Questions tagged [semi-simple-rings]
The semi-simple-rings tag has no usage guidance.
167
questions
1
vote
0
answers
83
views
Example of a isosimple module which is not J-semisimple.
Recall that a module $M$ is called isosimple if its each nonzero submodule is isomorphic to $M$. A module $M$ is called $J$-semisimple if $J(M)=0$, where $J(M)$ is the sum of all superfluous (small) ...
0
votes
1
answer
64
views
Linear combination of a character
Hi there I am trying to solve a problem about characters of a finite group $G$. If $\chi$ is a character such that $\langle \chi,\chi \rangle = 2$ and $\chi_1,\dotsc,\chi_n$ the irreducible characters ...
0
votes
1
answer
52
views
Semisimplicity implies separability for a perfect field
Let $k$ be a field, $A$ a finite-dimensional semisimple $k$-algebra. If $k$ is a perfect field (every finite field extension of $k$ is seperable), then $A$ is separable.
I know a proof that uses the ...
0
votes
0
answers
14
views
Inferring classification of Clifford algebras from classification of Clifford modules
Let $Cl_n$ be the Clifford algebra (over reals)
$$
Cl_n = T^{*}\mathbb{R}^n/\langle v\otimes v - q(v) \rangle.
$$
There is a periodic table of $K$-representations of $Cl_n$, i.e. $\mathbb{R}$-linear ...
0
votes
0
answers
29
views
Primitive idempotent in semisimple ring
I'm struggling to resolve an exercise in "Methods of Representation Theory" of Curtis & Reiner.
Let $A$ be a semisimple ring and let $e \in A$ an idempotent different from zero. Show ...
1
vote
1
answer
71
views
A PID is a semisimple ring iff it is a field
I am trying to prove that a PID $R$ is a semisimple ring iff it is a field. Clearly any field is semisimple. I am not sure about the converse. By Artin-Wedderburn, $R$ is a product of matrix rings ...
0
votes
1
answer
37
views
One sided ideals of a semisimple ring.
Let $A$ be a semisimple ring. I'm wondering whether all ideals of $A$ are two sided. I know that all semisimple rings are both left and right semisimple. And, since $A$ is a semisimple module over ...
1
vote
0
answers
33
views
stable range of stably free modules
This is part of exercise 1.1.5 of the K-book:
Notation: we say $R$ has stable range at most $n$ if every unimodular row $(r_0,\ldots, r_n)$ induces a unimodular row $(r_1',\ldots, r_n')$ with $r_i'=...
0
votes
1
answer
48
views
stably free modules over semisimple rings are free
This is exercise 1.1 of Chapter 1 in the K-book:
If $R$ is a semisimple ring, then $R$ is a direct sum of a finite number of simple modules. Furthermore, every stably free module over $R$ is free.
...
0
votes
0
answers
29
views
Cocenter is a free module over the center in a semisimple ring
Let $R$ be a semisimple (noncommutative) ring. Why is the cocenter $R/[R,R]$ of $R$ a free module over the center of $R$? (Here $[R,R]$ is the commutator of $R$.)
0
votes
0
answers
46
views
Isomorphism of matrix ring with elements from opposite ring
I want to show the map $\phi:M_\mathrm{n} (R) \rightarrow M_\mathrm{n} (R^{op})^{op}$ given by M$\rightarrow $$^t$M is an isomorphism of rings.
I have shown that it is injective and surjective but I ...
0
votes
1
answer
28
views
Subalgebra with semisimple centralizer itself semisimple?
Let $V$ be a finite-dimensional $K$-vector space and suppose that $\mathcal{B} \subseteq \operatorname{End}_K(V)$ is a $K$-subalgebra such that the centralizer $$\mathcal{Z}_{\operatorname{End}_K(V)}(\...
0
votes
1
answer
37
views
Product of semisimple rings is semisimple.
