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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 ...


3

$\text{Ext}^1(-, -)$ 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 ...


1

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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