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

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2
votes
1answer
170 views

Direct sum of submodules

I'm trying to prove that the following statements are equivalent for a commutative ring $R$ A: $_RR=_RN \oplus_RM$ for some submodules $_RN$, $_RM\subseteq_RR$ B: There exists an element $e=e^2\in ...
2
votes
1answer
143 views

For each $x$ in a Hilbert $A$-module $X$, there exists a unique $y\in X$ such that $x=y\langle y,y \rangle $.

How do I prove this lemma? Lemma: Suppose that $X$ is a Hilbert $A$-module. For each $x\in X$ there exists a unique $y\in X$ such that $x=y \langle y,y \rangle $. I don't know if this is needed but ...
0
votes
1answer
233 views

A question on finitely generated module over Noetherian ring

If $G$ is a finitely generated non-zero module over the non-trivial commutative Noetherian ring $R$ then is it possible that for all maximal ideal $M$ of $R$ we have $MG=G$ ? If $R$ is semi-local ...
4
votes
1answer
2k views

Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so ...
10
votes
3answers
653 views

Does an injective endomorphism of a finitely-generated free R-module have nonzero determinant?

Alternately, let $M$ be an $n \times n$ matrix with entries in a commutative ring $R$. If $M$ has trivial kernel, is it true that $\det(M) \neq 0$? This math.SE question deals with the case that ...
3
votes
2answers
317 views

Injective linear map between modules

If you have an injective linear map between two free modules of equal dimension, is the determinant of the matrix representing the map necessarily nonzero? If not is there an obvious counterexample? ...
4
votes
2answers
612 views

Isomorphism between quotient modules

Is it true for a commutative domain $R$ and its ideals $I$ and $J$ that if the quotient $R$-modules $R/I$ and $R/J$ are isomorphic then $I=J$?
4
votes
2answers
633 views

Module isomorphic to second dual

Is there a simple condition on a module $M$ over a ring $R$ which will ensure that $M$ is isomorphic to its double dual, $M^{**} = \operatorname{Hom} (\operatorname{Hom}(M,R),R)$? What about a ...
0
votes
1answer
49 views

Every right ideal of R is injective

How can I prove the following statement: Every right ideal of $R$ is injective iff $R$ is semisimple. It's a strange statement. If true, only from the condition satisfied by the ideals we can ...
1
vote
2answers
92 views

Notation for an indecomposable module.

If $V$ is a 21-dimensional indecomposable module for a group algebra $kG$ (21-dimensional when considered as a vector space over $k$), which has a single submodule of dimension 1, what is the most ...
2
votes
1answer
585 views

Cauchy-Schwarz Inequality proof (for semi-inner-product A-module).

I am reading a proof of the following Cauchy-Schwarz Inequality and I don't understand one part of the proof: Theorem: Let $A$ be a $C^*$-algebra and let $E$ be a semi-inner-product $A$-module. Then ...
3
votes
1answer
265 views

$C \otimes A \cong C \otimes B$ does not imply $A \cong B$

Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$. Today I proved both of the following: $$ S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ ...
4
votes
1answer
107 views

Taking fractions $S^{-1}$ commutes with taking intersection

Let $N,P$ be submodules of an $R$-module $M$ and let $S$ be a multiplicative subset of $R$. I think I proved $S^{-1}(N \cap P) = S^{-1}N \cap S^{-1} P$ but since my proof is not the same as the one ...
3
votes
1answer
131 views

The Frobenius-Nakayama Formula

I am currently reading a paper where it refers to the usual Frobenius-Nakayama formula describing quotients of an induced module. It is refering to the following result: If $k$ is a field, $P$ is ...
0
votes
1answer
56 views

The variety associated to a polynomial ring with a particular grading

We impose various grading on an algebra or a module to understand its homological properties. So I made up the following problem and I would like to understand its correspondence as a variety. ...
3
votes
2answers
285 views

About Rim's theorem for projective modules over group rings

Let $G$ be a finite group. A theorem of Rim (Proposition 4.9 here) states that a $\mathbb{Z}G$-module $M$ is projective if and only if $M$ is $\mathbb{Z}P$-projective for all Sylow subgroups $P$ of ...
4
votes
1answer
328 views

Is every proper nontrivial ideal in a Noetherian ring not flat?

I guess my general question is exactly what's in the title, but let me explain why I'm asking and how I came to it. Consider the ideal $I=\langle x,y \rangle \subset k[x,y]$ for a field $k$. Just to ...
11
votes
4answers
280 views

Show $\mathbb{Q}[x,y]/\langle x,y \rangle$ is Not Projective as a $\mathbb{Q}[x,y]$-Module.

