0
votes
0answers
42 views

Polynomial Ideals and Transverse modules

I have a given ideal and I want to find the "smallest" ideal so that when I add it to the original one, I get another certain ideal. Let $\mathcal{E}$ be the ring of smooth function germs ...
2
votes
3answers
125 views

Isomorphic quotient of a module over Noetherian commutative ring

I have a nice solution to the following problem and I thought of writing a paper about it but beforehand, I wanted to ask the problem here to see if this is an easy problem and if you people can solve ...
2
votes
3answers
46 views

Example of ring $R$ with ideals $I\neq J$ such that $R/I \cong R/J$ as modules

It's easy to prove that if $I$, $J$ are two-sided ideals and $R/I\cong R/J$ as modules over $R$, then $I=J$. What about left ideals? Is there a simple counterexample? I believe I've found an answer, ...
1
vote
1answer
21 views

ideal,ring,flat module,modules over R

Is there a characterization of modules (AND equivalent characterizations of rings R) over integral domains R with the property that each left ideal in R is flat?When all left ideals are ...
0
votes
2answers
37 views

Not so usual equivalence of maximal left ideal of a ring

I was reading Foundations of Module and Ring Theory and i found this equivalence of maximal left ideal as exercise in the the first chapter: A left ideal $I$ of a ring $R$ is a maximal if and only ...
1
vote
1answer
43 views

Annihilator of a quotient module

Let $J$ be an ideal of $R$, and $M$ a right $R$-module. Since $Jr \subseteq J$, $M / MJ$ is naturally a right $R$-module. Since it seems relevant to another problem, I am trying to determine ...
1
vote
1answer
47 views

Symmetric powers of ideal quotients in a local ring.

Let $R$ be a local ring and $I \subset R$ any ideal. When is it the case that $(I \: \backslash I^2)^n = I^n \: \backslash I^{n+1}$? Put another way, when is the natural map $\text{Sym}^n(I/I^2) ...
1
vote
1answer
30 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 ...
1
vote
1answer
68 views

Minimal primes and zero divisors

Let $R$ be a commutative local ring, $M$ a finitely generated $R$-module, and $x \in M$. Is it true that if for any $p \in$ $\operatorname{Min}(R)$ there exists $a_{p}\notin{p}$ such that $a_{p}x=0$, ...
2
votes
1answer
50 views

Let $k$ be a division ring, then the ring of upper triangular matrixes over $k$ is hereditary

I'm reading Ring Theory by Louis H. Rowen, and he claimed that The ring of upper triangular matrices over a division ring is hereditary (it's on page 196, Example 2.8.13 of the book). I think it ...
0
votes
0answers
68 views

Is every module a direct limit of cyclic modules?

I want to show that $M$ is $A$-flat is equivalent to $Tor_1^A(M,A/I)=0$ for every finitely generated ideal $I$. I want to show $Tor^A_1(M,N)=0$ for any $A$-module $N$. Is every module a direct ...
1
vote
1answer
35 views

The sum of all right ideals isomorphic as modules to a simple module is an ideal

I could use some help on the following problem. Let R be a ring. (a) If $r \in R$ and $U$ is a minimal right ideal of $R$, show that either $rU=0$, or that $rU$ and $U$ are isomorphic right ...
0
votes
0answers
118 views

When is $\operatorname{gr}_I (M)$ finite?

When is $\operatorname{gr}_I (R)$ (I mean associated graded ring of $I$) finite? When is $\operatorname{gr}_I (M)$ finite? ($R$ is Noetherian ring and $M$ finite $R$-module.)
0
votes
1answer
78 views

The ideal $I=(3,2+\sqrt {-5})$ is a projective module

Let $R=\mathbb Z[\sqrt{-5}]$ and $I=(3,2+\sqrt {-5})$ be the ideal generated by $3$ and $2+\sqrt{-5}$. I'm trying to prove that $I$ is a projective $R$-module. I'm using the lifting property ...
1
vote
1answer
62 views

Show that a finitely generated module is trivial if it's equal to a maximal ideal times itself

Let $R$ be any commutative ring which is local. $M$ a finitely generated $R$ module, and $ I \subset R$ a maximal ideal. How to show that $ M = IM$ $ \Longrightarrow$ $M = 0$ in an elementary way? And ...
1
vote
0answers
62 views

Is the universal enveloping algebra of the free Poisson algebra generated by finite set (left) noetherian?

