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13

$\def\id{\operatorname{id}}$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 ...


9

Note that $K\otimes_{R} M$ is isomorphic as an $R$-module to the localization of $M$ at the multiplicative subset of $R$ consisting of all non-zero elements. (I think this point of view would be helpful if you're familiar with the notion of localization of modules.) The kernel of the map $M\to K\otimes_{R} M$ is equal to the $R$-submodule of $M$ consisting ...


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

You just have to work with the definitions (i.e. universal properties) in order to answer this question. It is not an extra convention or something like that (unfortunately, many mathematicians believe this). If $(E_i)_{i \in I}$ is a family of $R$-modules with underlying sets $|E_i|$, and $F$ is some $R$-module, a map $\prod_i |E_i| \to |F|$ is called ...


9

Intro a right R-module is equivalent to a left R-module only when R is commutative This statement has a pair of serious problems to resolve. Because "equivalent" is undefined, it's unclear what the statement means. (Discussed briefly below.) It uses "only when," but that is the wrong logical direction: it should use just when. There are in fact ...


9

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 ...


8

I'm assuming that $\mathbb Z_n$ means $\mathbb Z/n\mathbb Z$. I'm also assuming that "subproduct" means "subgroup of the product". If this is the case then yes, the subgroup generated by $(1, 1, \ldots) \in \prod_{n > 1}\mathbb Z_n$ is isomorphic to $\mathbb Z$. To see that this is the case note that we can always define a homomorphism out of $\mathbb ...


8

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

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.


7

I guess you need $M$ to be torsion-free. If $0 \ne m \in M$ and $0 \ne r \in R$ are such that $r m = 0$, then $$ 1 \otimes m = (r^{-1} r) \otimes m = r^{-1} \otimes r m = 0. $$


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

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

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 ...


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

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, ...


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

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 $$\mathbb{Z}/m\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/{(m,n)}\mathbb{Z}$$


6

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 $\Bbb{Z}/2\Bbb{Z}$.)


6

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

This is not true, in general. The modules $M$ which occur as direct summands of a free module are precisely the projective modules, which are not in general free. The simplest counter-example is probably the ideal generated by $2$ and $1+\sqrt{-5}$ in the ring $\mathbf Z[\sqrt{-5}]$, which is projective over $\mathbf{Z}[\sqrt{-5}]$ but not free. If you ...


6

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


6

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 ...


6

In general, a left module over a ring $R$ is equivalent to a right module over the ring $R^{\operatorname{op}}$, whose multiplication is the same as that in $R$, but in the opposite direction (i.e. $a\cdot_{\operatorname{op}} b = b \cdot a$). This is because in a left module, say $M$, we multiply by elements on the left, so, if by $\lambda_a: M\mapsto M$ we ...


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 ...


6

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 ...


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

Hint $1$: If $R$ is a domain, then $K=\mathrm{Frac}(R)$ is a flat $R$-module. Hint $2$: If $N_r=\{m\in M\mid r\cdot m=0\}$, then $N_r\otimes_{R}K=0$ for $r\ne 0$. Now for any $r\ne 0$, consider the exact sequence of $R$-modules: $$0\to N_r\to M\to rM\to 0$$ Can you finish the argument from here?



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