# Tagged Questions

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### If $0$ is the zero-object $\Longrightarrow F(0)$ is the zero object when $F$ additive

Let $$F : \text{A-Mod} \to \text{A-mod}$$ be an additive functor. Then if $0$ is the zero-object $F(0)$ is the zero object. Why this is true ? The definition of additive functor that I know is ...
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### In a semi-simple module, any submodule is a direct factor?

I need help understanding the following : In a semi-simple module, any submodule is a direct factor (this is sometimes taken as the definition of semi-simple) (i) How is this equivalent to the ...
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### Defining algebras over noncommutative rings

A definition of "algebra" (that is, an associative algebra, in the sense of ring theory) generally requires a commutative base ring. But there are cases where it's reasonable to consider algebras ...
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### Commutativity of direct and inverse limits

In exercise 5.34(iv) of Homological Algebra book by Rotman one is asked to prove that direct limits and inverse limits do not necessarily commute. I have two questions : 1.) Is it true that ...
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### Applying an additive covariant functor to a sequence

Let $F$ be an additive covariant functor and $0→A_1→A_1⊕A_2→A_2→0$ be the usual exact sequence in the category of $R$-modules with the first map injection on the first, and the second map the ...
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### Should a left module be an enriched functor or enriched presheaf?

In this page http://ncatlab.org/nlab/show/module#InEnrichedCategory, They defined the left module over a monoid object A as an enriched presheaf. However, consider the modules over a ring, the left ...
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### Do covariant functors preserve direct sums?

Suppose $T$ is a covariant functor from the category $R$-mod to Ab (the category of abelian groups) Is is necessary that $T(B \oplus C) \cong T(B) \oplus T(C)$ ? Does the answer change if we ...
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### Extending morphism between derived functors

Let N,M be objects in a left exact functor $F:A\rightarrow B$ between abelian categoire's source, most importantly say there is an isomorpihsm $\psi: F(M)\rightarrow R^dF(N)$ is it possible to extend ...
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### equivalence in category

First gives some definitions, and then the property that I am confused. $A$, $B$ are both $R$-module, and $C$, $D$ an (additive) abelian group, consider the category $M(A,B)$ whose objects are all ...
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### Origin of the term “module” for profunctors

Why do they call profunctors "modules"? How do they exactly relate to modules over rings?
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### Intuitive Explanation of Hom Functor Property

I am new to category theory and keep running across it while studying algebra. Whenever I see the $\mathrm{Hom}$ functor mentioned (in the context of modules), two of its basic properties are listed: ...
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### Does “maximal submodule <=> simple quotient module” generalize to abelian categories?

Does the statement "If $A$, $B$ are modules over a commutative ring $R$, then $B$ is a maximal submodule of $A$ if and only if $A/B$ is a simple module" generalize to the setting of abelian ...
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### A example of a monoidal non symmetric category of $R$-bimodules

It is well know that if $R$ is a commutative ring, then the category of modules $_R\operatorname{Mod}$ is monoidal symmetric. (Edit. I wrote the category of $R$-bimodules, as F. Muro explained this ...
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### What about a module of rank $\frac{1}{2}$?

Let $R$ be a commutative ring. The possible ranks of free $R$-modules are $0,1,2,\dotsc$. But what about a generalized notion of an $R$-module where ranks may be rational numbers such as ...
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### Questions about Rickards proof that $D^b_\mathtt{sg}(A) \equiv \mathtt{stmod}(A)$

Setup: Let $A$ be a self-injective algebra (so projective = injective for modules) and let $D^b(A)$ and $K^b(A)$ be the bounded derived category and the full subcategory consisting of the perfect ...
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### $M\cong_{R\operatorname{-Mod}}N$ if and only if $M\cong_{\operatorname{Ab}}N$ (Warning: This statement is FALSE!)

I was rather surprised by the fact that two modules are isomorphic if and only if their abelian group structures are isomorphic. I might just sketch the proof here. Given a ring homomorhpism ...
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### Can we really understand $R$ by studying $R$-modules? [duplicate]

According to Algebra: Chapter 0, the category $R\operatorname{-Mod}$ reveals a lot about $R$. However after completing the first eight chapters, I still found no examples where this happens. Can ...
As a final exercise to VIII.1 in Algebra: Chapter 0, we are asked to prove If $\mathcal{F}\colon\operatorname{R-Mod}\to\operatorname{S-Mod}$ is a right-adjoint operator, then $\mathcal{F}$ is ...