Tag Info

Hot answers tagged

15

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


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

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

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

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

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


5

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. ...



Only top voted, non community-wiki answers of a minimum length are eligible