Disclaimer: Though I have been re-reading my notes, and have scanned the relevant texts, my commutative algebra is quite rusty, so I may be overlooking something basic. I want to show $\mathbb{Q} ...
4
votes
1answer
300 views

A formula for the minimum number of generators of a module over a semilocal ring

Let $R$ be a commutative ring with only finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_r$. Let $M$ be a finitely generated $R$-module. Then $$\mu_R(M)=\max\{\dim_{R/\mathfrak ...
6
votes
2answers
240 views

How can I find an element $x\not\in\mathfrak mM_{\mathfrak m}$ for every maximal ideal $\mathfrak m$

Let $R$ be a commutative ring with finitely many maximal ideals $\mathfrak m_1,\ldots,\mathfrak m_n$. Let $M$ be a finitely generated module. Then there exists an element $x\in M$ such that ...
2
votes
1answer
113 views

An equivalent condition for having finite length

Let $R$ be a commutative noetherian ring, $M$ a finitely generated $R$-module. How can I prove that $M$ has finite length if and only if $M_p=0$ for every non-maximal prime ideals $p$?
2
votes
3answers
73 views

If $Ra$ is free for $a\neq 0,$ is $a$ regular?

Let $R$ be a commutative ring with unity, and $0\neq a\in R.$ We will say that an element $x\in R$ is linearly independent if $\{x\}$ is a linearly independent set. A non-zero element of $R$ is called ...
3
votes
1answer
57 views

Free module, $\mathbb{Z}[a]$ over $\mathbb{Z}[(a+1)^2]$ for transcendental number a

I'm trying to prove that for a transcendental number $a$ the module $\mathbb{Z}[a]$ over $\mathbb{Z}[(a+1)^2]$ is free. For $\mathbb{Z}[a+1]$ over $\mathbb{Z}[(a+1)^2]$, the basis is $\{1,a+1\}$. What ...
1
vote
1answer
127 views

Proof needed for $\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$

I want a proof for $$\operatorname{Hom}_R(M,N) \otimes_RS \cong \operatorname{Hom}_S(M\otimes_R S,N\otimes_R S)$$ where $\phi\colon R \to S$ is a homomorphism and $M$ is finitely generated free ...
1
vote
1answer
91 views

Cokernel of $1+(P+N)(t-1)$

Let $V$ be a free $R$-module of finite rank (or, to start, even a finite-dimensional $K$-vector space), $N,P \in \mathrm{End}(V)$ such that $P,N$ commute, $P$ is idempotent, and $N$ is nilpotent. ...
6
votes
2answers
2k views

Finitely generated projective module

Would anyone can help me how to show that a finitely generated projective module over a local ring and PID are free? What I know about a finitely generated projective module $M$ over a PID $R$ ...
1
vote
1answer
359 views

Finding all simple $R$ modules of a ring.

I was hoping someone had an idea on how to go about solving the following; Find (up to isomorphism) all simple R-modules where i) $R = \begin{pmatrix} \mathbb{Z}/15 \mathbb{Z} & \mathbb{Z}/15 ...
8
votes
3answers
1k views

If $A$ an integral domain contains a field $K$ and $A$ over $K$ is a finite-dimensional vector space, then $A$ is a field. [duplicate]

Possible Duplicate: Proof that an integral domain that is a finite-dimensional $F$-vector space is in fact a field I need to prove this result, but the only starting point I think of is to ...
2
votes
2answers
215 views

On the sum of all the simple submodules of a module

$R$ ring and $M$ a left $R$-module. Call $\mathrm{Soc}\;M$ the sum of all the simple submodules of $M$. Then $M$ is artinian if and only if $\lambda_R(\mathrm{Soc}(M))<\infty$ and for very $0\neq ...
1
vote
2answers
186 views

If $M$ is a flat module then those two conditions are equivalent

Let $M$ be a flat $R$-module. Then the following are equivalent: 1) for every $R$-module $N$ we have $M\otimes_R N\neq0$ 2) for every maximal ideal $m$ of $R$ we have $M\neq mM$ I did 1 implies 2. ...
2
votes
1answer
143 views

On the length of a ring

Suppose that $R$ is a ring, and suppose that $\lambda_R(_RR)<\infty$ and $\lambda_R(R_R)<\infty$ (where $\lambda_R$ is the length of an $R$-module). Is it true that then ...
2
votes
0answers
63 views

Simple $R$-module where $R$ is a semisimple ring. Possible small improvement of a proof.

Reading through the proof of the following theorem (in Introduction to Group Rings, by Milies and Sehgal) Let $L$ be a minimal left ideal of a semisimple ring $R$ and let $M$ be a simple ...
2
votes
3answers
165 views

Let $R$ be a ring such that $R$ is a simple $R$-module. Show that $R$ is a division ring.

Let $R$ be a ring such that $R$ is a simple $R$-module. Show that $R$ is a division ring. I have an idea for this but I would like to make sure it is correct. My idea is that $R$-submodules of $R$ ...
0
votes
2answers
146 views

Let N be a submodule of the module M. Suppose M/N and N are semi-simple. Does it follow that M is semi simple?

