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## Hot answers tagged modules

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

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

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 $P = \mathbf C_4 ... 6$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$does not have any binary multiplication operation, and hence it does not make any sense to call it a field (or a ring.) 6 If$A$is any ring, then$A$and$M_n(A)$have equivalent categories of modules, and usually$A$and$M_n(A)$are not isomorphic. This is the simplest example of a Morita equivalence. 6 For the non-free part: Take any two nonzero elements$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$\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 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: $$(\ast)\quad\quad 0\to P\to R\frac{\partial}{\partial x}\oplus R\frac{\partial}{\partial y}\oplus R\frac{\partial}{\partial z}\xrightarrow{\alpha := ... 5 This is true with added generality, that is in the non commutative world. So, assume \phi\colon R\to S is a ring homomorphism between not necessarily commutative rings. If P_{R} is a projective right R-module, then, if P_{R} is a direct summand of the free right module R^{(X)} for some set X, we get that$$ P_{R}\otimes_{R} S $$is a direct ... 5 In fact all submodules of free \mathbb{Z}-modules are projective, even free, so the experimentation you're doing won't succeed. (This is because \mathbb{Z} is a principal ideal domain.) Taking (x,y) a submodule of k[x,y] for k some field will work better. (x,y) is not projective, which we can show by showing it's not flat, since projective ... 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 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 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 It is not true that if A is not finitely generated, you can find \mathbb Z-independent elements. For example, \mathbb Q is a non finitely generated \mathbb Z-module, but, clearly, you cannot find two \mathbb Z-independent rationals \frac ab, \frac cd, for ac = bc\times \frac ab = ad\frac cd. When we prove that \mathcal O_K is finitely ... 5 Let, for example R = \mathbb Z/(6). Then, as \def\Z{\mathbb Z}\Z/(6)-modules, \Z/(6) \cong \Z/(2) \oplus \Z/(3). But a free \Z/(6)-module cannot have two or three elements only. 5 Yes, it is. If a\in R, a\ne 0, then aR is free (as a submodule of the free R-module R), in particular torsion-free, so a is a non-zero divisor. This shows that R is necessarily an integral domain. Now let I be a non-zero ideal of R. Since I is free (as a submodule of the free R-module R) it must have a basis with a single element ... 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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