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Well this is certainly wrong as long as you don't assume M to be finitely generated (just take an infinite dimensional vector space). If M is finitely generated this should be true, even without the assumption that supp(M) is finite (which will rather be a consequence). First note that this is obviously true if R is artinian,since M is a quotient of some ...


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The direct limit of $\Bbb Z$-modules: $$\Bbb Z/p\to \Bbb Z/p^2\to \Bbb Z/p^3\to\cdots$$ is not finitely generated as a $\Bbb Z$ module but every proper submodule is isomorphic to $\Bbb Z/p^k$ for some $k$. Edit: For more information, please see this wiki page.


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This is not true. For example, $\{1\}$ and $\{2,3\}$ are minimal generating sets of the $\mathbb{Z}$-module $\mathbb{Z}$.


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Based on what I remember from playing such games twenty-five years ago, The $3^3$ version is a guaranteed win for the first player, by going in the middle square. There are so many lines through it that the first player can always force moves. After the 2nd player places his irrelevant O, the first player chooses a plane through the middle X that doesn't ...


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$4^3$ ("Qubic") is a win for the first player. According to this link, it was first proved by Oren Patashnik in 1980. The proof is complicated. It took 12 years for this proof to be converted into a practical computer algorithm; I was present at the 1992 Computer Olympiad where the program of Victor Allis and Patrick Schoo romped to victory.


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I assume (as rschwieb points out) that the question concerns the category of $\mathbb{Z}$-modules. 1) $\mathbb{Z}_{p^\infty}$ is the injective hull in the case when $n=p^k$ as well. Basically because we have $\mathbb{Z}_p\subseteq \mathbb{Z}_{p^k} \subseteq \mathbb{Z}_{p^\infty} $ and the extension $\mathbb{Z}_{p} \subseteq \mathbb{Z}_{p^\infty}$ is ...


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The same idea for $\mathbb{Z}$ extends to a polynomial ring over a field: for the ideal $(x) \subseteq k[x]$, $\{x\}$ and $\{x^2, x + x^2\}$ are both minimal generating sets. In general, given a generating set for an ideal $I$, say $I = (a_1, \ldots, a_n)$, one cannot conclude that $I$ can be generated by a proper subset of the $a_i$, even if $I$ is known ...


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Apologies in advance for this answer. The problem with it is that it is too advanced, and also relies on a lemma that is very similar to your question. I will continue to seek a more elementary answer. Lemma: Every left $R$ module over a left Artinian ring $R$ has a projective cover. Lemma: Every nonzero projective module has a maximal submodule. Lemma: ...


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For every $i$ we have $C_i=B_i/A$. Taking direct limit, when $B'$ denotes the limit of $(B_i)$, we get $$C=B'/A,$$ and $B'$ is a submodule of $B$. But we know that $C=B/A$, hence $B'=B$.


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Given any ascending sequence of sets $X_1\subset X_2\subset X_3\cdots\subset X$ with $\bigcup_{n\ge1}X_n=X$ and any collection of functions $g_i:X_i\to Y$ such that $g_i|_{X_{i-1}}=g_{i-1}$, we can form the map $g:X\to Y$ defined by the relation $g(x)=g_i(x)$ whenever $x\in X_i$. Each function $g_i$ is a bigger and bigger "glimpse" of $g$. One can check ...


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Let $N_n=\sum_{i=1}^{k}c_iR$; then $D$ is the union of the increasing family of submodules $N_n$ and as such it is its direct limit with inclusions as transition maps. If you consider the restriction $h_n$ of $g_n$ to $N_n$, the given condition translates into the fact that $h_n\colon N_n\to B$ is a family of morphisms compatible with the inclusion maps, so ...



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