# Tag Info

1

Apologies in advance for this answer. The problem with it is that it is too advanced, and also relies on a lemma that is very similar to your question. I will continue to seek a more elementary answer. Lemma: Every left $R$ module over a left Artinian ring $R$ has a projective cover. Lemma: Every nonzero projective module has a maximal submodule. Lemma: ...

1

The same idea for $\mathbb{Z}$ extends to a polynomial ring over a field: for the ideal $(x) \subseteq k[x]$, $\{x\}$ and $\{x^2, x + x^2\}$ are both minimal generating sets. In general, given a generating set for an ideal $I$, say $I = (a_1, \ldots, a_n)$, one cannot conclude that $I$ can be generated by a proper subset of the $a_i$, even if $I$ is known ...

1

I assume (as rschwieb points out) that the question concerns the category of $\mathbb{Z}$-modules. 1) $\mathbb{Z}_{p^\infty}$ is the injective hull in the case when $n=p^k$ as well. Basically because we have $\mathbb{Z}_p\subseteq \mathbb{Z}_{p^k} \subseteq \mathbb{Z}_{p^\infty}$ and the extension $\mathbb{Z}_{p} \subseteq \mathbb{Z}_{p^\infty}$ is ...

2

This is not true. For example, $\{1\}$ and $\{2,3\}$ are minimal generating sets of the $\mathbb{Z}$-module $\mathbb{Z}$.

1

Let $N_n=\sum_{i=1}^{k}c_iR$; then $D$ is the union of the increasing family of submodules $N_n$ and as such it is its direct limit with inclusions as transition maps. If you consider the restriction $h_n$ of $g_n$ to $N_n$, the given condition translates into the fact that $h_n\colon N_n\to B$ is a family of morphisms compatible with the inclusion maps, so ...

1

Given any ascending sequence of sets $X_1\subset X_2\subset X_3\cdots\subset X$ with $\bigcup_{n\ge1}X_n=X$ and any collection of functions $g_i:X_i\to Y$ such that $g_i|_{X_{i-1}}=g_{i-1}$, we can form the map $g:X\to Y$ defined by the relation $g(x)=g_i(x)$ whenever $x\in X_i$. Each function $g_i$ is a bigger and bigger "glimpse" of $g$. One can check ...

1

What you're asking is if $S^{-1}R$ is a faithfully flat $R$-module. It isn't, take $R=\mathbb{Z}$, $A=B=C=\mathbb{Z}/2\mathbb{Z}$, and the maps $A\to B$ and $B\to C$ being the zero maps, and consider $S=\mathbb{Z}-\{0\}$.

1

When I google "absolutely irreducible module" the second hit I get is a passage in Lam's Exercises in classical ring theory which explains that an irreducible module $M$ over a $k$ algebra $R$ ($k$ a field) is called absolutely irreducible if $M\otimes_k K$ is irreducible over $R\otimes_k K$ for every extension field K of k. I would guess the FL part is a ...

0

Expanding my former comment, first we create a cartesian product of ranges: gap> list1:=Cartesian([3..4],[7..8]); [ [ 3, 7 ], [ 3, 8 ], [ 4, 7 ], [ 4, 8 ] ] Then we may enumerate all non-empty combinations of k elements of list1 as follows: gap> for k in [1..Length(list1)] do > iter:=IteratorOfCombinations(list1,k); > for c in iter ...

2

Hint: every element is associate to a power of $7$.

1

If the isomorphism $M/L\cong M/N$ is canonical it is easy to show that $M=N$. But in general is not true see the example: $M=\Bbb R[X]$ and $N=\Bbb R_n[X]$ and $L=\Bbb R_{n+1}[X]$ (as $\Bbb R$ module $=$ $\Bbb R$ vector space), it is clear that $L$ and $N$ are not isomorph, but $M/N\cong (x^k, k\geq n+1)\cong (x^k,k\geq n+2)\cong M/L$. $(x^k, k\geq n+1)$ ...