Let $P$ be the free Poisson algebra over $k$ (a field) generated by finite set $x_1,...,x_n$. Let's consider the universal enveloping algebra $P^e$ of the free Poisson algebra $P$. Hence a Poisson ...
6
votes
0answers
80 views

Two ideals both alike in dignity, in fair Paris where we lay our scene.

Let $A$ be an integral domain. I have to show that two ideals $\mathfrak a$ and $\mathfrak b$ are isomorphic as $A$-modules if and only if there exist $a$ and $b$ such that $a\mathfrak b=b\mathfrak ...
4
votes
2answers
65 views

Special basis of an ideal as a $\mathbb{Z}$-module in number fields

I was speculating that the following may be true (but do not see any easy way to settle it; it must be known, I suppose): Given a (say, prime) ideal $\mathfrak{p}$ of the ring of integers ...
2
votes
1answer
87 views

An equivalence for $\operatorname{grade}(I,M)\ge 2$. [duplicate]

Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $I$ an ideal of $R$. Show that $\operatorname{grade}(I,M)\ge 2$ iff the canonical homomorphism $M \mapsto\operatorname{Hom}_R(I,M)$ is an ...
0
votes
2answers
69 views

$\operatorname{Ann}(\operatorname{Ann}(N))=N$ for a submodule $N$ of an $R$-module?

Suppose $R$ is a commutative and unitary ring and $M$ is an $R$-module. Is it generally true that for any submodule $N$ in $M$, $\operatorname{Ann}(\operatorname{Ann}(N))=N$? If it's not true in ...
8
votes
2answers
159 views

Counterexample: multiplying modules by elements of an ideal vs. taking linear combinations

Let $R$ be a ring (commutative, unital) and $M$ an $R$-module. Let $I \subset R$ be an ideal. We make the following definitions: $$ A := \{ am \ | \ a \in I,\ m \in M \} $$ $$ B := \left\{ ...
3
votes
1answer
105 views

Some questions about Fitting ideals

Let $R$ be a ring and $M$ a finitely presented $R$-module. Given a free presentation $$ R^{\oplus m} \to R^{\oplus n} \to M \to 0 $$ we define $Fitt_k(M)$, the $k$-th Fitting ideal of $M$, to be the ...
2
votes
1answer
51 views

Let $R$ be an Euclidean domain. Let $p$ be irreducible. Let $a\ge1$. Show that every non-zero submodule of $R/Rp^a$ contains $Rp^{a-1}/Rp^a.$

Let $R$ be a Euclidean domain. Let $p$ be irreducible. Let $a\geq 1$. Show that every non-zero submodule of $R/Rp^a$ contains $Rp^{a-1}/Rp^a$. I bet the answer is stupidly obvious, but I just ...
2
votes
0answers
42 views

Mistake in the proof that a domain is flat as a module over any subring

Where is the mistake in the following argument? I feel that there has to be one, for example by the very existence of this article. Let $R$ be an integral domain and $S \subseteq R$ be a subring ...
2
votes
1answer
34 views

$(\bigoplus_{A}M_{\alpha })/ I(\bigoplus_{A}M_{\alpha }) \cong \bigoplus _{A}M_{\alpha }/IM_{\alpha }$

Let $(M_\alpha )_{\alpha \in A}$ be an indexed set of left R-modules and let $I$ be a left ideal of $R$. Prove that $I(\bigoplus_{A}M_{\alpha })= \bigoplus_{A} IM_{\alpha }$ and ...
2
votes
1answer
86 views

Ideals in the polynomial ring over a division ring are free

Let $k$ be a division ring. I want to show that every (right) ideal in $k[x]$ is free considered as a right $k[x]$-module. That means if $I$ is an ideal in $k[x]$ we have to show that $I=f\cdot ...
0
votes
0answers
82 views

Codimension of ideals in polynomial rings over PIDs

Let $R$ be a (commutative) principal ideal domain and let $J$ be an ideal in $R[x_1,\dots,x_n]$. Is it possible to make a general statement about the codimension (=rank/height) of $J$? As $J$ does ...
5
votes
1answer
226 views

Ideal generated by two elements that is not free

I am having problems understanding an example constructed by M. Ojanguren and R. Sridharan showing that over the polynomial ring in two variables over a division ring (which is not a field) there ...
-1
votes
1answer
103 views

Show that if an ideal is free as a module then it is principal.

