For questions about modules over rings, concerning either their properties in general or regarding specific cases.

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14 views

Constructing nontrivial $\mathbb{Z}$-bilinear map from $\mathbb{R} \times (\mathbb{R} / \mathbb{Z})$

I'm trying to show $\mathbb{R} \otimes_\mathbb{Z} (\mathbb{R} / \mathbb{Z})$ is nontrivial, where my tensor product is defined using the universal property. This problem can be easily reduced ...
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8 views

Sylvester domains

I'm an undergrad mathematics student and I'd like to request some books about Sylvester domains. Specifically I'd like to understand the fact that not all modules are Sylvester domains. I just proved ...
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0answers
42 views

Global dimension of $\mathbb Q [x]$

I'm trying to show that the global dimension of $\mathbb Q [x]$ is 1. I have shown that $D(\mathbb Q [x]) \leq 1$ as follows. One can reduce to the case of showing that $$sup_{\{B\}}\; \text{pd}\; ...
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1answer
40 views

question about typical proof of Krull Intersection Theorem

In Atiyah-Macdonald, in the proof of Theorem 10.17 (Krull's intersection theorem), the authors go through a 4-line chain of arguments to show that that the kernel $E=\bigcap_{n=1}^{\infty}\mathfrak ...
2
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1answer
74 views

Union of Associated Primes being finite.

Let $R$ be a commutative Noetherian ring with unit. Let $I=(x_1,x_2,\dots,x_t)$ be a nonzero ideal of $R$. Define $I_n=(x_1^n, x_2^n,...,x_t^n)$. Are there known results about $\cup_n ...
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2answers
33 views

Sub-modules of free modules

I'm going back through basic module theory notes, and I've come across a paragraph explaining that a sub-module of a finitely generated free module may not itself be free. In my course a free module ...
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1answer
29 views

Show that Q, as a Z module, is a direct summand in a direct product of copies of Q/Z.

Prove:Q, as a Z module, is a direct summand in a direct product of copies of Q/Z. This is a problem from P.J.Hilton&Stammbach's Homological Algebra. If this is true, then there exists a ...
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1answer
46 views

ring and module problem

Let $$F=\mathbb{R}$$ $$V=\mathbb{R}^{4}$$ consider two matrices $$S1=\begin{vmatrix} 0&-1& 0& 0\\ 1& 0& 0& 0\\ 0& 0& 0& -1\\ 0&0 & 1 & 0 ...
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1answer
58 views

Why we can consider both modules as modules over $R_{(p)}$? (Bruns and Herzog, Theorem 1.5.9)

I'm reading Bruns-Herzog's book Cohen Macaulay rings and have a probably elementary question. Why we may consider both modules as modules over $R_{(p)}$ in this theorem? ... i know that ...
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51 views

Tensoring over a direct sum of R

Is tensoring over a ring R same as tensoring over a finite direct sum of R. I mean is $A \otimes_R B$ isomorphic to $A \otimes_{R^n} B$? And if it doesnt hold for any ring $R$, Does it hold for ...
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1answer
36 views

Show that the image or the kernel are submodule of R-module.

Let $R$ be Commutative ring and $M$ be an $R$-module. Show that $im(H)$ or $ker(H)$ are submodule of $R$-module $M$, where $$H\in Hom(M,-)$$ First, I think by lemma: If $M$ is an $R$-module and $N$ is ...
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2answers
32 views

The definition “module of finite type”.

I know the definition of "module of finite type" Is that different from finitely generated module ? Thanks.
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2answers
42 views

Can a the field of fractions or quotient field F of an integral domain R be free over some set as an R-module?

We know any integral domain R when extended to a quotient field F, then F is free as an F-module on the set {1}. Can this field be free over some set as an R-module.
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0answers
15 views

Identifying the abelian group with a presentation matrix

I am doing problems from Artin: \begin{bmatrix} 2 \\ 1\\ \end{bmatrix} and \begin{bmatrix} 2 & 4\\ 1 & 4\\ \end{bmatrix} For the First one after manipulating rows I ...
2
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1answer
45 views

equivalence in category

First gives some definitions, and then the property that I am confused. $A$, $B$ are both $R$-module, and $C$, $D$ an (additive) abelian group, consider the category $M(A,B)$ whose objects are all ...
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0answers
30 views

Tensor product of two simple modules

Let $k$ be a field of arbitrary characteristic. Suppose that $A$ and $B$ are finite dimensional $k$-algebras. Let $S$ be a finite dimensional simple $A$-module and let $T$ be a finite dimensional ...
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55 views

If $\alpha$ and $\beta$ are algebraic integers then the roots of $x^2+\alpha x+\beta$ are algebraic integers

(This question is a dupplicate from If $\alpha$ and $\beta$ are algebraic integers then any solution to $x^2+\alpha x + \beta = 0$ is also an algebraic integer.) I'm trying to solve this problem with ...
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2answers
40 views

$R$ -module homomorphism

Let $M,N$ be $R$-modules. Suppose there exists $R$-module homomorphism $\phi:M \to N$ and $\psi:N \to M$ such that $(\psi \ \circ \ \phi)(m) =m, \forall m \in M. $ Could anyone advise me on how to ...
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0answers
68 views

What are the consequences of presentation of an algebra by generators and relations?

