# Tagged Questions

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### Problem with Wedderburn Theorem proof.

This is a Wedderburn Theorem proof in Frank W. Anderson, Kent R. Fuller: Rings and Categories of Modules Please explain that: "Therefore $_RR$ has a composition series of length $n$". Exercise ...
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### Weak flat condition?

Let $R$ be a unit ring (not necessarily commutative). Then it is clear that for a right $R$-module $M$ we have: $M$ is flat $R$-module $\Rightarrow$ for any left $R$-module $E$ with $E\otimes_{R}M=0$ ...
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### Is there a proof of this that does not use idempotents?

I am going to present a statement and a proof. The proof makes use of idempotents which makes it a little cumbersome. Is there a proof that does not use idempotents? (using well-known theorems ...
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### Multiplicity of the simple $R$-module $M$ in the semisimple ring $R$

I'm confused about the conclusion of Wedderburn's structure theorem for semisimple rings. Let's consider the special case where $R=M^n$ as modules for some simple module $M$. Wedderburn's theorem says ...
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### Question about Noncommutative Rings by I. N. Herstein

Here's an example that I do not quite understand it fully, it's on page 6 of the book. And here's what it says: Let $F$ be a field, and $F_n$ be a ring of all $n\times n$ matrices over $F$. We ...
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### Partial cycles in projective resolutions of square-free algebra

Short version: Over a square-free algebra must every projective resolution of a simple module eventually terminate or contain a shift of itself as a direct summand? I suspect not, but have not ...
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### Defining the Rank of a Projective Module

I am trying to understand the definition of rank for a projective module over a noncommutative ring. The definition I am using is: A sufficient condition for the rank of a free module over a ring ...
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### An ideal generated by a nontrivial idempotent is not a free module

Let $R = \Bbb QG$ with $G$ isomorphic to a cyclic group of order $2$, and $x$ its generator. I'm trying to show that $$\frac{1}{2}(1+x)R$$ is not a free $R$-module. I found out that ...
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### When do finiteness conditions coincide on modules?

Let $R$ be a (possibly noncommutative, but unital) ring. There are several finiteness conditions on a left $R$-module that we can consider. The ones I have come across include finitely generated ...
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### Total divisor in a Principal Ideal Domain.

Let $R$ be a right and left principal ideal domain. An element $a\in R$ is said to be a right divisor of $b\in R$ if there exists $x \in R$ such that $xa=b$ . And similarly define left divisor. $a$ ...
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### How to prove finite dimensionality of Hom-spaces between modules of finite length?

Let $R$ be a ring and $k$ be a field such that $k\hookrightarrow R$. Thus, given two $R$-modules $M$ and $N$, we can regard $\operatorname{Hom}(M,N)=\operatorname{Hom}_R(M,N)$ as a vector space over ...
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### Natural homomorphism $\mathfrak{a} \prod E_\lambda \to \prod \mathfrak{a} E_\lambda$

Let $A$ be a ring (we don't assume that $A$ is commutative), and $\frak a$ a left ideal of $A$. Let $(E_\lambda)$ be a system of left $A$-modules. There is a natural homomorphism of modules  \phi ...
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### On the Nakayama functor

Let $A$ be a finite dimensional $k$-algebra with 1. Denote by $_AP$ the category of projective left $A$-modules finite dimensional. And with $_AI$ the category of injective left $A$-modules finite ...
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### Trivial extension of an algebra

Suppose that $A$ is a finite dimensional $k$-algebra. Call $Q=\mathrm{Hom}_k(A,k)$. $Q$ admits an $A$-$A$-bimodule structire in the obvious way. The trivial extension of $A$ is defined as follows: ...
Suppose that $R$ is a finite dimensional $k$-algebra. I say that $R$ is Frobenius if it is locally bounded (see this question for a definition) and indecomposable projectives and injectives coincide. ...
Let $k$ be a field and $R$ an associative $k$-algebra suppose $R^2=R$. We say that $R$ is locally bounded if there exists a complete set of pairwise orthogonal primitive idempotents $\{e_x:x\in I\}$ ...