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1 answer
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Are finitely generated subrings of $\mathbb{H}$ Noetherian?

Let $R$ be a finitely generated subring of the ring of real quaternions $\mathbb{H}$, that is, $R$ is the subring generated by a finite subset of $\mathbb{H}$. I want to show (or find a counterexample)...
user avatar
2 votes
1 answer
129 views

(Left) Noetherian domains and Torsion submodules

By a domain I mean a non trivial ring without any zero-divisors (not necessarily commutative). Let $R$ be a ring and $M$ be a left $R$-module. We say an element $m\in M$ is a torsion element iff ...
Dilemian's user avatar
  • 1,057
0 votes
1 answer
80 views

Noetherian modules and Noetherian rings

I want to show that if $R$ is a Noetherian ring then $Mat_n(R)$ is also a Noetherian ring. It is obvious that $Mat_n(R)$ is a finitely generated $R$-module. So $Mat_n(R)$ is a Noetherian R-module. ...
S_Alex's user avatar
  • 991
3 votes
0 answers
95 views

Simplicity of Noetherian $B$, $A \subseteq B\subseteq C$, where $A$ and $C$ are simple Noetherian domains

After receiving important comments, which show that my original question has already been asked and answered, I now change my question to the following non-commutative setting: Let $A \subseteq B \...
user237522's user avatar
  • 6,937
4 votes
1 answer
312 views

In a left noetherian ring, does having a left inverse for an element guarantee the existence of right inverse for that element? [closed]

In a left noetherian ring, does having a left inverse for an element guarantee the existence of right inverse for that element?
user470412's user avatar
  • 1,009
0 votes
0 answers
163 views

Intersection of all prime ideals of weakly Noetherian ring

Prove that intersection of all prime ideals is nilpotent in weakly Noetherian associative ring R. Hello, I've got stuck with this question. Could you please give any advice or reference to some ...
Sleppy's user avatar
  • 137
2 votes
0 answers
73 views

$R$ right Noetherian. Is it true that $R(x)\otimes_{R[x]}R(x)\cong R(x)$?

Let $R$ be a right Noetherian ring (actually it is left Noetherian as well) and $S=R[x]$ the polynomial ring in one (commuting) variable. If $X$ is the set of all monic polynomials then $X$ is a right ...
Sam Williams's user avatar
  • 1,431
3 votes
1 answer
193 views

Is $R\langle x_1,\ldots,x_n\rangle$ Noetherian?

Let $R$ be a Noetherian integral domain of characteristic $0$, and let $R\langle x_1,\ldots,x_n\rangle$ be the free $R$-ring. Is $R\langle x_1,\ldots,x_n\rangle$ necessarily Noetherian? 1) The proof ...
user457187's user avatar
0 votes
0 answers
308 views

The ring of quotients of the first Weyl algebra

Since there are no comments to this question, I now restrict it to the following question: It is known that: A ring $R$ is a prime left Goldie ring if and only if $R$ has a left quotient ring which ...
user237522's user avatar
  • 6,937
1 vote
0 answers
117 views

A simple Artinian left quotient ring of a left Noetherian domain

Recall the following known result, Theorem 6.1: A ring $R$ is a prime left Goldie ring if and only if $R$ has a left quotient ring which is a matrix ring over a division ring (= simple Artinian). Now,...
user237522's user avatar
  • 6,937
4 votes
1 answer
183 views

When does an integral group ring have finite global dimension?

Let $G$ be a finite group and $R=\mathbb{Z}[G]$ the integral group ring. If $G$ is such that $R$ is Noetherian (so $G$ polycyclic-by-finite) when does $R$ have finite global dimension? Another way of ...
BillScroggs's user avatar
4 votes
0 answers
164 views

When is $RS^{-1}$ a local ring?

Suppose we have a noncommutative ring $R$ and multiplicatively closed set that is both right Ore, and right reversible, i.e. it is a right denominator set. Now, we can localize $R$ at $S$ to form $RS^{...
Sam Williams's user avatar
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5 votes
1 answer
286 views

Ore extensions of noetherian $k$-algebras are again noetherian (proof explanation)

Let $k$ be a field. All occuring $k$-algebras are required to be associative and unital. By noetherian I always mean left noetherian. In a lecture I’m currently taking the notion of an Ore extension ...
Jendrik Stelzner's user avatar
1 vote
1 answer
81 views

Uncountably many left ideals?

Let $R$ be a following subring of $M_2(\mathbb{C}):$ \begin{equation*} R = \left\{ \begin{bmatrix} a & r \\ 0 & s \end{bmatrix} ~:~ a\in \mathbb{Q} ~\mbox{...
Learner's user avatar
  • 544
2 votes
1 answer
1k views

matrix ring Noetherian, Artinian, semisimple?

Let $k$ be a field and $\Lambda=\begin{bmatrix} k & 0 \\ k^2 & k[x]/(x^2) \end{bmatrix}$. This ring is an algebra over $k$. (a) What is $\dim_k \Lambda$? (b) Is $\Lambda$ a left ...
user160919's user avatar
3 votes
1 answer
438 views

Literature on noncommutative rings

I am looking for books or notes about non commutative rings with with a maximum of data exposed without the help of modules (because I have many references which deal with the subject but modules are ...
Carré rond's user avatar
2 votes
1 answer
283 views

Non-commutative noetherian integral domain-Ore condition

Let $R$ be a non-commutative integral domain with unity which is also a right Noetherian ring. By integral domain I mean that the product of nonzero elements is always nonzero. I am trying to show ...
Questions-Math's user avatar
1 vote
1 answer
526 views

Noncommutative finitely generated algebras need not be noetherian

I would like to understand an example (of the title) given in the book "An Introduction to Noncommutative Noetherian Rings" by K. R. Goodearl, R. B. Warfield... On page 8, Exercise 1E, an example of ...
Kristina's user avatar
  • 521
1 vote
1 answer
303 views

Simple Noetherian domain which is not a division ring

I need a simple Noetherian domain which is not a division ring. I do know that this ring must not be Artinian, since otherwise it would be a division ring. Thanks in advance!
karparvar's user avatar
  • 5,848
3 votes
1 answer
144 views

Coker of powers of an endomorphism

Let $F\in\operatorname{End}_R(M)$, where $M$ is a Noetherian $R$-module. If $\operatorname{Coker}F$ is of finite length, is Coker and Ker of all powers of $F$ of finite length? Is the condition of ...
karparvar's user avatar
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