Here $\mathfrak a$ is an ideal of a commutative ring $A$. Show that $\mathfrak a$ is principal if it is free as an $A$-module.
2
votes
1answer
44 views

Definition of “invariant in a module”

What does it mean if someone say that the class of an ideal $I$ in a ring $R$ is an invariant of a module $M$?
3
votes
1answer
86 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 ...
0
votes
1answer
82 views

$G/G' \cong I/I^2$ where $I$ is the augmentation ideal [duplicate]

Possible Duplicate: Isomorphism between $I_G/I_G^2$ and $G/G'$ Let $G$ be a finite group. Let $I\unlhd\mathbb{Z}[G]$ be the augmentation ideal of the integral group ring $\mathbb{Z}[G]$. ...
5
votes
1answer
162 views

Idempotent and Hermitian vectors in Group Algebra

Let $C$ be the field of complex number and $G$ a finite group, then define $C[G]$ be a vector space over $C$, with elemnts of $G$ as the basis. Then any element in $C[G]$ can be written as $\sum_{g ...
1
vote
1answer
291 views

Annihilator of a simple module

Let $R$ be a finitely generated commutative ring and $C$ an $R$-algebra ($C$ is not necessarily commutative). Assume that $C$ is a finitely generated $R$-module. If $S$ is a simple $C$-module, then ...
2
votes
1answer
95 views

A question on artinian semi-primitive rings

So the question is as follows: Suppose $U$ is an ideal of artinian ring $R$, then show that there is an ideal $V$ such that $U+V=R$ and $U\cap V \subseteq J(R)$ . Let me describe my approach. I took ...
2
votes
1answer
125 views

A problem on the Jacobson radical, from Isaacs Graduate Algebra

This is problem 14.10 from Isaacs Graduate Algebra. Let $U$ and $V$ be ideals of a ring $R$ and assume $U+V$ = $R$, and $U \cap V \subseteq J(R)$ . Suppose that $v \in V$ and that $U + v$ is ...
1
vote
2answers
80 views

Question about factor rings

Assume $m_i$ are maximal ideals in a ring $R$. Then I have $m_1 \cdot \dots m_{k}$ is an ideal in $m_1 \cdot \dots m_{k-1}$ hence I can quotient to get a factor ring $m_1 \cdot \dots m_{k-1} / m_1 ...
1
vote
1answer
93 views

A question on the submodule given by action of an ideal

Suppose $M$ is a completely reducible left $R$-module for a ring $R$ and $I$ is an ideal of $R$. Then prove that the following are equivalent: $M=IM$ If $I$ annihilates an element $x \in M$, then ...
4
votes
2answers
156 views

Insight of some concepts in commutative algebra

I really enjoyed the basic algebra course and wanted to teach myself a little more. So I am trying to learn commutative algebra from Atiyah-MacDonald and Eisenbud. The department in our university ...
4
votes
1answer
281 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 ...
2
votes
0answers
54 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 ...
4
votes
2answers
150 views

Does totally flat commutative ring imply all ideals are idempotent?

From reading Atiyah and MacDonald, I know of the result that a absolutely flat commutative ring has all principal ideals idempotent. Reading around on math reference, I think that if a commutative ...
3
votes
2answers
117 views

Set of associated primes of direct sum

Let $M$ be a module over the ring $R$. Let $\operatorname{Ass}(M)$ be the set of annihilator ideals $\operatorname{Ann}(x)$, which are prime, so $$\operatorname{Ass}(M) = \{\operatorname{Ann}(x) ...
2
votes
1answer
155 views

Extending an ideal of a polynomial ring to a polynomial ring with more indeterminates. Is it a tensor product?

Let $\mathbb{k}$ be a field, let $S'=\mathbb{k}[x_1,x_2,\dots,x_m]$, and let $I'\subseteq S'$ be an ideal. For some $n>m$, let $$S=\mathbb{k}[x_1,x_2,\dots,x_n]\ \ \ ...
4
votes
1answer
298 views

The ideal $(x,y)$ is not a free $K[x,y]$-module

Given a field $K$ we have the polynomial ring $K[x,y]$ in $2$ variables, which is also a left module (over itself). How can we prove that the ideal $(x,y)$ is not a free module?
6
votes
1answer
396 views

Annihilator of quotient module M/IM

Let $A$ be a commutative ring, $I$ an ideal of $A$ and $M$ an module over $A$. Is it true that $\operatorname{Ann}(M/IM) = \operatorname{Ann}(M) + I$? One inclusion is certainly true, but I ...
1
vote
1answer
280 views

On the grade of an ideal

I need to prove the following statment (actually a special case of it). Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $I$ an ideal of $R$. Then $\operatorname{grade}(I,M)\geq 2$ if ...