All Questions
290
questions
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26
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Primitive idempotent and bilateral ideals
I'm trying to show for my algebra class that in a semisimple ring with unity $R$ (not necessarily commutative), every primitive idempotent element must belong to a minimal two-sided ideal.
Here, by ...
- 321
0
votes
1
answer
30
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If $I$ is a two-sided ideal of $R$ and $M$ is a left module over $R$, why is $IM$ also a left module over $R$? There is also a related question.
Let $R$ be a ring with unity. If $I$ is a two-sided ideal of $R$ and $M$ is a left module over $R$, then $IM$ is also a left module over $R$, where $IM = \lbrace
{\sum_{i=1}^n a_im_i|a_i\in I, m_i\in ...
- 923
2
votes
0
answers
43
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Irreducible modules and composition series of an algebra.
Let $A = \left\{ \left(\begin{smallmatrix} a & b \\ 0 & c \end{smallmatrix}\right) \in M_2(\mathbb{C}) \;\middle|\; c \in \mathbb{R} \right\}$.
Consider $A$ as an $A$-module. Show that
$I = \...
- 195
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votes
0
answers
22
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Composition series of $M_2(\mathbb{R})$ as an $A$-module over itself
Let $A = M_2(\mathbb{R})$ and consider $A$ as the regular $A$-module. Show that $A$ has infinitely many composition series.
The submodules of $A$ as the regular $A$-module are the left ideals of $A$, ...
- 195
1
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0
answers
31
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example of a $B$-bimodule $X$ that is not an $A$-bimodule.
Suppose that $(A, ∥ · ∥_A)$ and $(B, ∥ · ∥_B)$ are two Banach algebras such that $B$
is an ideal in $A$ and $∥·∥_A ≤ ∥·∥_B$. We know that each Banach $A$-bimodule $X$
is also a Banach $B$-bimodule. ...
- 11
1
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0
answers
36
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cyclic $R$-module correspondence in $R$ which is PID
I encountered the following question:
(This is taken the book Module Theory - an approach to linear algebra, by T.S Blyth. This is question is from exercise 4.1)
Let $M=Rx$ be a cyclic $R$-module. ...
- 876
2
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1
answer
58
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Is the inverse of a finitely generated $A$-submodule of $\mathrm{Frac}(A)$ still finitely generated?
Context. I recently realized that the definition of a fractional ideal of an integral domain $A$ I knew was probably the wrong one.
To give some context, let me give two different possible definitions ...
- 15.3k
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votes
1
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65
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Why Are these Elements Not in the Ideal?
Let an ideal $I = (2,x)$ be generated by 2 and $x$ in the ring $R = \mathbb{Z}[x]$. It seems both 1 and $-1$ are not in the ideal, but I don't know why. I would appreciate a simple explanation for ...
- 451
2
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58
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Finitely generated ideals of integer valued polynomials
Let $\operatorname{Int}(\mathbb{Z}):= \{f(x) \in \mathbb{Q}[x] \mid f(\mathbb{Z}) \subseteq \mathbb{Z}\}$. Any element of the integer valued polynomials can be written as a $\mathbb{Z}$ linear ...
- 605
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59
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Is this solution to show that the Ideal given by the kernel of$ f \in R \mapsto f(0,0) \in \mathbb C[X,Y]$ is not finitely generated correct?
Let $R \subseteq \mathbb C[X,Y]$ be the subring of all polynomials $f \in \mathbb C[X,Y]$ that can be written as $f = g(X)+X ·h(X,Y)$
a. Let $I \subset R$ be the kernel of the evaluation map $R \...
- 3,162
2
votes
1
answer
91
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Alternative proof for Structure theorem for finitely generated modules over a principal ideal domain
I'm thinking of an alternative proof for Structure theorem for finitely generated modules over a principal ideal domain (also called Fundamental theorem of finitely
generated modules over a PID in ...
- 488
1
vote
2
answers
193
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Seeing the real numbers as a $\mathbb{C}$-module
I recently bumped into this question on the community and the accepted answer left me thinking. I always thought that we couldn't view the real numbers as a $\mathbb{C}$-module for the following ...
- 2,067
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votes
1
answer
44
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What is $Hom_A(A/I,M)$?
Let $A$ be a commutative ring. Let $I$ be an ideal in $A$. Let $M$ be an $A$-module. It seems like $Hom_A(A/I,M)$ should measure $I$-torsion in $M$. Does $Hom_A(A/I,M)$ have a nice description or ...
- 89
3
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1
answer
75
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What $R$-modules $M$ and ideals $I$ satisfy $IM=M$?
Let $R$ be a unital associative ring and $I\subseteq R$ be an ideal. Let $M$ be a (left) $R$-module. What unital associative rings $R$, $R$-modules $M$ and ideals $I\subseteq R$ satisfy $IM=M$? Do ...
- 1,769
0
votes
0
answers
50
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A Noetherian-as-a-module ring is Noetherian as a ring?
This SE answer claims [see the edit] (in the beginning paragraph) that if $A$, $B$ are rings with $B$ being Noetherian as an $A$-module, then $B$ is Noetherian as a ring because
any chain of $B$-...
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