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## Hot answers tagged modules

7

$\mathbb{Z}$-modules are precisely abelian groups. As every ring is an abelian group, it is a $\mathbb{Z}$-module. It is entirely possible to be a module over more than one ring. For example, if $M$ is an $R$-module then it is also an $S$-module for any subring $S$ of $R$ (you seem to be interested in the case where $M = R$). Another example is given by ...

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$\text{Ext}^1(-, -)$ sends direct sums in the first variable to direct products, and as mentioned in the comments, as an abstract abelian group $$S^1 \cong \left( \bigoplus_X \mathbb{Q} \right) \oplus \mathbb{Q}/\mathbb{Z}$$ where $X$ is uncountable. So it suffices to compute $\text{Ext}^1(\mathbb{Q}, \mathbb{Z})$ and $\text{Ext}^1(\mathbb{Q}/\mathbb{Z}, ... 2 Assuming we are additionally given that$0\le m<N$, then, yes, it is possible to reveal$m$: Just try the candidates one by one and check if$(m+r)^e\bmod B$turns out to be$C$. Mathematically, this solves the problem. Also, we have found an explicit algorithm, albeit with exponential running time ($O(N)$, which is linear in$N$, but exponential in the ... 2 Let$A = \mathbb Z$and$M = \mathbb Z/2\mathbb Z$. 1 I think the following should work, but take it with a grain of salt. Let$\mathbb Z(p^\infty)$denote the$p$-Prüfer group, that is, the injective envelope of$\mathbb Z(p)$. The module$I=\prod_p \mathbb Z(p^\infty)$is an injective module containing your module$M=\prod_p \mathbb Z(p)$as a submodule. Now consider the module$$E = M + \bigoplus_p \mathbb ... 1 Yes, unless$V$is trivial. I'll help you unpack the definitions, but I'll leave the meat of the problem to you. Observation. Let$V$denote a$k$-module. Then$V$is a simple$\text{End}_k V$-module iff:$V$has at least two distinct$k$-submodules, Every non-trivial$k$-submodule of$V$that is closed under the action of$\text{End}_k V$... 1 Hint: The answer is yes (as is often the case for finite dimensional vector spaces). Note that for any$v \in V \setminus \{0\}$, we may select maps$T_1,\dots,T_n \in A$so that$\{T_j(v)\}$forms a basis (or a spanning set, if you prefer). 1 Yes, the inclusion of a factor into a direct product$M \to M \oplus N$defined by$m \mapsto (m, 0)$is always an injective homomorphism no matter what modules you choose for$M$and$N$. In particular, you can inject$M \to M \oplus R[G]^n$not just for some$n$, but in fact for any$n$. If$M$and$N$are$G$-modules then it would be a good exercise for ... 1 Yes, they are.$A$is a simple Artinian ring. Therefore it has, up to isomorphism, a unique simple right$A$-module$S$and every$A$-module is isomorphic to a direct sum$S^{(I)}$for some index set$I$. Moreover, the module$S$has some finite dimension$n$over$k$. (Explicitly,$A$is isomorphic to$M_r(D)$for a division ring$D$which is ... 1 Take$A = M = \mathbb{Z}_{(p)}$, localized at a prime$p$, and$N = \mathbb{Q}$. Take$f: M \to N$to be the inclusion$\mathbb{Z}_{(p)} \hookrightarrow \mathbb{Q}$. Take the ideal$\mathfrak{a}$to be the maximal ideal of$A$. In fact take any local ring$A$and take$N\$ to be its field of fractions. Incidentally (amusing) together with your result ...

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Yes, it can be confusing talking about modules one moment and then talking about representations the next. In your case there isn't much difference. In fact, the confusion shouldn't confuse you. A Lie algebra module is a vector space such that ... A Lie algebra module is irreducible is there are no invariant (non trivial) proper submodules. Since the ...

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