Let $A$ be a finite dimensional associative $K$-algebra, where $K$ is a field. I wonder how the presentation of $ A $ by generators and relations helps in the study of structure of the algebra ...
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1answer
50 views

$\operatorname{Hom}(R\times R,M)$ isomorphic to $\operatorname{Hom}(R,M) \times\operatorname{ Hom}(R,M)$

If $M$ is a right $R$-module, then $\operatorname{Hom}(R \times R,M)$ is isomorphic to $\operatorname{Hom}(R,M) \times \operatorname{Hom}(R,M)$ My questions are: What is the isomorphism map? (I ...
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1answer
30 views

How to show Not a Free Module

Let $\mathbb K$ be a field, $A= \mathbb K [x,y]$ and $ M = Ax + Ay$. prove that $M$ is NOT a free module!
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19 views

Divisible Direct Sum or Direct Product

We know that direct sum and direct product of divisible R-modules are divisible when R is a domain. Does there exist non-divisible modules with their product or sum being divisible? Or, if R is not a ...
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1answer
32 views

About the definition of the tensor product of modules

I was reading something about tensor product (balanced product) of modules (over an arbitrary ring $R$), but I cannot realize why we need a left and a right module. Would it be the same with two right ...
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1answer
97 views

exercise of Matsumura about CM

I have 2 question about this exercise of Matsumura: question 1- why $y^3$ is $R/(x^3)$ regular? question 2- I hardly (in 20 lines) can prove is there a short way or intuition for this part ? ...
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1answer
49 views

Localization of a module as direct limit

Let $A$ be a commutative ring, $S \subset A $ a multiplicatively closed set and $M$ an $A$-module. For every $s \in S$ we denote by $M_{s}$ the localization of $M$ with respect to $\{ 1, s, s^2, ...
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53 views

Prove this morphism is a derivation

Let $A$ a ring, $B$ a $A$-algebra, $I=\ker (B\otimes_A B \to B)$ given by multiplication, i wanna see why is $$d:B \to I/I^2$$ $$b \mapsto b\otimes 1 -1 \otimes b$$ well defined (in fact, why mod ...
5
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1answer
114 views

How to calculate $Ext(M,N)$?

I am confused about the calculation of $\text{Ext}(M,N)$. If $N$ is a fixed module and if we consider the projective resolution $$\cdots \to C_1 \to C_0 \to M \to 0,$$ then $\text{Ext}_n(M,N)$ is the ...
2
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2answers
54 views

Tensor product is o?

$K$ generated by $(ar,rb)$ for every $a\in A$, $b\in B$, $r\in R$. $F$ generated by $r(a,b)$ for every $a\in A$, $b\in B$, $r\in R$. $(a,rb)=(ar,b)=r(a,b)$,then $K=F$, $F/K=0$ Apparently I ...
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0answers
49 views

Is the length of the composition series of a free module identical to the number of its bases?

Let $A_0$ be an Artinian ring, $M$ a free $A_0$-module. Then, is the length of the composition series of $M$ identical to the number of its bases? It seems to me that it is not. If $\mathfrak a$ ...
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26 views

Tensoring and retaining projectiveness

Let $A$be a unital associative ring, If $A$ is not a projective $A$-bimodule, however $A\otimes A$ is a projective $A$-bimodule, can we conclude that $A\otimes A \otimes A$ is also projective?
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1answer
60 views

Can a ring isomorphism change the structure of a module?

Let $M$ be an $R$-module, where $R$ is a ring with unit. Given a ring automorphism $\phi: R \rightarrow R$, we can define a new $R$-module structure on $M$ by $r \cdot x = \phi(r) x$ for all $r \in ...
2
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1answer
38 views

A Sort of Exact Sequence

I have not given a lot of thought to this question: It may be very easy or very hard or somewhere in between. Suppose we have a sequence of modules and morphisms which looks like $ \ldots \to A_1 ...
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19 views

projective dimension of modules over triangular matrix rings

Let $R$ and $S$ be rings and $M$ be an $S$-$R$- bimodule. Then consider the triangular matrix ring constructed from these data (as, e.g., in Auslander-Reiten-Smalo). what one can say about the ...
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2answers
34 views

Free $A$-module isomorphic to a direct sum of copies of $A$?

