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$\def\id{\operatorname{id}}$Suppose $M\otimes N$ is isomorphic to $R^n$. Pick a basis $\{x_1,\dots,x_n\}$ of $M\otimes N$, with $x_i=\sum_{j=1}^{r_i}m_{i,j}\otimes n_{i,j}$ for each $i\in\{1,\dots,n\}$. Let $r=r_1+\cdots+r_n$, let $\{e_{i,j}:1\leq i\leq n, 1\leq j\leq r_i\}$ be a basis of $R^r$, and consider the map $f:R^r\to M$ which maps $e_{i,j}$ to ...

10

I posted this example earlier on MathOverflow. Let $R=k[x,y]$ for a field $k$, and let $$M=\frac{k[x,y,y^{-1}]}{k[x,y]}\oplus\frac{k[x,x^{-1},y]}{k[x,y]}.$$ Then $M$ is a direct sum $M_1\oplus M_2$ of two modules for which $M_1\otimes M_1=0$, $M_2\otimes M_2=0$, but $M_1\otimes M_2\neq0$, so that $M\otimes M\cong(M_1\otimes M_2)\oplus (M_1\otimes ... 9 Let$M, N$be$R-$modules. Then the following holds. If$M$and$N$is flat, then so is$M\otimes_{R}N$: see related question here. If$M$and$N$are projective, then so is$M\otimes_{R} N$. Indeed, writing$M\oplus M'=F,\ N\oplus N'=F'$, for free$R-$modules$F,\ F'$, one has that $$F'':=F\otimes_{R}F'$$ is free (tensor product of free modules) and ... 9 There is a notion of injective cogenerator of the category$\mathrm{Mod}$-$R$: An injective cogenerator is an injective$R$-module$C$such that for every nonzero$R$-module$M$, there is a nonzero homomorphism$M\rightarrow C$, i.e.$\mathrm{Hom}_R(M,C)\neq 0$. As it turns out, injective cogenerators are precisely the modules satisfying the property you ... 8 Let$0_M$and${\rm id}_M$denote the zero map and identity map on an$A$-module$M$. We have $$M=0\iff 0_M={\rm id}_M.$$ Since$F$is a functor,$F({\rm id}_M)={\rm id}_{FM}$. Since it's also additive,$F(0_M)=0_{FM}$. 8 If$R$is a ring,$M$is a right$R$-module and$N$is a left$R$-module, then the functor of balanced maps$F : \mathsf{Ab} \to \mathsf{Set}, A \mapsto \{\beta: |M| \times |N| \to |A| \text{ balanced}\}$satisfies the assumptions of Freyd's criterion for representability: It is easy to check that it preserves limits. For the solution set condition, let ... 8 Starting from all of$\Bbb Q/\Bbb Z$, you can successively forbid all fractions whose denominator contains a factor$2$, then factors$3$,$5$, and so forth. This gives an infinite decreasing chain, and the module is not Artinian. 8 Consider the exact sequence$M \to N \to C \to 0$, where$C$is the cokernel of$v$. Then for any$A$-module$P$we have an exact sequence$\DeclareMathOperator{\h}{Hom} 0 \to \h(C,P) \to \h(N,P) \to \h(M,P)$. By assumption the last map is injective, so we have$\h(C,P) = 0$. Since this holds for all$P$it holds especially for$P = C$, so we conclude$C=0$... 7 This is true if$R$is commutative. Otherwise, say that you are dealing with left$R$-modules, for instance. If you attempt to define multiplication by$r$by$(rh)(m) = rh(m)$for any homomorphism$h \colon M \to N$, then you run into the problem that the mapping$rh$may not be$R$-linear. For example, let$h \colon R \to R$be the identity map. Then ... 7 Let$Q$be a finitely generated injective$R$-module. Suppose that$R$is not a field and let$\mathfrak m$be a maximal ideal of$R$. Since$\mathfrak m\ne 0$there is$a\in\mathfrak m$,$a\ne 0$. Then we have$aQ=Q$(injective modules are divisible), and therefore$\mathfrak mQ=Q$. Localizing we get$\mathfrak mQ_{\mathfrak m}=Q_{\mathfrak m}$, and by ... 7 Modules, rings:$A=\Bbb Q^{\oplus\omega}$,$B=A\oplus\Bbb Z$. To see$A\not\cong B$consider additive divisibility. Fields: For every char$p\ge0$and cardinal$\kappa\ge{\frak c}$there exists a unique algebraically closed field of characteristic$p$and cardinality$\kappa$. If$F$is an infinite field then$|\overline{F}|=|F|$. Let$F$be an ... 7 Hint: There is a very easy free resolution of$\mathbb{Z}/p\mathbb{Z}$: $$0\to\mathbb{Z}\xrightarrow{\times p}\mathbb{Z}\to\mathbb{Z}/p\mathbb{Z}\to 0$$ Combing this with the fact that $$\mathbb{Z}/m\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/{(m,n)}\mathbb{Z}$$ 7 Algebra objects or monoid objects can be defined in any monoidal category. When$R$is a commutative ring, then the category of left$R$-modules has a monoidal structure given by$\otimes_R$, and algebras in that category coincide with$R$-algebras. But the category of left$R$-modules has no "natural" monoidal structure when$R$is not commutative - this is ... 7 Here's a counterexample. Let$R = k[x]$for$k$a field. Then $$k[x]/(x - a) \otimes_{k[x]} k[x]/(x - b) \cong 0$$ for$a \neq b$because the supports of the two modules are disjoint, but $$k[x]/(x - a) \otimes_{\mathbb{Z}} k[x]/(x - b) \cong k \otimes_{\mathbb{Z}} k$$ can be quite large, e.g. if$k = \mathbb{Q}$then it is$\mathbb{Q}$. The mistake ... 7 From the definitions, there's no reason to expect projective/injective objects of a subcategory to be projective/injective in the ambient category. If$P, B \in \mathcal{A}$, then any map$P \to B$factors through any epi$A \twoheadrightarrow B$with$A \in \mathcal{A}$, but if$A, B$are not in$\mathcal{A}$, there is no reason to expect a lift. For a ... 7 Here is the case for abelian groups: Theorem: If$M$is an abelian group with$M \otimes M \otimes M = 0$, then$M$is a divisible torsion abelian group and$M \otimes M = 0$. The proof is pretty standard abelian group theory. Basic subgroups are probably not well known outside of that theory (at least I never learned about them over rings that weren't ... 7 Yes, it may even happen that$N'=0$, i.e. that$M$is isomorphic to$M/N$but$N \neq 0$. Take$M=R \oplus R \oplus \dotsc$and$N = R \oplus 0 \oplus 0 \oplus \dotsc$. 7 It holds when$R$is a PID (here injective$\Leftrightarrow$divisible). It is not true in general. The following papers study and characterize this property of$R$that tensor products of injectives are injective. Ishikawa, Takeshi. "On injective modules and flat modules." Journal of the Mathematical Society of Japan 17.3 (1965): 291-296. Enochs, ... 6 Here's one (class of) example(s). Let us restrict our attention to$\mathbb Z$-modules, i.e. abelian groups. Note that two abelian groups of order$p$are always isomorphic, thus it is sufficient to find a group$P$of order$p^i$(for any$i$) with two subgroups$A$and$B$of order$p^{i-1}$such that$A\not\cong B$. For instance, if$P = \mathbf C_4 ...

