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4

If $m = n$, the $0 \to \mathbb{Z}/n\mathbb{Z} \xrightarrow{\operatorname{id}} \mathbb{Z}/n\mathbb{Z} \to 0$ is a free resolution. If $m < n$, it will be useful to write $n = km$, where $k > 1$. The first step of the resolution looks like $F_0 \xrightarrow{\epsilon} \mathbb{Z}/m\mathbb{Z} \to 0$ for some free $\mathbb{Z}/n\mathbb{Z}$-module $F_0$. ...


3

Take $R=\mathbb{Z}$, $M=\mathbb{Q}\oplus \mathbb{Z}/2\mathbb{Z}$ and $N= \mathbb{Z}/2\mathbb{Z}$, with inclusion $R\to M:n\mapsto (n,\overline{0})$. Then, in $M\otimes_\mathbb{Z} N$, we have that $$ (1,\overline{0})\otimes \overline{1} = (\frac{1}{2},\overline{0})\otimes 2\cdot\overline{1} = (\frac{1}{2},\overline{0})\otimes \overline{0} = 0.$$ Therefore ...


3

If $x$ is a non-unit, then so is $ux$ for any $u$. This is true in any commutative ring. Otherwise we'd have $$(ux)^{-1}ux=((ux)^{-1}u)x=1$$ This contradicts the assumption that $x$ is not a unit. If the ring is noncommutative we can still obtain the result that $ux$ is not a unit from the fact that the nonunits form an ideal in a local ring.


3

For the second question, the finite rings are precisely those for which every finitely generated module has finitely many submodules. For a ring $R$ and $r\in R$, let $M_r$ be the submodule of $R\oplus R$ generated by $(1,r)$. Then $(1,r)$ is the only element of $M_r$ whose first coordinate is $1$, and so $M_r\neq M_s$ for $r\neq s$, and so if $R$ is ...


2

$1-ux$ can not be in the maximal ideal as $x$ is and hence $ux$ is. And since we have a local ring, any element not in the maximal ideal must be a unit.


2

The symbol $\bar r$ denotes the equivalence class of $r$ in $\text{Ann}_R(M)$. The point here is that you want to define the action of the quotient on the module. This is well-defined because if $\bar{r_1}=\bar{r_2}$, this means that $r_1-r_2\in\text{Ann}_R(M)$, so for any $m$ we have $(r_1-r_2)m=0$, i.e. $r_1m=r_2m$.


2

The matrix $A$ corresponds to an endomorphism $\varphi_A$ of the $K$-vector space $K^n$. Then, by Fitting's Lemma, there is a decomposition $K^n=V_1\oplus V_2$ such that $\varphi_A$ restricted to $V_1$ is nilpotent, and $\varphi_A$ restricted to $V_2$ is invertible, that is, there is a $K$-basis of $K^n$ such that the matrix associated to $\varphi_A$ in this ...


2

A submodule that is generated by two elements. (This isn't exactly a standard or well-known term, but in that context there's nothing else it could sensibly mean.)


1

I want to add a categorical phrasing of this problem. Consider the quotient category $R\text{-Mod}/\langle N\rangle_\oplus$ of the category $R\text{-Mod}$ at the additive closure $\langle N\rangle_\oplus$ of $N$: Its objects are again the $R$-modules, but for two $R$-modules $A,B$ we define $\text{Hom}_{R\text{-Mod}/\langle N\rangle_\oplus}(A,B)$ to be the ...


1

Since $I$ is a principal ideal, it is isomorphic to $R$ as an $R$-module. Hence $$I \otimes_R I \cong R \otimes_R R \cong R$$ which is obviously torsionfree.


1

Note that an $R$-module is the same as an abelian group $M$ together with a ring homomorphism $R\longrightarrow\operatorname{End}_\mathbb{Z}(M)$. The kernel of this homomorphism is precisely $\operatorname{Ann}_R(M)$ (actually this is how one should define the annihilator, so one does not have to check it is an ideal). Now the homomorphism theorem yields a ...


1

When a projection map $R\to S$ is in play, the notation $\bar{r}$ often denotes the image of $r$ in $S$ under the projection map. E.g. the elements of $\Bbb Z/n\Bbb Z$ are the residues $\overline{0},\overline{1},\overline{2},\cdots,\overline{n-1}$. A priori one would have to wonder if $\overline{r}m:=rm$ is well-defined, since $\overline{r}=\overline{s}$ is ...


1

It seems like your best bet for both questions will be to consider finite rings and their finitely generated modules. These at least will be closed under products.



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