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Representations over reals

I know that the regular representation over complex field contains all its non-isomorphic irreducible representations with multiplicity isomorphic to its dimension. Is the result still true if we work ...
Quantum Horizon's user avatar
1 vote
1 answer
92 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) ...
bipin's user avatar
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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 ...
Mathematician's user avatar
1 vote
1 answer
79 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 ...
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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 ...
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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 ...
newuser's user avatar
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1 answer
99 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 ...
Margaret's user avatar
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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 ...
Ty Perkins's user avatar
1 vote
0 answers
48 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'=...
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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. ...
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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 $R$.)
Yellow Pig's user avatar
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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 ...
RØDIN's user avatar
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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)}(\...
Sebastian A. Spindler's user avatar
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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 $R$...
Micheal Brain Hurts's user avatar
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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 ...
Fiktor's user avatar
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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 ...
permutation_matrix's user avatar
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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)=\...
Milly Moo's user avatar
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$\mathbb{R}(n)$ 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 ...
marc's user avatar
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1 answer
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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 ...
philip's user avatar
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5 votes
2 answers
422 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 ...
Infinity_hunter's user avatar
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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 ...
zyy's user avatar
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3 votes
0 answers
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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 ...
Shrugs's user avatar
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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 ...
MZG's user avatar
  • 1
2 votes
1 answer
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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 ...
Connorlindquist's user avatar
5 votes
1 answer
192 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 (...
toxic's user avatar
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0 answers
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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, \...
Hussein Eid's user avatar
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1 answer
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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 ...
Laila Serge's user avatar
2 votes
0 answers
112 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 \...
D.R.'s user avatar
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1 vote
1 answer
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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 ...
Yos's user avatar
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1 vote
1 answer
92 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[...
Subham Jaiswal's user avatar
3 votes
1 answer
366 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 ...
Sic Vis's user avatar
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0 answers
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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 ...
PlatoEinsYu's user avatar
1 vote
1 answer
150 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 ...
math seeker's user avatar
2 votes
1 answer
304 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 ...
Dr. Nirbhay Kumar's user avatar
6 votes
1 answer
232 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 ...
Martin Brandenburg's user avatar
2 votes
0 answers
240 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 ...
math seeker's user avatar
4 votes
1 answer
210 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"/&...
sigma's user avatar
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1 vote
0 answers
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$D \otimes_k K \cong M_p(K)$ 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 ...
Erlike's user avatar
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3 votes
0 answers
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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 $\mathbb{Q}$-...
Charuvinda's user avatar
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0 answers
115 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 ...
PAB's user avatar
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0 answers
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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 ...
Divide1918's user avatar
  • 2,175
2 votes
1 answer
69 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 ...
mnr's user avatar
  • 177
0 votes
1 answer
184 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 ...
user avatar
1 vote
1 answer
121 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 ...
user avatar
1 vote
1 answer
239 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 ...
user avatar
0 votes
1 answer
85 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 $M = \oplus_{i\in I} M_i$ ...
user avatar
-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) ...
user avatar
2 votes
1 answer
571 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 ...
user avatar
7 votes
1 answer
499 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 ...
ChrisNick92's user avatar
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1 vote
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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 ...
Promit Mukherjee's user avatar