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We know that a module is semisimple if and only if every submodule is a direct summand. Therefore, what you are asking is whether $\mathcal M _2 (\Bbb Z)$ is a semisimple $\Bbb Z$-module. Note that $\mathcal M _2 (\Bbb Z)$ is a free $\Bbb Z$-module (of rank $4$), having for basis the subset $$\left\{ \left( \begin{array}{cc} 1 && 0 \\ 0 && ... 5 One way to proceed is to notice that \mathbb Z_8 is a local ring, so that its finitely generated projective modules are in fact free. Of course, this implies that finitely generated modules have at least 8 elements. 2 \newcommand{\Z}{\mathbb{Z}}\Z_4 is not projective over \Z_8. Indeed, consider the following exact sequence:$$0 \to \Z_2 \to \Z_8 \to \Z_4 \to 0.$$The map i : \Z_2 \to \Z_8 maps 1 to 4, and the map p : \Z_8 \to \Z_4 is the quotient map. Then this exact sequence is not split, i.e. there's no s : \Z_4 \to \Z_8 such that p \circ s = ... 0 As Tobias Kildetoft points out, \mathbb{Q} is both a (\mathbb{Q},\mathbb{Z})-bimodule and a (\mathbb{Z},\mathbb{Q})-bimodule making \mathbb{Q}\otimes_{\mathbb{Z}}\mathbb{Q} a (\mathbb{Q},\mathbb{Q})-bimodule. Also, \mathbb{Q} is a (\mathbb{Q},\mathbb{Q})-bimodule, making \mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Q} a ... 5 It does. If C is any chain complex in your additive category and F = 0 the zero functor, then H_n(FC) \cong F\bigl(H_n(C)\bigr) for each n \in \mathbf Z, as both sides are zero. 0 You're asking for a resolution of k[x,y]/(x,y) \simeq k. Note that we have an exact sequence (as you write yourself)$$ 0 \to (x,y) \to k[x,y] \to k \to 0. But as Najib Idrissi says in his comment, (x,y) is not a free k[x,y]-module. For example, we have the relation y \cdot x - x \cdot y = 0. What is true however, is that this is essentially the ... 4 The problem with your argument is quite subtle: you can't say \pi f(m/1)\in N, because the canonical homomorphism i:N\to S^{-1}N may not be injective (because elements of S might annihilate elements of N). That is, there is always some element n\in N such that i(n)=\pi f(m/1), but that n might not be unique, and it is not clear that you can ... 3 You're making things too complicate. ;-) Let f\colon R/I\to R/J be a homomorphism. Consider the composition map g=f\circ \pi\colon R\to R/J and write g(1)=x+J. Then, for every r\in R, g(r)=g(1r)=g(1)r=xr+J. Since g(r)=0, for every r\in I, we know that xI\subseteq J, so that x\in (J\mathbin:I). Now, ... -1 The mistake is in the sentence: 'and is clearly a homomorphism' (there is a choice of the r_i, they are not unique). 5 The eigenvalues of a matrix A over a field k are the roots of the characteristic polynomial \chi(A)=\det(\lambda I-A). But what is this thing we're taking the determinant of? \lambda is not an element of k, but an indeterminate, so to properly describe \lambda I-A we ought to say its coefficients, e.g. 2-\lambda, are elements of the polynomial ... 3 To explain what I said on MO (and also explaining what goes on in knsam's answer): The way to think about this is that since the group permutes the given basis vectors, it fixes the sum of all the given basis vectors. This gives a 1-dimensional invariant submodule. 2 Hint. Arguably, the simplest invariant subspace would be one of dimension 1. What would such a thing be? Do you see any such subspace in this case? 2 (x,y) is a regular sequence in R=\mathbf Q[x,y] , hence the Koszul complex: \begin{alignat*}{3}0\longrightarrow R&\xrightarrow{\begin{bmatrix}x\!&\!\!y\end{bmatrix}} R^2&\xrightarrow{\smash[t]{\begin{bmatrix}-y\\x\end{bmatrix}}}&R\longrightarrow R/(x,y)\longrightarrow 0 \\ t&\longmapsto \rlap{(xt, yt)}\\ &(u,v)&\longmapsto ... 0 The standard convention is that the trivial submodules are the zero submodule and the whole module. The fact that every module has those as submodules makes them "trivial." In contrast "proper" is used to refer to submodules that aren't the whole. It is the counterpart of "nonzero." So, the phrasing is ok. Adding "proper" does not hurt of course, and is ... 2 It seems like, by K-module, you are really talking about vector spaces, and they are isomorphic if and only if they each have a basis of the same cardinality. The size of the basis for K[x] is easily seen to be the cardinality of the natural numbers. Can the basis of K[[x]] be the cardinality of the natural numbers? It's not clear why you are ... 0 This follows from the observations below: Every subgroup of \mathbb Z/n\mathbb Z  corresponds to a subgroup of \mathbb Z that contains n \mathbb Z. The subgroups of \mathbb Z that contain n \mathbb Z are precisely d\mathbb Z, with d a divisor of n. Every subgroup of \mathbb Z is an ideal. The canonical group homomorphism \mathbb Z \to ... 