2

No. If $V$ is any vector space with underlying set $S$, then it is a quotient of $K^{\oplus S}$, which has $S$ as a basis. But the existence of bases in arbitrary vector spaces is equivalent to AC (by a result by Andreas Blass).

1

Hints (1) Let $v\in V$, $v\ne0$; then, for each $w\in V$ there is $f\in\operatorname{End}(V)$ such that $vf=w$. Thus $vE=V$, which means that the only proper $E$-submodule of $V$ is $\{0\}$. (2) If $B$ is a basis for $Ve$, then $B\cup\{v\}$ is linearly independent and so it can be completed to a basis for $V$.

3

(2) is incomplete. Recall that not every element in the tensor product is a pure tensor. In order to construct the algebra structure, I suggest the following (well-known) reformulation: If $A$ is some $R$-module, then an $R$-algebra structure on $A$ is the same as $R$-linear maps $\eta : R \to A$ and $\mu : A \otimes_R A \to A$ such that certain diagrams ...

0

Let us informally write $$A/A'=\{a+a',a\in A,a'\in A'\},B/B'=\{b+b',b\in B,b'\in B'\}$$ where $a,b$ are class representatives and $a',b'$ can be any elements in $A', B'$. We now see the tensor product $$(a+a')\otimes (b+b')=a\otimes b+a'\otimes b+a\otimes b'+a'\otimes b'$$ But $a'\otimes b, a\otimes b',a'\otimes b'\equiv 0$ in $A/A'\otimes B/B'$. ...

1

$\Bbb Z$ is a free, hence projective, hence flat $\Bbb Z$ module. $\Bbb Z$ modules are flat iff they are torsion-free, and $\Bbb Q_\Bbb Z$ is torsion-free. (And this is another reason $\Bbb Z_\Bbb Z$ is flat.) $\Bbb Z$ modules are injective iff they are divisible, so to produce a nonflat, noninjective module, it suffices to think of a nontorsion-free module ...

0

What you have stated here is that the tensor product $(\_ \otimes \_)$ is a covariant bifunctor from the category of bimodules to itself. Thus it makes the category into a Monoidal Category. If you are worried about isomorphisms, or the functor being fully faithful: as you know, this functor need not be right exact in either variable, in general case. But ...

0

Couldn't there be a principal ideal domain $R$ and $c \in R$ such that for every $R$-module $A$, there is no $a \in A$ having $\mathcal{O}_a = (c)$? No, because $A:=R/(c)$ is cyclic (generated by $\bar{1}$) and ${\cal O}_{\bar{1}}=(c)$. Check for yourself! (A scalar element $r\in R$ annihilates $1+(c)$ in $R/(c)$ means that .... [write stuff down, ...

2

Hint: If we let $D=N$, there is a special homomorphism in $\mathrm{Hom}_R(N,N)$, namely the identity morphism on $N$. Since the sequence of Hom sets is exact, there must be some $R$-module map $N\to M$ which maps to the identity morphism in $\mathrm{Hom}_R(N,N)$. What does this mean with respect to the map $M\to N$ in the original sequence?

2

If you want an example where the modules are free over $\mathbb{Z}$, then let $G=\langle g\rangle$ be a cyclic group of order $2$, let $M=\mathbb{Z}G$ be the regular $\mathbb{Z}G$-module, and let $N=U\oplus V$ be a direct sum of two copies of $\mathbb{Z}$, where $g$ acts trivially on $U$ and by multiplication by $-1$ on $V$.

0

The answer is no. One example to keep in mind is the 2 dimensional representation of $\mathbb{Z}/p$ over $\mathbb{F}_p$ where 1 acts by the matrix $$\left( \begin{array}{ccc} 1 & 1 \\ 0 & 1\end{array} \right)$$ This representation has trivial character.

1

Hint: Write $\Phi=(\Phi_1,\Phi_2)$ for $\Phi_1$ and $\Phi_2$ functions from $M$ to $N$. Choose appropriate elements of $M_2(R)$ and use the fact that $\Phi$ commutes with the action of $M_2(R)$ to show that: (1) $\Phi_i:M \rightarrow N$ is an $R$-module homomorphism. (2) $\Phi_1=\Phi_2$.

