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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 $r(a \otimes p)=0$).

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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 ...

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One of the properties of the tensor product is that if $r \in \mathbb{Z}$ is an integer, and we're tensoring over $\mathbb{Z}$, then $$mr \otimes n = m \otimes rn$$ for all $m \in M$ and $n \in N$. Since $N = \mathbb{Z} / 2 \mathbb{Z}$ is the group with two elements, twice any element is the identity; that is, $2n = 0$ for every $n \in N$. As a result, we ...

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Exactness of localization can indeed make the problem simpler. The inclusions $B_i \subseteq \sum B_i$ localize to give $S^{-1}B_i \subseteq S^{-1}(\sum B_i)$ for every $i$, so $\sum S^{-1}B_i \subseteq S^{-1}(\sum B_i)$. Conversely, an element of $S^{-1}(\sum B_i)$ is of the form $\frac{b_1 + \ldots + b_n}{s}$, where $b_i \in B_i$. But $\frac{b_1 + \ldots + ... 1 This can't be proven as stated: it is false. Consider the field of two elements$F_2$, and look at the ring$R=F_2\times F_2$. The ideal$I=F_2\times\{0\}$is maximal, and if we set$I=M$we have a finitely generated$R$module$M$. Now$IM=M$, but$M\neq \{0\}$. Edit That was a huge omission indeed! If we add the additional condition that$R$is local, ... 1 Take some inspiration from linear algebra. If$R$were a field then$A=P(r_1,\cdots,r_n)$would be some affine subspace$-$a translate, or coset, of the vector subspace$U$comprised of points orthogonal to the vector$(r_1,\cdots,r_n)$. Forming a vector operation on this affine subspace would come down to choosing an origin$a\in A$: once you do that, you ... 1$\psi$could also precompose with any homomorphism$V \to V$, which is likely to ruin any chances of it being an$f_*$. For example, let$V = R^2$,$M = N = R$. Let$\psi = s^*$, where$s(\langle x,y \rangle) = \langle y,x \rangle$. Now let$\pi_1$be the first projection,$\pi_1(\langle x, y \rangle) = x$: I claim$\pi_1 \in \mathrm{Hom}_R(R^2, R)$. Then ... 1 I guess you defined a map$g\colon M/\mathop{\mathrm{Ker}} f\to\mathop{\mathrm{Im}} f$,$g([m]):=f(m)$. The first step is to prove that this is well-defined and linear. Surjectivity is obvious. For injectivity, assume$g([m])=g([m'])$. You have to show that this implies$[m]=[m']$, and as soon as you write down what that means, this is also obvious. 1 As you know that for any$N$you get an exact sequence for the$Hom$modules, try putting different$N$s. In particular$N=M'$,$N=M''$and$N=M$together with their identity homomorphisms. More details: My strategy was to use the following. Taking$N=M'$to obtain a map$r:M \to M'$such that$r \circ f =id_{M'}$. Together with the given$g:M \to ...

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Any abelian group can be split as $G=D\oplus M$, where $D$ is divisible and $M$ is reduced (that is, $\bigcap_{n>0}nM=\{0\}$). Of course, $G$ is artinian if and only if both $D$ and $M$ are artinian. Now, suppose $D$ is divisible artinian: then $D=t(D)\oplus D/t(D)$, where $t(D)$ is the torsion part of $D$. If $t(D)\ne D$, then $D$ contains an isomorphic ...

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