1 An ideal has to be a subgroup (of the additive group of the ring) to begin with. So if you prove that all subgroups are already ideals, you are done. PS Just correct the statement every sub-group is mZ_n for 0<m<n to every subgroup is of the form m Z_{n} for m \mid n. 1 In general, the answer is no. See here for example. There it is shown that \mathbb {Z}^{\mathbb {N}} is not free as \mathbb {Z} -module. 1 A normalized vector has unit norm, that's the definition of being normalized, and it also follows from your definition. Let's check (in symbols): \def\abs#1{\left|#1\right|}\abs{\def\n{\mathord{\rm normalized}}\n(a)} = \sqrt{\n(a).x^2 + \n(a).y^2} = \sqrt{\frac{a.x^2}{\abs a^2} + \frac{a.y^2}{\abs a^2}} = \sqrt{\frac{a.x^2 + a.y^2}{\abs a^2}} = 1 The ... 1 A normalized vector (more commonly known as a unit vector) has norm 1. The reason why you don't have exactly one is that you're using finite precision on a computer (\sqrt{25+49} isn't 8.6. It's 8.6023252670...). 1 Since \dim M=0 we have \dim A/\operatorname{Ann}M=0, so A/\operatorname{Ann}M is a local artinian ring and you are done. 1 As an example why this does not work, consider the \mathbb{Z}-module M = \mathbb{Z}/4\mathbb{Z}. It is generated by S = \{1 + 4 \mathbb{Z}\} and as there is only one element in S we cannot do any elimination. But then (S) = M which is definitely not free. From a more general point, note that you are trying to take a submodule N = (S) of M, ... 2 Let \mathcal{S} be a category with finite products and let R be a ring in \mathcal{S}. There is an \mathcal{S}-enriched Lawvere theory \mathcal{T}_R where \mathcal{T}_R (n, m) = R^{m \times n} with composition defined by matrix multiplcation. We can define models of \mathcal{T}_R in \mathcal{S} to be an object M together with morphisms ... 0 This is essentially a corollary of Nakayama's Lemma. This states that IM=M is equivalent to (1-a)M=0 for some a \in I. Now, if M' is any submodule of M, it follows (1-a)M'=0 and hence IM'=M'. 1 Since M=IM, we can write m_1=a_1m_1+a_2m_2+\dots+a_nm_n, for a_i\in I. Rearranging terms, we find (1-a_1)m_1\in IM'. Now use this to show that (1-a_1)m_i\in IM' for i=2,\dots,n as well. Since a_1\in I, conclude that in fact m_i\in IM' for i=2,\dots,n. 1 It's no surprise you're having trouble proving that (2) implies (1), because it isn't true! For a counterexample, let G be the free abelian group on the first uncountable ordinal \omega_1 (or more generally, any ordinal with uncountable cofinality). Make G into a totally ordered group by ordering it reverse-lexicographically (that is, if ... 1 I will assume that actually, B is projective as a right R-module, not a left S-module. I think it's a typo... Suppose you have an inclusion i : X \to Y of left R-modules, and a map of R-modules f : X \to \hom_S(B,E). You want to construct a factorization g : Y \to \hom_S(B,E) such that g \circ i = f. The map f is equivalent to a map of ... 1 I don't know exactly what your question is. You seem to be unclear on both what O(3, \mathbb{C}) is and what an \mathbb{F}G-module is (which is fine). Are you clear on what a group representation is? I'll give some details, and you can tell me if you are looking for something else. I'm assuming O(3,\mathbb{C}) is referring to the orthogonal group ... 2 Let m_1, \ldots, m_n be generators of M and define an R-module homomorphism f: R^n \rightarrow M by (r_1, \ldots, r_n) \mapsto r_1m_1 + \ldots + r_nm_n. Since the m_i generate M, f is surjective. Therefore M \cong R^n/\ker f. Recall that an R-module is called Noetherian if every submodule is finitely generated. A ring R is called ... 1 The one that comes to mind is - \begin{align} &R = \mathbb{Z}, V = \mathbb{Z}, M = \mathbb{Z}/2\mathbb{Z}, D= End_{\mathbb{Z}}(M)\\ &\implies Hom_{R}(M,V) = Hom_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}) = 0\\ &\implies Hom_{R}(M,V) \bigotimes_{D}M = Hom_{\mathbb{Z}}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}) \bigotimes_{D}\mathbb{Z}/2\mathbb{Z} ... 1 Tensor each side with A/m, where m \in Spec(A). By noting that A/m \otimes A^n \simeq A^n/mA^n  is a n-dimensional vector space over A/m, compare the dimensions on both sides. 0 Although it may seem intuitively correct, M might not be isomorphic to rM for any non-zero divisor r\in R. But fortunately, this generalisation is not needed in this case. Note that M is an injective module, thus by Baer's criteria (A standard principle I might add for injective modules, and can be found in any basic textbook), any morphism from an ... 0 I'm turning darij's comment into an answer, since it solved my problem. For each b \in B we definde a function \delta_b\colon B \to K (where K is the commutative ring over which we are working) by \delta_b(b)=1 and \delta_c(b) = 0 for all c \neq b. Extend this \delta_b to a K-linear map M \to K. Now, if you have a linear dependency ... 