2

Hint: for all matrices $A \in M_2(R)$ and all elements of $x \in M^2$, we have $$\Phi(Ax) = A\Phi(x)$$ Assuming $R$ is a ring with unity, consider $x = (m,0)^T \in M^2$ and $A = \pmatrix{0&1\\1&0}$.

0

(To make sure this question doesn't hang around unanswered...) In every definition of UFD that I've seen, the fact the product is finite is always explicitly mentioned (although it is implicit because infinite products are not defined in general rings.) To state uniqueness of a decomposition, an author is usually obliged to write this: If ...

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

1

the way you write it it is not correct, because you could choose $x=y=p$, with $p$ irreducible and not a unit. then you would have counterexample As far as i know there is a definition for "prime" in integral domains: $p \mbox{ is prime } :\Leftrightarrow \forall x,y \in R : p|xy \Rightarrow p|x \mbox{ or } p|y$. "prime" and "irreducible" are equivalent ind ...

0

When you form the free Abelian group $F$ generated by the set $X$ (we will use the notation $Free(X)$ from here on, as introduced in Paulo Aluffi's text $Algebra:$ $Chapter$ $0$) is the set of formal combinations of "letters" $$\sum\limits_{x \in X} a_x$$ where $a_x + b_x = (a+b)_x$. In this way the set $X$ acts as a basis for $Free(X)$; in fact this is ...

0

Please forgive if its a double post, I'm really a beginner in this Module Theory. So please mention if there is any flaw in argument. Lemma: Let $f:V\to W$ be a module homomorphism, where $V,W$ are both $R$ modules. Let $V',W'$ be respectively submodules of $V,W$. Then image of $V'$ and pre-image of $W'$ are also submodules of their repective space. Proof: ...

0

Indeed but the "proof" can be transpose to our case. For example in the second outline of a proof, M/tM is a finitely generated module with a generating set $(e_{1},...,e_{n})$ of homogenous element (we can always find such a set). Then in the proof $R^{n}$ can be replaced by $({\bigoplus\limits_{i=1}^n e_{i}D})$. Thus you can make a correspondance between ...

0

The sequence $0 \to N \to M \to M/N \to 0$ of $R$-modules splits, since $M/N$ is free. Hence we have the desired property. To help you with your actual problem: you certainly don't need the freeness for the injectivity. But however: note that you need the freeness to define the map! Edit (as a response to your edit where you raised the question about ...

2

The inclusion $E\to E+F$ induces a map $\phi: E\to (E+F)/F$ given by $x\mapsto x+F$. This defines a homomorphism with kernel $E\cap F$. That is, $$E/E\cap F \cong Im \ \phi.$$ To show $\phi$ is surjective, we let $x+y+F\in (E+F)/F$ where $x\in E$ and $y\in F$. Observe that $x+y+F=x+F$ in $(E+F)/F$ since $y\in F$ and notice that $$\phi(x)=x+F=x+y+F.$$ Thus, ...

1

The first definition is what you might call a "ring" bimodule. It only asks for $A$ to be a ring, and it doesn't mind if $A$ has any $k$ algebra structure. I would call your second definition an algebra-bimodule, but I think you're missing an axiom. Not only does $M$ have a $k$-module structure, but this structure is compatible with $A$'s $k$-structure! You ...

2

Just as a heads-up, take any non-Noetherian ring $R$. Then $R$ is itself a finitely-generated $R$-module (generated by $1$), and so is $R/\mathfrak a$ for any ideal $\mathfrak a \trianglelefteq R$ (since it is again generated by $1$), but your ideal $\mathfrak a$ can be picked non-finitely generated if $R$ is picked not Noetherian. About your example, ...

1

As you say in your reasoning, the $A$-module structure on $M$ induces a $k$-module structure. So does the $B$-module structure. But without added conditions these two $k$-module structures could be different. For example, take $A=B=M=\mathbb{C}$ with the bimodule structure $$a\cdot m\cdot b=am\overline{b},$$ where $\overline{b}$ is the complex conjugate of ...