A semisimple ring $R$ with $1$ (but not necessarily a commutative one) considered as left-$R$ module is a direct sum
$$R\cong L_1\oplus L_2 \oplus \cdots\oplus L_n$$
such that for some $e_i$ are in $R$...
1
vote
0
answers
44
views
Isn't any algebra with finite composition length and non-isomorphic simple modules semisimple?
Assume $A$ is an algebra over a field $K$ with composition length $n$ and $A$ has $n$ pairwise non-isomorphic simple modules. Is it true that this implies that $A$ is semisimple?
The answer seems to ...
0
votes
1
answer
59
views
proving that a ring is not semisimple
proving that a ring is not semisimple.
A question asks me explicitly that the ring of matrices
$M_{a,b}=\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}$ is not semisimple by showing that the ...
0
votes
0
answers
52
views
Computing the nilradical of a ring
Let $R=\begin{pmatrix}
\mathbb{C} & \mathbb{C} & \mathbb{C}\\
0 & \mathbb{C} & \mathbb{C}\\
0 & \mathbb{C} & \mathbb{C}
\end{pmatrix}$. I want to find the nilradical: $$N(R)=\...
0
votes
1
answer
46
views
$\mathbb{R}(n)$ is a simple algebra
Let $\mathbb{R}(n)$ be the set of n by n real matrices.
An algebra $\mathcal{A}$ is said to be simple if (Lang pag 653):
$\mathcal{A}=\bigoplus_{i=1}^n I_i$ $\quad$ with $I_i$ being simple left ...
1
vote
1
answer
79
views
If eRe is a division ring, then Re is a simple ideal.
I am currently working on showing this:
Let $R$ be a semisimple ring, and $e\in R$ be idempotent, then if $eRe$ is a division ring, $Re$ is a simple ideal.
I am unsure if my working so far is ...
5
votes
2
answers
400
views
How Zorn's lemma is used here?
I am studying the following theorem in Advanced modern algebra/ Joseph J. Rotman. - Third edition,(Graduate studies in mathematics ; volume 165),
A left $R$ module $M$ over a ring $R$ is semisimple ...
2
votes
0
answers
33
views
Is there some connection between these two methods counting isomorphic irreducible submodules for a decomposition of regular $A$-module?
Let $A$ be a semisimple algebra over $\mathbb{C}$. Given a decomposition $A^{\circ} = \oplus W_i$ of the regular $A$-module $A^{\circ}$ and an irreducible $A$-submodule $M$, I have seen two ways to ...
3
votes
0
answers
73
views
Simple modules over a matrix ring and their the dimension of their tensor product over the base field.
I am trying to solve the following problem:
Let $k$ be a field, and let $R=M_{n}(k)$, the non-commutative ring of
$n \times n$ matrices over $k$.
(a) Give examples of a simple left $R$-module $M$ and ...
0
votes
0
answers
75
views
A semisimple ring with a finite number of left maximal ideals
Let $R$ be a semisimple ring with a finite number of left maximal ideals. (Here "semisimple" means that the Jacobson radical is zero.)
Show that $R \cong R_1 \times ... \times R_n$
Such ...
2
votes
1
answer
86
views
Why is this subset finite? [duplicate]
We are given that $R$ is a ring with identity, and that $R$ is left semisimple, i.e. $R$ can be decomposed into a sum of minimal left ideals of $R$ ($R=\bigoplus_{n\in S} I_n$). What I'm confused ...
4
votes
1
answer
113
views
Module $k[x]/(x-a)^2$ is not semisimple, elegant proof?
Let $k$ be a field and $k[x]$ polynomial ring, and take the module $k[x]/(x-a)^2$ for arbitrary $a\in k$. How to show that this module is not semisimple?
I was thinking the easiest way is to use this (...
