorname{im}(f^{2n}). Let m\in M. Since \operatorname{im}(f^n)=\operatorname{im}(f^{2n}), there is some t\in M such that f^n(m)=f^{2n}(t). Write ... 3 Let \mathrm{CFM}_\mathbb{N}(R) denote the ring of "column finite matrices", where R is some ring.. Then, one can show that \mathrm{CFM}_\mathbb{N}(R)\to\mathrm{CFM}_\mathbb{N}(R)^2 defined by$$M\mapsto (\text{odd indexed columns of }M,\text{even indexed columns of }M)$$is an isomorphism of \mathrm{CFM}_\mathbb{N}(R)-modules. It clearly then ... 3 For the first part, the general definition is that a sequence \cdots \to L \xrightarrow{f} M \xrightarrow{g} N \to \cdots is exact at M if and only if the image of f is the kernel of g. A short exact sequence is an sequence consisting of five terms whose endpoints are zero, and is exact at every (internal) point. i.e.$$ 0 \to A \xrightarrow{f} B ... 3 So for such$x,y\neq 0$, you want to show$x\otimes y=y\otimes x$in$V\otimes_F V$if$x=ay$for some$a\in F$. Then$x\otimes y=ay\otimes y=y\otimes ay=y\otimes x$. Since this seems by far to be the easier half of the problem, I am beginning to wonder if you meant to ask about the other direction. Suppose$x,y$are linearly independent. As such, this ... 3 Suppose$\eta$is a natural tranformation. Its value at the left$R$-module$R$is a map$\eta_R:R\to R$of left$R$-modules. You can easily check that$c=\eta_R(1)$, the image of$1\in R$, belongs to the center$Z(R)$of$R$. If now$M$is any$R$-module, and$m\in M$, there is a unique map$f:R\to M$such that$f(1)=m$. Using naturality of$\eta$, we see ... 2 By the snake lemma there is an exact sequence $$0\rightarrow\mathrm{Ker} \ \alpha\rightarrow 0 \rightarrow \mathrm{Ker} \ \gamma \rightarrow \mathrm{Coker} \ \alpha\rightarrow 0\rightarrow\mathrm{Coker} \ \gamma\rightarrow 0,$$ whence$\mathrm{Ker} \ \alpha=0$and$\mathrm{Coker} \ \gamma=0$; also$\mathrm{Ker} \ \gamma\cong\mathrm{Coker} \ \alpha$so that ... 2 We are given$N$and that will give us the prime factors$p$and$q$as: $$N = 91 = p \times q = 7 \times 13$$ We need the Euler Totient Function of the modulus, hence we get: $$\varphi(N) = \varphi(91) = (p-1)(q-1) = 6 \times 12 = 72$$ Now, we choose an encryption exponent$1 \lt e \lt \varphi(N) = 72$. We were told to pick an an$e \lt 6$, so lets ... 2 The answer is that such matrices are always similar over$\def\Z{\Bbb Z}\Z$(conjugate in$GL_2(\Bbb Z)$). The question is deeper however than it might look at first, and as far as I can see any solution requires some somewhat subtle arithmetic considerations. A few things are easy:$A,B$always have determinant$~1$(from the constant coefficients of the ... 2 I don't know what$A$is, but hopefully it is a group. If that's not true, it's safe to ignore everything I write below! Even though$A$is probably abelian, I'll write it multiplicatively to make the below easier. This isn't a rigorous definition but is hopefully enough to let you search. The ring$\mathbb{F}_pA$, sometimes written$\mathbb{F}_p[A]$, is ... 2 Consider the ring$R=F_2[Z]/(Z^2)=M$where$F_2$is the field of two elements. This is a self-injective ring, so$M$is an injective$R$-module. But now consider$x=1$and$r=Z$, where I abuse notation for the images of$1$and$Z$in this ring. Saying that there exists$y\in R$such that$yZ=1$implies that$Z$is a unit, but it is clearly not since it is ... 2 Take a set of basis$e_1,\ldots,e_n$of$V$and let$x=\sum_ {i}x^ie_i,~y=\sum_ {j}y^je_j$, then $$x\otimes y=\sum_ {i,j}(x^ie_i)\otimes(y^je_j)=\sum_ {i,j}x^iy^je_i\otimes e_j$$ $$y\otimes x=\sum_ {i,j}y^jx^ie_j\otimes e_i$$ The symmetry implies $$x^iy^j=x^jy^i$$ That is, $$\frac{x^i}{y^i}=\frac{x^j}{y^j}=a$$ for some constant$a$. 2 About your first question: yes, that is the definition of a short exact sequence. For the second question, note that if$\phi: B \rightarrow C$is any$R$-map then $$0 \rightarrow \text{ker}(\phi) \rightarrow B \rightarrow \text{im} (\phi) \rightarrow 0$$ is short exact. In particular,$\text{im}(\phi) \cong B / \text{ker}(\phi)$. Now, suppose that$$... 2 It should be straightforward. We want to prove$\beta$is surjective, so start out from an arbitrary element$b'\in B'$. We can do one thing: consider$c':=\phi'(b')\in C'$. Since$\gamma$is surjective, we get$c$with$\gamma(c)=c'$. That the pair$\phi,0$of maps is exact means nothing else but that$\phi$is surjective. It yields an element$b\in B$, ... 2 The notation$R \cap \mathfrak m$is sometimes used to denote the inverse image of$\mathfrak m$in$R$. It is a prime ideal$\mathfrak p$of$A$. There is a canonical morphism$R_{\mathfrak p} \to A_{\mathfrak m}$, and the$A_{\mathfrak m}$-module$M_{\mathfrak m}$acquires an$R_{\mathfrak p}$-module structure via this map. 2 Joseph Rotman - An Introduction to Homological Algebra (2nd edition) is a great book about homological algebra, but it contains many sections about modules and rings. I advise you to give it a look since it is a wonderfully written book! 2 We might get the idea to send$f:V\to V$to$f|_S:S\to V$. Actually we can say more: if$f(S)\neq 0$, then$S\subseteq f(S)$, but since$f(S)$is a nonzero image of a simple module, it is also simple, so$f(S)=S$. If$f(S)=\{0\}$then$f(S)\subseteq S$trivially. Thus$f|_S:S\to S$. Is it a ring homomorphism? Yes: check it out. The assignment$f\mapsto ...

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