0 Hint: There are two maps fitting into an appropriate diagram for 0: A'\to B. 1 In general \ker \varepsilon is not projective, otherwise every module would have a projective resolution of length one. Consider the following counter example: Let R=K[x,y] where K is a field and let M=R/(x,y). Let \epsilon: R \to M be the map that sends an element to its residue class. The kernel is the ideal (x,y) which is not projective. ... 2 For noetherian rings these properties are indeed equal, but not in general. Take your favorite non-noetherian ring R. It has an ideal I which is not finitely generated. Then R/I is finitely generated but not finitely presented. 1 This is not true in general. For instance, suppose R is a Boolean ring (a ring in which every element is idempotent; any such ring is automatically commutative). If M is an R-module, then for any r\in R, M=rM\oplus (1-r)M. If M is indecomposable, let I be the annihilator of M; then for any r\in R, either r\in I or 1-r\in I. It ... 2 Such modules are called cyclic. 1 If M is flat, then it must be torsion-free (see here), but this is not the case: xy=0 and y\ne 0 in M. 1 Theorem. Let R be a commutative ring. If an R-module X is scale-homogeneous, then R is pair-homogeneous. Proof. Let X be a scale-homogeneous R-module. We need to show that X is pair-homogeneous. First, assume that R is not an integral domain. Then, there are two nonzero elements a and b of R such that ab = 0. Consider these a and ... 1 Proposition: If M is nonzero, M is scale-homogeneous iff R is a domain and M is a torsion-free, divisible module. Proof: Obviously the condition that multiplication by nonzero ring elements is injective implies that no element of M has a nonzero annihilator. It also implies that the composition of scaling is injective, so ab\neq 0 whenever ... 1 Yes, you can use the fundamental theorem of isomorphism. In order to do this consider the following composition of maps: R\to(x) given by a\mapsto ax, and (x)\twoheadrightarrow (x)/(xy), the canonical projection. The first map is an isomorphism, and the second is surjective, so their composition is surjective. The kernel is \{a\in R:ax\in(xy)\} and ... 0 Consider the exact sequence 0 \to (y) \to R \to R/(y) \to 0. $$and note that the map x:R \to xR given by multiplication by x is an isomorphism of R-modules. Apply this map to each term in the sequence. What do you get? 2 They are perhaps the easiest type of module to understand. They are the most vector space-like modules, in a sense. For example, the endomorphism rings of finitely generated free R modules are just square matrix rings over R. They contain enough information to recover all Morita equivalent rings. To do that, you just need to know all the finitely ... 3 I only have the algebro-geometric "commutative" intuition, and from that point of view modules are generalizations of vector bundles over Spec A, and flat modules are actual vector bundles --- localisation of M at a prime ideal of A is free, and this is a sufficient condition for flatness too (for rigour, one probably has to throw in some ... 0 "\Rightarrow": Let \Phi: V_\varphi \to V_\Omega be a module isomorphism. Then by definition of module isomorphisms$$ \forall p, q \in K[x], v, w \in V : \Phi(p \cdot_\varphi v + q \cdot_\varphi w) = p \cdot_\varphi \Phi(v) + q \cdot_\varphi \Phi(w). $$In particular, for w = 0 and p(x) = x:$$ \Phi(\varphi(v)) = \theta(\Phi(v)), $$and thus ... 0 Let M=k[x]/(x^2), N=k[x]/(x) and f:M\to N the map induced by the identity of k[x]. Let K be the kernel of f, so that we have an exact sequence$$0\to K\to M\to N\to 0$$What happens if you tensor this with k[x,y]/(x^2,xy)? 2 The universal property of R^J for a set J is that there is a natural isomorphism$$ \hom_R(R^J,M)\simeq \hom_{\rm Set}(J,M)$$Now, by the usual adjunctions in {}_R\,\mathbb {mod} and \mathrm{Set} we have natural isomorphisms$$\hom_R(R^I\otimes_R R^J,M)\simeq \hom_R(R^J,\hom_R(R^I,M))\\\simeq \hom_{\rm Set}(J,\hom_{\rm Set}(I,M))\simeq \hom_{\rm ...

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I assume $R$ is a domain. We call its fraction field $K$. (1) If $G$ is a torsion-free $R$-module it embeds into $G\otimes_RK$, which is divisible. (2) If $G$ is a torsion-free, divisible $R$-module then it is a vector space over $K$.

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By definition, we have that $$R^{mn}\cong Re_{11}\oplus...\oplus Re_{1n} \oplus Re_{21}\oplus ... \oplus Re_{2n} \oplus ... \oplus Re_{m1} \oplus ... \oplus Re_{mn}$$ That is, $R^{mn}$ has a basis consisting on $mn$ elements which we have chosen to denote by $\{e_{ij}\}$, where $i=1,...,m$ and $j=1,...,n$. In the same fashion, pick $\{a_1,...,a_m\}$ a ...

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