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The simplest case is when the group is equipped with a topology with respect to which it is compact (Hausdorff), so the proof of Maschke's theorem still works and the Peter-Weyl theorem is available. In particular, the representation theory of compact Lie groups is very well understood. The representation theory of noncompact Lie groups is still a major area ...

6

To expand on Zhen Lin's comment: Let $R$ be an integral domain, $A$ a finitely presented $R$-algebra, $Q$ the fraction field of $R$. Then: to say that $A$ is smooth over $R$ can be reinterpreted geometrically as saying that the morphism of affine schemes $\operatorname{Spec} A \rightarrow \operatorname{Spec }R$ has nonsingular schemes as fibres over ...

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The set $\{1\}$ isn't independent, because $2(1)=1+1=0$. (The $2$ here lives in $\Bbb{Z}$, so the "multiplication" is the module action; all $1$s and $0$s live in $\Bbb{Z}/2\Bbb{Z}$.)

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Any natural definition of $R$-action works only, when $R$ is commutative. For example, if you try to define $$(rf)(m)=r(f(m))$$ for all $r\in R$, $f\in Hom_R(M,N)$, $m\in M$, then the mapping $rf$ fails to be homomorphism of $R$-modules in general. If $s\in S$ is such that $sr=rs$, then $$(rf)(sm)=r(f(sm))=r(sf(m))=(rs)(f(m))\neq s((rf)(m))$$ in general. ...

6

Presumably, you'll want to define $\psi=r.\phi$ by $$\psi(m)=r.\phi(m)$$ (where $r\in R, \phi\in\mathrm{Hom}_R(M,N)$ and $m\in M$.) However, this map, which is a morphism of abelian groups, need not be $R$-linear when $R$ isn't commutative : $R$-linearity would imply that for all $r'\in R$ (and all $m\in M$) $\psi(r'.m)=r'.\psi(m)$, i.e., by $R$-linearity of ...

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Hint: An endomorphism of $\mathbb Q$ is determined by the image of $1$. Here are some details:

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Let $R$ be ring for which $F$ is a free module. Then we have isomorphisms $R^n \rightarrow F$ and $F \rightarrow R^{n+1}$ which gives us an isomorphism $R^n \rightarrow R^{n+1}$. Thus for any $m \geq n$ we have an isomorphism $R^m \cong R^n \oplus R^{m-n} \cong R^{n+1} \oplus R^{m-n} \cong R^{m+1}$. Composing these isomorphisms we get an isomorphism $R^n ... 5 The Tate module of a product$A\times B$of abelian varieties over$k$is naturally isomorphic, as a$G_k$-module, to$T_\ell A\times T_\ell B$. This follows directly from the universal property of a direct product: $$(A\times B)(k^\text{sep}) \simeq A(k^\text{sep})\times B(k^\text{sep}) \text{.}$$ 5 For groups, you may consider $$\mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots \hookrightarrow \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots \hookrightarrow \mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \cdots.$$ They are not isomorphic: in the second group, any element of order two is divisible by 2. Another example, but ... 5 Take just a numerable set of entries, and forget the other for the moment. Now consider the following fact: there's a continuos chain in the part of$\mathbb{N}$(this is clear once you consider$(-\infty,x) \bigcap \mathbb{Q}, x \in \mathbb{R}$). So call F this family. To each A in F, you associates$1_A$(that is you make a bjection with the numerable ... 5 The equality$\mathfrak a\left(\bigcap_{n\ge1}\mathfrak a^nM\right)=\bigcap_{n\ge2}\mathfrak a^nM$assumes that multiplication by ideals and intersection of submodules commute. This is unfortunately not the case. In general we only have$\mathfrak{a} \left(\bigcap_{n\ge1} M_n\right) \subset \left(\bigcap_{n\ge1}\mathfrak{a} M_n\right)\$, but not equality. ...

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