Tagged Questions

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 ...
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 ...
49 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, ...
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 ...
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 ...
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 ...
48 views

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 ...
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 ...
75 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 ...
160 views

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]\ \ \ ...
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?
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 ...
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 ...