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Step 1. For a commutative ring $R$, we have $\mathrm{GL}_{n}(R) / \mathrm{SL}_{n}(R) \cong R^{*}$. Step 2. $|\mathrm{GL}_{n}(\mathbb{F}_{q})| = (q^{n}-1)(q^{n}-q)\cdots (q^{n}-q^{n-1})$. Step 3. Let $R$ be a commutative local ring with maximal ideal $\mathfrak{m}$ and residue field $K = R / \mathfrak{m}$ . Then $|\mathrm{GL}_{n}(R)| = ... 4 I must say the argument seems fine to me. Assuming no module in$S$is maximal, we can construct a sequence$N_0\subsetneq N_1\subsetneq\ldots$which contradicts the ACC condition. But note you need the axiom of choice to construct the sequence, as it's not obvious at all how to choose the modules. In fact, only the dependent choice is required. 4 Let$R=k[x,y]$, and let $$M_1=(x,y)/(y),\qquad M_2=k[x,y]/(y),\qquad M_3=k[x,y]/(x,y).$$ Then$M_1$is the kernel of the obvious quotient map$M_2\to M_3$(third isomorphism theorem), so we do have an exact sequence $$0\longrightarrow M_1 \xrightarrow{\text{inclusion}}M_2\xrightarrow{\text{quotient}}M_3\longrightarrow 0$$ but $$\mathrm{Ann}(M_1)=(y),\qquad ... 4 A module is an abelian group. (It's more useful to think of a module as the analog of a vector space, but with the set of scalars coming from a ring instead of a field. Usually, one arrives at this notion of a module in terms of "the action of a ring on a set" where the set is a module.) A monoid is a relaxation of the definition of a group. A monoid has ... 4 Yes. A module M over A is just a module M over the ring A; the additional structure of A as a k-algebra plays no role. (and 3.) That amounts to the same: since A is a k-algebra, you already have a map k \to A which turns M into a k-module as well. M, ultimately being a k-module as well, is a k-vector space. I'm note sure what ... 4 The direct sum of modules is their coproduct. In general the underlying set of a coproduct of modules is not the coproduct (i.e. disjoint union) of their underlying sets. (It doesn't make sense to talk about \coprod_{i \in I} M_i as being a module until you endow it with the structure of a module.) 3 If \mathfrak{a} and \mathfrak b are ideals in a ring R, with elements x and y respectively, the product xy makes sense: Multiplication is certainly defined for ring elements. Then (finite) sums of the form \sum_i x_i y_i certainly make sense because x_i y_i \in R, and we can add ring elements. This is not true, however, for modules: There is ... 3 Generic modules can be so ill-behaved compared to vector spaces that I never recommend thinking about them as the same kind of beast. Spaces are free, divisible, torsionfree, semisimple, and the only simple module (the atoms of the semisimple universe) over a field (up to iso) is the field itself. I cannot for the life of me figure out what you think your ... 3 There is a chain of forgetful functors which progressively forget the various operations in the structure:$$\mathrm{Mod_R}\to\mathrm{Ab}\to\mathrm{AbMon}\to\mathrm{Set}$$The interesting thing is that you can go in the opposite direction too with free functors$$\mathrm{Set}\to\mathrm{AbMon}\to\mathrm{Ab}\to\mathrm{Mod_R}$$Each forgetful functor U is ... 3 The statement is not correct^1. Here are two correct ones (which you have already proven): 1) Let M_1,...,M_n be submodules of M. Let us denote by \iota_i : M_i \to M the inclusion. Then M = \oplus_i M_i (internal direct sum) iff there are \pi_i : M \to M_i such that (i),(ii),(iii) hoold. 2) Let M_1,\dotsc,M_n be modules. Then M \cong ... 3 For 1 Yes, it's true. The trick is to remember that the simple modules of A are the same as the simple modules of A/J(A), where J(A) is the Jacobson radical of A. Since A is a finite dimensional algebra, it is a right and left Artinian and Noetherian ring. As such, it has a composition as a left module over itself (and as a right module over ... 3 Every monomorphism f:A \to B induces an exact sequence 0 \to A \to B \to \text{coker}(f) \to 0. As tensoring is right-exact, a module M is flat iff tensoring with M preserves monomorphisms. Now use that \Bbb Z \otimes_{\Bbb Z}\Bbb Z /n \Bbb Z = \Bbb Z/n \Bbb Z to see why the map x \to nx is not injective on the tensors. 3 Guessing that n:\mathbb{Z}\to\mathbb{Z} is multiplication by n? If so, this is certainly injective, so starts an exact sequence$$ 0 \to \mathbb{Z} \stackrel{n}{\to} \mathbb{Z} $$and the kernel of the next map will have to be the image, n\mathbb{Z}, so we take the canonical map from p:\mathbb{Z}\to \mathbb{Z}/n\mathbb{Z} (which is surjective) to get ... 3 Choose a projective resolution P_\bullet \to B \to 0. Then \mathrm{Tor}_n(A,B) \stackrel{\mathrm{def}}{=} H_n(A \otimes P_\bullet). This is a quotient of a submodule of A \otimes P_n, so that it suffices to observe that A \otimes P_n is torsion, which is obvious (if ra=0 then r(a \otimes p)=0). 3 The general strategy here is to take the exact sequence$$ 0\to \ker \pi\to R^2\overset{\pi}\to I\to 0 $$and find a splitting for \pi. Then we'd have R^2\cong \ker\pi\oplus I, so I is projective as a direct summand of a free module. You have to implicitly use fractional ideals of R, so if you know what that is do some Googling, but if you don't then ... 