Does this proposition hold even if it's not finitely generated? I think it does, since $M$ isomorphic to the direct sum of $M_i$, $M_i$ isomorphic to the direct sum of ...
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1answer
21 views

Annihilation and localization

Can somebody help proofing the following lemma. Let $x$ be an element of a module $M$, and let $\mathfrak{a}$ be its annihilator. Let $\mathfrak{p}$ be a prime ideal of $A$. Then $(Ax)_{\mathfrak{p}} ...
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1answer
26 views

Free modules and ideals

I am trying to show that an ideal I of R=$\mathbb{C}[x_1,x_2]$ generated by $x_1, x_2$ is free R-module. I am trying to show that I has a basis of the two generators given above. But I am not able to ...
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0answers
8 views

Topological modules with enough continuous linear functionals.

Context: I'm trying to find out which topological (unital) modules are "good enough" for generalizing results from real or complex functional analysis. For example, I say that a module, in order to be ...
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1answer
63 views

The meaning of a symbol in the proposition

The question is what's the meaning of the symbol $\phi$? If it just a mapping,what's the mean of the equation? I guess it's $\phi(x)$, $x$ is the element of $M$. Then $\phi(x)$ is the element of ...
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1answer
33 views

What is the difference between a module of finite rank and finitely generated module.

R is an integral domain and every module we talk about is an R-module. If a module is finitely generated then obviously every element of the module can be written as finite R-linear combination of the ...
2
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1answer
63 views

help in an example.

In page 226 of David Eisenbud's book Commutative Algebra with a View Toward Algebraic Geometry there is an example which I need help in some parts of it: why $codim I= 1$? why $dim M = dim R = ...
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1answer
132 views

Support of $\operatorname{Hom}(R/I, M)$

Let $R$ be a Noetherian ring, $I$ be an ideal of $R$ and $M$ be an $R$-module. Is the following formula true? $\operatorname{Supp}\operatorname{Hom}(R/I, M)=\operatorname{Supp}(M) \cap V(I)$ If ...
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45 views

Associated primes of $M/IM$

Let $R$ be a Noetherian ring, $I$ an ideal of $R$ and $M$ an $R$-module. Is the following formula true? $$Ass(M/IM)=Ass(M)∩V(I)$$ Thanks.
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1answer
90 views

Asociated prime ideals of Hom (Bruns and Herzog, exercise 1.2.27)

Let $R$ be a Noetherian ring and $M,N$ finitely generated modules. I want to show that $$\mathrm{Ass}_R(\mathrm{Hom}_R(M,N)) = \mathrm{Ass}_R(N) \cap \mathrm{Supp}(M).$$ I don't understand what ...
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1answer
19 views

Show polynomials $I$ is not finitely generated as $R$-module

Let $R=\{a_0+a_1X+\cdots+a_nX^n\;|\;a_0\in\mathbb{Z},a_1,a_2,\cdots ,a_n\in\mathbb{Q}, n\in\mathbb{Z}_{\geq 0}\}$ and $I=\{a_1X+\cdots+a_nX^n\;|\;a_1,a_2,\cdots ,a_n\in\mathbb{Q}, n\in\mathbb{Z}^+\}$. ...
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1answer
20 views

Prove module $M$ is finitely generated if $N$ and $M/N$ are finitely generated

Let $R$ be a ring with $1$ and $N$ be a submodule of an $R$-module $M$. Prove that $M$ is finitely generated if $N$ and $M/N$ are finitely generated. There are two definitions of "finitely ...
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1answer
24 views

If $N$ and $M/N$ are free modules of finite rank, so is $M$

Definiton: A $R$-module $M$ is said to be free of finite rank if $M\cong R^k = R\times R\times\cdots \times R$ (k times). Let $R$ be a ring with $1$ and $N$ be a submodule of an $R$-module $M$. ...
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142 views

If $M\otimes N=R^n$ need $M$ be projective?

So if over a commutive ring, $R$, we have that $M\otimes N=R^n$, $n\neq 0$, need we have that $M$ and $N$ be finitely generated projective? We have finite generation, because if $M\otimes N$ is ...
2
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1answer
62 views

Is any right exact sequence of modules induced by free modules?

Let $R$ be a ring and let $M \to N \to K \to 0$ be an exact sequence of $R$-modules. Is there an exact sequence of free modules $A \to B \to C \to 0$ and a commutative diagram $$\begin{array}{c} M ...
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1answer
35 views

Tor of submodule

Let $R$ be a $CRing$. If $i:A \rightarrow B$ is the inclusion of a $R$-subalgebra A into an $R$-algebra $B$, then what is ther relationship between: $Tor_{A^e}$ and $Tor_{B^e}$?
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31 views

Question about singular homology

in order to prove that $H_0(X)\simeq \mathbb{F}$, $\mathbb{F}$ is the unitary commutative ring we have to prove that $C_0(X)/B_0(X)\simeq \mathbb{F}$ since we have that $C_0(X)$ is generated by the ...