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11

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 $M\otimes M\cong(M_1\otimes M_2)\oplus (M_1\otimes ... 9 There is a notion of injective cogenerator of the category$\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$. As it turns out, injective cogenerators are precisely the modules satisfying the property you ... 9 Let$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 ... 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 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$C=0$... 8 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 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 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 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 ... 6 Given a finitely generated module$M$over a noetherian ring$A$, there exists a filtration of$M$by submodules$M=M_0\supset M_1\cdots \supset M_n=0$such that$M_i/M_{i+1}\cong A/\mathfrak p_i$for some prime ideals$\mathfrak p_i\subset A$(Bourbaki, Commutative Algebra, Chapter IV, §1, Theorem 1, page 261) Now$M$has finite length iff all the ... 6 No, if the ring is commutative, yes, if the ring is noncommutative. The easiest counterexample is the endomorphism ring$R$of$V=F^{(\omega)}$, where$F$is a field and$F^{(\omega)}$denotes a direct sum of countably many copies of$F$(as vector space). Let homomorphisms act on the left, so$V$becomes a left$R$-module. Then$R\cong R^2$as left ... 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

If $\mathfrak a$ is an ideal in a ring $A$ and $M$ is an $A$-module, then the map $$\begin{array}{ccc} M\otimes_A A/\mathfrak a &\to& M/\mathfrak aM \\ m\otimes (a+\mathfrak a)&\mapsto& ma+\mathfrak a M \end{array}$$ is an isomorphism, which we will call the quotienting isomorphism. Proving this map is an isomorphism is exactly Exercise 2.2 ...

6

$M$ is module isomorphic to $R/I$, but that doesn't mean it has suddenly become a field. Now, you can treat the module $M$ as a ring by transferring the structure of $R/I$ back to $M$ through the bijection, but until you do that, $M$ does not have any binary multiplication operation, and hence it does not make any sense to call it a field (or a ring.)

6

For the non-free part: Take any two nonzero elements $x, y ∈ ℚ$ and show they satisfy $λx + μy = 0$ for some nonzero $λ, μ ∈ ℤ$, hence any two elements are linearly dependent. Thus, since $ℚ$ is not cyclic, it cannot have a basis. For the non-finitely-generated part: If $ℚ$ was finitely generated then without loss of generality (by finding the common ...

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

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

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

6

Yes, there is such a finite subset. Because $X$ is a generating set for the module, given any $m \in M$ there exist $x_1,\dots, x_k \in X$ and scalars $r_1,\dots,r_k \in R$ such that $m = r_1 x_1 + \cdots + r_k x_k$. Now suppose $m_1,\dots, m_n \in M$ is a finite generating set. Then for each $m_i$ there exist $x^{(i)}_1, \dots, x^{(i)}_{k(i)} \in X$ such ...

6

In general , there is no one big theorem about monoidal closed categories that justifies or illuminates them. The internal hom does what it says on the tin: it lets you represent your homsets as objects of your category. This allows you to state questions and theorems that simply wouldn't make sense in a non-closed setting. My advice would simply be to keep ...

6

Your idea is right: If $x$ has finite order and your group homomorphism is $\phi$, then $\phi(x)$ has an order that is less than or equal to that of $x$. This is true regardless of whether $\phi$ is 1-1 or not. Thus you can't map onto elements of infinite order. When I read the original post, I misread that the indexing was over $p$! But of course, if you ...

5

A prominent counter-example is the following: Take $R := {\mathbb R}[x,y,z]/(x^2+y^2+z^2-1)$, the ring of real-valued polynomial functions on the $2$-sphere, and consider the following short exact sequence: $$(\ast)\quad\quad 0\to P\to R\frac{\partial}{\partial x}\oplus R\frac{\partial}{\partial y}\oplus R\frac{\partial}{\partial z}\xrightarrow{\alpha := ... 5 \mathbb R  over \mathbb Q is a vector space with dimension 2^{\mathbb N}. and also \mathbb R^n  over \mathbb Q is a vector space with dimension 2^{\mathbb N}. So the additive group of \mathbb R is isomorphic with additive group of \mathbb R^n. for example consider f:\mathbb R\rightarrow \mathbb R^n be such isomorphism , now define new ... 5 This is true with added generality, that is in the non commutative world. So, assume \phi\colon R\to S is a ring homomorphism between not necessarily commutative rings. If P_{R} is a projective right R-module, then, if P_{R} is a direct summand of the free right module R^{(X)} for some set X, we get that$$ P_{R}\otimes_{R} S $$is a direct ... 5 Freyd proved in his paper Concrenteness (1973) that an abelian category which admits a faithful exact functor to \mathsf{Ab} is well-powered; actually also the converse. There are abelian categories which are not well-powered, see MO/93853. Another example is mentioned in the foreword of the tac reprint of Freyd's book Abelian categories (2003) with ... 5 The answer is negative since A\subset B flat and B regular implies A regular; see Bruns and Herzog, Theorem 2.2.12. But in this case A\simeq k[a,b,c]/(ac-b^2), so A is not regular. Edit. A simpler approach: let I=(x^2,xy) and A/I\to A/I be the multiplication by y^2. Since A/I\simeq k[y^2] this is injective, but on A/I\otimes_AB\to ... 5 Let R be ring for which F is a free module. Then we have isomorphisms R^n \rightarrow F and F \rightarrow R^{n+1} which gives us an isomorphism R^n \rightarrow R^{n+1}. Thus for any m \geq n we have an isomorphism R^m \cong R^n \oplus R^{m-n} \cong R^{n+1} \oplus R^{m-n} \cong R^{m+1}. Composing these isomorphisms we get an isomorphism R^n ... 5 For groups, you may consider$$\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 ...

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