3 Ok, so for p\mid n first do as @rschwieb suggests and prove that you can write$$ U = \langle v_1 - v_2 , \dots , v_1 - v_n\rangle. $$That is, you have to convince yourself that \{v_1 - v_2 , \dots , v_1 - v_n\} is a basis. That is, you want to show that this set spans all of U and that the set is linearly independent. The linear independence should ... 2 Hint for hands-on approach: Show that it's enough to take s = 1, i.e., given an element of the form n/1 \in \ker g, show that n/1 \in \text{im } f_m, and then show that this implies that \ker g \subseteq \text{im } f_m (for this last part, notice that n/s = (n/1) \cdot (1/s)). High-brow approach: localization is an exact functor. This implies ... 2 Assume that A \neq 0. Since pA=0 we can consider it as a vector space over F_p, and by taking the G-submodule generated by some nonzero element, we can assume that it is finite dimensional. In particular an action of G is just a homomorphism \varphi:G \to GL_m(F_p). Suppose first that G is cyclic generated by g. Since g^{p^m}=e where ... 2 In general, if T is any K[G]-module, then \hom_{K[G]}(K[X],T) identifies with \hom_G(X,U(T)), where U(-) is the forgetful functor from K[G]-modules to G-sets. If X = \coprod_i X_i as G-sets, then \hom_G(X,U(T)) \cong \prod_i \hom_G(X_i,U(T)). Therefore, we may reduce the question to the case that X is transitive, hence isomorphic to ... 2 Suppose U : \mathcal{C} \to \mathcal{D} is a comonadic functor. Then every object in \mathcal{C} is a subobject (indeed, regular subobject) of a cofree object. In particular this is true when \mathcal{C} is the category of comodules for a coalgebra. In more detail, let A be a commutative ring, let C be an A-coalgebra, and for an A-module M, ... 2 This is a particularly delicate use of the axiom of choice. Historically the principle of dependent choice, which is used exactly for these sort of inductive constructions, was overlooked by some of the greatest people that opposed to the axiom (e.g. Lebesgue). When we construct a sequence by induction, we can only assure that for every finite number n, ... 2 The determinant can be determined for every finitely generated projective module, because these are precisely the locally free modules of finite rank (which doesn't have to be constant, but it is locally constant, and on each constant piece we take the corresponding exterior power). It is additive on short exact sequences (see for example Daniel Murfet's ... 2 Answer 1. The closest thing to this construction I have seen is the Eilenberg-Watts theorem, which says that for any right exact functor F\colon R-mod\to Ab that commutes with arbitrary direct sums, we have a natural isomorphism F(-)\cong F(R)\otimes_R-, where F(R) is given its natural structure as a right R-module. The key observation to ... 2 An instructive example for the group case is$$ 0\to \mathbb Z/3\stackrel f\longrightarrow S_3\stackrel g\longrightarrow \mathbb Z/2\to 0$$where$f(1+\mathbb Z)=(1\ 2\ 3)$. Then we have$u\colon \mathbb Z/2\to S_3$,$1+\mathbb Z\mapsto (1\ 2)$is a right split, but there does not exist a left split (and the action of$\mathbb Z/2$on$\mathbb Z/3$is the ... 2 A good choice for a type of ring "sharing many properties with$\Bbb Z$" would be any principal ideal domain, and there is a simple classification theorem for finitely generated modules over such rings. You've picked an even narrower subclass of finitely generated$\Bbb Z$modules: that of the finitely generated free$\Bbb Z$-modules. Quite nicely, the ... 2 There is no way of defining the inclusion maps (what is$1 \in M$?). When$M,N$are free of ranks$n,m$, we have that also$M \otimes_R N$is free of rank$n \cdot m$. But the direct sum i.e. coproduct$M \oplus N$is free of rank$n+m$. So they are not isomorphic when$n,m>0$. In fact, there is a big difference between them. There is the following ... 2 If$R$is a local normal domain with$\dim R=2$, then every MCM is reflexive. First prove that$M$is torsion-free. This shows that$M_p$is free over$R_p$for any prime$p$with height$\le 1$. Next, if$p$is a prime of height$2$it's obvious that$M_p$satisfies Serre's condition$(S_2)$. In the end, use Proposition 1.4.1(b) from Bruns and ... 2 You actually didn't use that$M$is finitely generated. If$x_1,\dots,x_n$is a system of generators of$M$, then$x_i/1=0/1$in$S^{-1}M$, so there exists$s_i\in S$such that$s_ix_i=0$. Set$s=s_1\cdots s_n$and note that$sx_i=0$for all$i=1,\dots,n$. Then$sM=0$since every element of$M$is a linear combination of the generators. 2 No. For example, let$k$be a ground field,$S=M_2(k)$and$A=k(t)$, the field of rational functions in the variable$t$. Then$S\otimes S\otimes A\cong M_4(k(t))$and$S\otimes A\cong M_2(k(t))$. There two algebras are not isomorphic (not even as rings) For example, each of them is a direct sum of minimal left ideals, and the numbers of summands are ... 2 If$S$is a simple$R$-module, then$S \cong R/M$for a maximal (one-sided) ideal$M$. Hence every simple module$S$is isomorphic to a quotient module of$R/J(R)$for$J(R)$the Jacobson radical (which is the intersection of the maximal (one-sided) ideals$M$). In particular, if$R$is a nice ring where$R/J(R)\$ has a composition series, then the ...