1
vote
0
answers
37
views
The matrix ring $M_n(R)$ of a semisimple ring $R$
Suppose that $R$ is a semisimple ring with unity. Let $S=M_n(R)$ be the matrix ring. For simplicity, we proceed with the special case $n=2$. Then, as easily seen,
\begin{align}
S=e_{11}S + e_{22}S,
\...
0
votes
1
answer
129
views
Why simple factorization of semisimple modules is unique?
The Ring $A$ is commutative or not. I'm trying to understand the proof of the theorem of unicity of simple factorization of semisimple modules. I already know Shur's lemma, and that, as a corollary ...
2
votes
0
answers
87
views
Correspondence between central orthogonal idempotents, direct sums of two-sided ideals, and product of rings?
In this article https://mathstrek.blog/2015/03/02/idempotents-and-decomposition/, there is the following theorem:
Theorem. Let $R$ be any ring. There is a bijection between:
(1) an isomorphism $R \...
1
vote
1
answer
32
views
Relationship semisimple matrices and semisimple modules.
For a commutative ring (if necessary field) $R$, we say a square matrix $X$ over $R $is semisimple iff $X$ is diagonalizable.
On the other hand, R-Module $M$ is semisimple iff $M$ is direct sum of ...
1
vote
1
answer
81
views
Semi simplicity over commutative algebras over non-algebraically closed fields
I have a question:
If k is an arbitrary field then is it true that if $M$ a finite dimensional $k[x, y]$ is semisimple as a $k[x]$ module and also as a $k[y]$ module then it is semisimple (as a $k[...
3
votes
1
answer
264
views
Tensor product of finite-dimensional semisimple algebras over algebraically closed field is semisimple
Let $K$ be an algebraically closed field, and let $A$ and $B$ be semisimple finite-dimensional $K$-algebras. I've seen a claim that the tensor product $A \otimes_K B$ is also a semisimple ring.
To ...
1
vote
0
answers
37
views
Irreducible representations of a finite group over different algebraically closed fields.
It's an exercise from "Advanced Modern Algebra" Rotman. Exercise 8.44 on page 574.
The problem is, prove that the degrees of the irreducible representations of $G$ over $K_1$ are the same ...
1
vote
1
answer
122
views
Simple modules in the decomposition of modular group algebra KG
Suppose F is a field such that $char(F) = p \ \nmid \ |G|.$ Then we know that in this case (i.e. semisimple), there is a bijection between the irreducible representations of G and the simple ...
2
votes
1
answer
214
views
For a module $M$ one has $rad(M)=0$ if and only if $M$ is isomorphic to a submodule of a direct product of simple modules
Definition: For a module $M$, the intersection of all the maximal submodules and module $M$ is called as radical of the module and denoted by $rad(M)$.
For a module $M$ one has $rad(M)=0$ if and only ...
6
votes
1
answer
201
views
Rings such that every module is a direct sum of generator modules
Is there a classification of those rings $R$ for which the category of left $R$-modules $\mathbf{Mod}(R)$ is generated by a small set of left $R$-modules under direct sums?
For example, every ...
2
votes
0
answers
189
views
Representations of $GL_n(F_q)$ over a finite field
If $F_q$ denotes a finite field of characteristic $p,$ then I want to learn about the representations of $G = GL_n(F_q)$ over a finite field $K$ such that $char(K) \ \nmid |G|.$ Any reference ...
3
votes
1
answer
180
views
Why is the element in the radical of a C* algebra nilpotent?
I am reading this book and in the 2nd chapter (II.1.6.4 Corrollary) the author proved the following:
But in the book the author didn't defined explicitly the terms "semisimple algebra"/&...
1
vote
0
answers
41
views
$D \otimes_k K \cong M_p(K)$ with $D$ a central simple division algebra of dimension $p^2$
Let $p$ be a prime number, and let $R$ be a central simple division algebra of dimension $p^2$ over a field $K$. Let $\alpha\in R$ be an element not in the center, and define $K:=k(\alpha)$. I am ...