Let N be a submodule of the module M. Suppose M/N and N are semi-simple. Does it follow that M is semi simple? I think the answer is yes but I am not sure how to prove it. Any help would be ...
8
votes
4answers
486 views

What exactly is an $R$-module?

From Wikipedia: "If $R$ is commutative, then left $R$-modules are the same as right $R$-modules and are simply called $R$-modules." The definition of left $R$-module: $M$ is a left $R$-module if $M$ ...
4
votes
1answer
403 views

Why is this ideal projective but not free?

Let $R=\mathbb{Z}[\sqrt{-5}]$ and $I=(2,1+\sqrt{-5})$. How can I prove that $I$ is projective but not free?
9
votes
2answers
644 views

Motivation behind the ingredients of First Cohomology group $H^1$

I started reading the Cohomology theory of groups. But I am not able to get any intuition or motivation behind the following : It is concerned with the formal definitions of crossed and principal ...
6
votes
2answers
402 views

Artinian if and only if Noetherian

Let $R$ be a ring (commutative, with identity), $m$ a maximal ideal and $M$ an $R$-module. Suppose $m^nM=0$ for some $n>0$. Then $M$ is Noetherian if and only if $M$ is Artinian Do you have any ...
3
votes
0answers
84 views

Decompose a module $M$ of the form $N \times N$, where $N$ is simple

Let $M$ be a $\mathbb{C}[G]$-module of the form $M=N\times N$, where $N$ is simple. How to conclude that $M$ has infinitly many direct sum decompositions into two copies of $N$ ? This is what I have ...
4
votes
2answers
140 views

An equivalent condition for an element to be integral

Let $R$ be a noetherian domain, $Q$ its field of fraction and $u\in Q$. Could you help me to prove that $u$ is integral over $R$ if and only if there exists $r\in R$ $r\neq0$ and $ru^n\in R$ for ...
2
votes
1answer
118 views

Why is this homomorphism an isomorphism?

Let $R$ be a commutative ring with identity. Suppose $R=(r_1,\ldots,r_k)$. Take an homomorphism of $R$-modules: $f:M\rightarrow N$. Suppose that the function $\frac{f}{1}:M_{r_i}\rightarrow N_{r_i}$ ...
5
votes
2answers
823 views

Noetherian and Artinian modules

I don't understand clearly what is meant by Noetherian and Artinian modules. I tried explaining it to myself using the definitions but it is still not clear. Can someone please help explain it to me ...
4
votes
1answer
410 views

Isomorphic $R$-modules

How can i show that two $R$-modules of finite rank are isomorphic if and only if they have the same rank. i.e $R^n \cong R^m$ iff $n=m$ .
3
votes
1answer
119 views

A module decomposition problem

Let $A$ be a finite dimensional semisimple algebra over $\mathbb{C}$ and $M$ is a finitely generated A-module. Prove that $M$ has only finitely many submodules iff $M$ is a direct sum of pairwise ...
2
votes
0answers
263 views

Inductive proof of a version of Nakayama's lemma

In Matsumura's 'Commutative Ring Theory', the following version of NAK is shown: If $M$ is a finitely generated $A$-module, $I\subseteq A$ an ideal s.t. $IM=M$, then there exists an $a\in A$ with ...
3
votes
3answers
259 views

Why is this ring semisimple?

Let $R$ be a simple ring (i.e. a ring with no nontrivial two-sided ideals) which contains a left ideal which is simple as a left $R$-module. How can I prove that $R$ is semisimple?
2
votes
1answer
132 views

Field homomorphism $\varphi:\mathbb{Q}(\sqrt[3]{2})\to\mathbb{K}$ where $\mathbb{K}$ is the splitting field of $x^3-2$

Denote $\mathbb{K}$ as the splitting field of $x^3-2$ . I wish to find $\Gamma:=\left\{ \varphi:\mathbb{Q}(\sqrt[3]{2})\to\mathbb{K}|\forall q\in\mathbb{Q}:\varphi(q)=q\right\} $. Sind $\varphi$ is ...
1
vote
1answer
219 views

If $R$ is a finite dimensional algebra over a field then $R$ is simple as a ring if and only if it has a faithful simple left $R$-module

How can I prove that if $R$ is a finite dimensional algebra over a field then $R$ is simple as a ring if and only if it has a faithful simple left $R$-module?
1
vote
1answer
350 views

Tensor product of a finitely generated modul and a finite length module is finite length

Let $R$ be a commutative ring and $M,N$ $R$-modules finitely generated with $M$ of finite length. How can I prove that $M\otimes_R N$ is of finite length?
1
vote
1answer
567 views

Quotient rings and free modules

This is a question related to this: "...if $R$ is a commutative ring and $I$ a nontrivial ideal of it, then $R/I$ is never free. Thus the only commutative rings $R$ for which every finitely generated ...