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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 ...
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 $$\ker \psi = \psi^{-1}(0) = ... 5 The answer is yes. Hint: You can easily see that considering the kernel, image resp. of an R-homomorphism. Conversely, if assumptions ii) or iii) hold for such R-homomorphisms, then M has to be simple, where you can consider the possible submodules of M to check. 5 Note first that for any ring R and any R-module M, if the cardinality \lvert M\rvert is infinite and greater than \lvert R \rvert, then any generating set of M (in particular, any basis) must have cardinality equal to \lvert M\rvert (this is simple combinatorics). In particular, any basis of \bf C over \bf Q must have cardinality of the ... 4 As P is f. g. we have an exact sequence 0\rightarrow Q\rightarrow A^n\rightarrow P\rightarrow 0, Q denoting the kernel of the map A^n\rightarrow P. As P is projective, this exact sequence splits, A^n\cong Q\oplus P. The exact sequence A^n\cong Q\oplus P\rightarrow A^n\rightarrow P\rightarrow 0 shows P to be f. p. (where Q\oplus P\rightarrow ... 4 You know the rank of a free module, right? It is the cardinality of a basis (well-defined since R is commutative and R \neq 0). This is already well-known from linear algebra (R is a field), where it is called the dimension (but as you see, this is really the same concept). Now it turns out that finitely generated projective R-modules are "not far" ... 4 Hint: You could use the universal property of the tensor product. Let's look at \ell_s:M\times N\to M\otimes_R N where \ell_s(m,n):=(sm)\otimes n. It's definitely well-defined since the module operation on M is well-defined. What other properties does this map have? Of course after all is said and done we're going to relabel \ell_s(m,n) as ... 4 Nope. Consider two 4 \times 4 matrices:$$\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 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 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 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 ... 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 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 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 ... 3 The ordinary approach is to prove R[s] is a finitely generated R-module if and only if s is integral over R and that the elements of S that are integral over R form a subring of S. (See, for instance, Atiyah-Macdonald 5.1, 5.3 or Eisenbud 4.2, 4.6). This settles your problem: s and t are integral over R and since the integral elements of ... 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 ... 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 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 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 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 $$... 2 Just some hints: What you've just done is to make M into an R-module. You could equally look at M as an abelian group (= \Bbb Z-module) with an action of the ring R, defined by R \to Aut(M), but the main problem is, that not every image of \mu is an automorphism. So, in fact, you could have \varphi \in End(M) \setminus Aut(M) as an image of ... 2 Let r_1, \dots, r_n be elements of R such that \sum_i r_if_i = 0. Our goal is to show that r_i = 0 for all i. Write \vec{r} = \langle r_1, \dots, r_n \rangle^t (so a column vector), one quickly sees (using that e_1, \dots, e_n is a basis) that this is equivalent to A\vec{r} = \vec{0}. Even though \det(A) may not be invertible and hence we ... 2 Consider the ring R=F_2[Z]/(Z^2)=M where F_2 is the field of two elements. This is a self-injective ring, so M is an injective R-module. But now consider x=1 and r=Z, where I abuse notation for the images of 1 and Z in this ring. Saying that there exists y\in R such that yZ=1 implies that Z is a unit, but it is clearly not since it is ... 2 The notation R \cap \mathfrak m is sometimes used to denote the inverse image of \mathfrak m in R. It is a prime ideal \mathfrak p of A. There is a canonical morphism R_{\mathfrak p} \to A_{\mathfrak m}, and the A_{\mathfrak m}-module M_{\mathfrak m} acquires an R_{\mathfrak p}-module structure via this map. 2 For b), you were right. To look at a, we say x = a + m\cdot q and see if the result is independent of m, i.e. if f(a+m\cdot q) \equiv f(a):$$f(a+m\cdot q) = (a + m\cdot q)^2 = a^2 + 2m\cdot q + m^2 q^2 \equiv a^2 = f(a) \qquad (\text{mod }m)$$This concludes the proof, that f is well-defined 2 By the snake lemma there is an exact sequence$$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 ...