3
votes
0
answers
87
views
Galois group of the Galois closure of a subfield.
Here's a problem from the Spring 2020 UCLA Algebra Area Exam:
If $K\neq \mathbb{Q}$ appears as a subfield (sharing the identity) of some central simple algebra over $\mathbb{Q}$ of $\mathbb{Q}$-...
0
votes
0
answers
87
views
Semisimplicity and global dimension
I know this could be a dumb question, but I've been studying for hours and I might be too tired to see why:
A ring R is semisimple if and only if its global dimension is zero.
We define the global ...
3
votes
0
answers
65
views
Multiplication by an element in semisimple subalgebra of endomorphism
Serge Lang Algebra, Sec. XVII, Exercise 9:
Let $E$ be a finite-dimensional vector space over a field $k$. Let $R$ be a semisimple sub-algebra of $\operatorname{End}_k(E)$. Let $a, b \in R$. Assume ...
2
votes
1
answer
66
views
Please help, getting a contradiction with non-commutative tensor products
Let $D/{\mathbb Q}$ be a quaternion algebra and $K / {\mathbb Q}$ a quadratic field extension that is contained in $D$.
Then I would like to see $D$ as a module over $K \otimes_{\mathbb Q} D$ via the ...
0
votes
1
answer
159
views
Intersection of all maximal ideal in a semi-simple ring.
I have met the statement below many times, either here on this site or while reading through books, but I am for my life is unable to prove why it is correct. Here is the statement:
If $R$(commutative ...
1
vote
1
answer
112
views
how to prove finitely many? which route is easier?
I want to prove that: If $N$ is finitely generated semi-simple $R-$module, then $N$ is a sum of finitely many simple submodules.
I know that if $N$ is a finitely generated $R-$module, then that the ...
0
votes
1
answer
152
views
$R$ is finitely generated?
I have seen many books using the idea that a commutative semi-simple ring with unity is finitely generated as an $R-$module but I do not understand why this is correct. Any elaboration will be ...
0
votes
1
answer
77
views
Proof of " $M$ is semisimple implies every submodule is a direct summand" step clarification.
Here is the proof of the statement as written in Rotman "An introduction to homological algebra"(but S changed to M and the role of $I,J$ is reversed).
Assume that $M = \oplus_{i\in I} M_i$ ...
-1
votes
1
answer
60
views
Using finitely generated in proof (2).
I was reading the proof of $(c) \implies (a)$ i.e., (Given any submodule $M \subset N,$ there exists a submodule $M' \subset N$ such that $N = M \oplus M'$) implies ($N$ is a sum of simple modules)
...
1
vote
1
answer
440
views
Semi-simple rings and fields.
I want to show that:
$R$ is semi-simple iff $R$ is isomorphic (as a ring isomorphism)to a direct product of a finite number of fields.
Definition: $R$ is a semi-simple ring if it is a direct sum of ...
5
votes
1
answer
369
views
A simple ring which is not semisimple
Let $V$ be an $\mathbb{F}$ - vector space with a countably infinite basis. Let $R=\text{End}_R V$ the ring of all linear functions $\phi:V\to V$ and $I=\{f\in R:\, \text{dim}\, f<\infty\}$ the two ...
1
vote
0
answers
122
views
Let V be an infinite dimensional vector space over a division ring D. The Set F={θ:V→V: Im(θ) is a finite dimensional subspace of V} is a simple ring.
I have been able to prove that F is a proper ideal of End(V) but however stuck to show that F is a simple ring. My idea is to start with a two sided ideal of F, I assuming I≠0, then using a nonzero ...
1
vote
1
answer
72
views
Jacobson Radical of integer matrices
Let $R = M_2(\mathbb Z)$
Find $J(R)$
The reason I'm asking is that I know by Artin Wedderburn that $M_n(D)$ is semisimple for any division ring D and hence $J(D)=0$. But here $\mathbb Z$ is of ...