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(2) is incomplete. Recall that not every element in the tensor product is a pure tensor. In order to construct the algebra structure, I suggest the following (well-known) reformulation: If $A$ is some $R$-module, then an $R$-algebra structure on $A$ is the same as $R$-linear maps $\eta : R \to A$ and $\mu : A \otimes_R A \to A$ such that certain diagrams ...

2

Hint: every element is associate to a power of $7$.

2

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

2

No. If $V$ is any vector space with underlying set $S$, then it is a quotient of $K^{\oplus S}$, which has $S$ as a basis. But the existence of bases in arbitrary vector spaces is equivalent to AC (by a result by Andreas Blass).

1

If the isomorphism $M/L\cong M/N$ is canonical it is easy to show that $M=N$. But in general is not true see the example: $M=\Bbb R[X]$ and $N=\Bbb R_n[X]$ and $L=\Bbb R_{n+1}[X]$ (as $\Bbb R$ module $=$ $\Bbb R$ vector space), it is clear that $L$ and $N$ are not isomorph, but $M/N\cong (x^k, k\geq n+1)\cong (x^k,k\geq n+2)\cong M/L$. $(x^k, k\geq n+1)$ ...

1

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 ...

1

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 ...

1

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: ...

1

Hints (1) Let $v\in V$, $v\ne0$; then, for each $w\in V$ there is $f\in\operatorname{End}(V)$ such that $vf=w$. Thus $vE=V$, which means that the only proper $E$-submodule of $V$ is $\{0\}$. (2) If $B$ is a basis for $Ve$, then $B\cup\{v\}$ is linearly independent and so it can be completed to a basis for $V$.

1

What you're asking is if $S^{-1}R$ is a faithfully flat $R$-module. It isn't, take $R=\mathbb{Z}$, $A=B=C=\mathbb{Z}/2\mathbb{Z}$, and the maps $A\to B$ and $B\to C$ being the zero maps, and consider $S=\mathbb{Z}-\{0\}$.

1

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 ...

1

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 ...

1

When I google "absolutely irreducible module" the second hit I get is a passage in Lam's Exercises in classical ring theory which explains that an irreducible module $M$ over a $k$ algebra $R$ ($k$ a field) is called absolutely irreducible if $M\otimes_k K$ is irreducible over $R\otimes_k K$ for every extension field K of k. I would guess the FL part is a ...

1

What about the $\mathbb Z$-module $\oplus_{n\ge1}\mathbb Z/2^n\mathbb Z$?

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