2

In Definition 2 the action of $k$ on $M$ induced by the action of $k$ on $A$ can be different from the given $k$-action on $M$.

4

$\mathbb R$ 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 ...

0

I am 100% sure that Fact A and Fact B cannot be proven from another. [So in some sense, I don't give an answer here, but rather the meta-answer that there is no answer.]

1

As I was also looking for the answer I found that the shift operator $\sum^{\alpha}$ simply moves an element of grading $i$ into an element of grading $i\ + \alpha$ for every element in D (as D is a graded PID and each homogenous element of D has a grade). That's what I understand so far from this book http://bit.ly/1srXIs0. I hope it will help

1

I'll assume $R$ is commutative so $M\otimes_RN$ is an $R$-module. The elements of $M\otimes_RN$ look like sums of pure tensors of the form $m\otimes n$ with $m\in M,n\in N$, modulo the relations that describe bilinearity of the $\otimes$ symbol. The $R$-module structure of the tensor product is defined by $r(m\otimes n)=rm\otimes n$ (which equals $m\otimes ... 0 This was my answer to the question (which turns out to be wrong). Will be glad to hear any interesting comments :) My solution, in a couple of steps: 1) If$M$is generated by 1 member m then if$\exists s\in A-p$such that$sm=0$then$\forall a\in A : sam=0$and thus$sM=0$and thus$s\in Ann(M)$in contradiction to the fact that$p \supset ...

1

$\DeclareMathOperator{\Hom}{\operatorname{Hom}}$There is a well-known criterion for $\Hom$ to vanish: Proposition: Let $A$ be a Noetherian ring, $M, N$ f.g. $A$-modules. Then $\Hom_A(M, N) = 0$ iff $\text{ann}_A(M)$ contains a nonzerodivisor on $N$. Taking $N = A/p$ for $p \in V(\text{ann}_A(M))$ gives that $\Hom_A(M, A/p) \ne 0$, i.e. some nonzero ...

2

The claim is not true. If $A$ is an integral domain which has a non-trivial Picard group, there is some invertible $A$-module $M$ which is not free. But then $M$ has no quotient isomorphic to $A$, since any epimorphism $M \to A$ is an isomorphism (using that $M$ is locally free of rank $1$). However, we can prove the following: Let $M$ be a finitely ...

2

Năstăsescu and van Oystaeyen wrote two books, "Graded and Filtered Rings and Modules" and the more recent "Methods of Graded Rings". Even if you don't want to read the more specialised material in them, the first chapter of the former and the first two of the latter contain a fairly comprehensive introduction for most purposes.

1

The paper that I am reading assumes rank of a module always exists. Hence, it may be using a different definition of rank, which may (or may not) coincide with the definition provided in the previous answer, when $M\otimes_RQ$ is free over $Q$. After I posted my question I found the following definition of rank in the book Syzygies (By E. Graham Evans, ...

1

A proof of your question can be found in this old paper.

0

Suposse that $\{M_\lambda\}_{\lambda\in\Lambda}$ is family of divisible $R$-modules and consider $Z=\prod_{\lambda\in\Lambda} M_\lambda$. Take $r\in R$ and $(x_\lambda)_{\lambda\in\Lambda}\in Z$, we can chose for each $\lambda$ an element $y_\lambda$ such that $r y_{\lambda}=x_\lambda$ then we have that \$r ...

2

Yes, this is very old work. In 1935, Koethe proved that the modules of Artinian principal ideal rings are all direct sums of cyclic submodules. Of course, your example falls into this category. In fact, all of the proper quotients of a principal ideal domain are Artinian principal ideal rings (in fact they are also self-injective, hence quasi-Frobenius.) If ...

1

This is a remark concerning points (e-g): your result is mentioned in the article R. B. WARFIELD, COUNTABLY GENERATED MODULES OVER COMMUTATIVE ARTINIAN RINGS, PACIFIC JOURNAL OF MATHEMATICS Vol. 60, No 2, 1975 right at the beginning, but without proof. However some pointers are given to articles about certain non-commutative rings in which the problem ...

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