Questions tagged [projective-module]

For questions related to projective modules, their structures, and properties.

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Modules of an infinite-dimensional simple algebra

Let $A$ be an infinite-dimensional algebra (over a closed field of characteristic zero) that is simple (that is, it is simple as an $A-A$-bimodule). Then what can we say about its finitely generated ...
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Prove that $M/Rx$ is a finitely presented module, if $Rx,x\in M$ of $M$ is a direct summand of $M$. [closed]

Let $R$ be a commutative ring with unity and $M$ be a right $R$-module. Suppose that a cyclic submodule $Rx,x\in M$ is a direct summand of $M$. Prove that $M/Rx$ is a finitely presented module. My ...
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Projective modules over Artinian ring.

Let R be an Artinian ring with 1. I know that as a (right) R-module it decomposes into a finite direct sum of projective indecomposable R-submodules. Let P be a projective R-module. Is it true that P ...
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If $R$ is a principal integer domain, then any finitely generated projective $R$-module is isomorphic to $R$

I am trying to prove that if $R$ is a principal integer domain, then any finitely generated projective $R$-module is isomorphic to $R$. To do so, I've fixed $P$ as a finitely generated projective and ...
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What are the projective submodules of $\Bbb Z/n\Bbb Z$ as a module over itself?

Let $R$ be the ring $\mathbb{Z}/n\mathbb{Z}$ with $n \in \mathbb{N}$ composite and consider $R$ as a right module over itself. I'd like to determine the projective submodules of $R$ if there are any. ...
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Show that $R$ is self-injective iff every finitely generated projective right $R$-module is injective

While studying for an upcoming exam, I've crossed the following problem: A ring $R$ is self-injective if, and only if, every finitely generated projective right $R$-module is an injective right $R$-...
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Finitely generated projective $\mathbb{Z}/n\mathbb{Z}$-Modules

I have to prove the following statement: Let $d$ be a divisor of $n$. Then $\mathbb{Z}/d\mathbb{Z}$ is a projective $\mathbb{Z}/n\mathbb{Z}$-Module $\iff$ $\text{gcd}(d, \frac{n}{d}) = 1$. The ...
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Proof Verification: $R[S]$ is a projective $R-$module (S is an arbitrary nonempty set)

I am asking if my proof is correct as follows: $\textbf{Theorem}$ Let $R$ be a ring with unity and $S$ a nonempty set. Then the polynomial ring $R[S]$ is projective. $\textit{proof }$ First of all, $R$...
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37 views

Decomposition of morphisms of projective module with its extension of scalars

Let $R$ be a commutative ring, $G$ a finite group, $P$ be a projective, finitely generated, $R[G]$-module, let $P_0$ be $P$ considered as a $R$-module. Serre states in Linear Representation of finite ...
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27 views

Projectivity over subrings

Let $R$ be a ring, $S \subseteq R$ a subring, and $M$ is a (left) $R$-module. If $M$ is projective as an $S$ module, then does it follow that $M$ is projective as an $R$-module? I guess yes, since $S$-...
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64 views

Cohomology of Heisenberg Group

If we let $G :=$ the Heisenberg group over the integers, then I can show that the center of this group is isomorphic to the integers $\mathbb{Z}$ and that the abelianization of G is isomorphic to $\...
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Question about simple module over semisimple Artinian ring

Let $R$ be a semisimple Artinian ring and $M$ be a simple left $R$-module. I need to show that there exists some left ideal $I$ of $R$ such that $M \oplus I \cong R$. I am not really sure how to do ...
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27 views

$r \mathbb{Z}_n$ is projective as $\mathbb{Z}_n$-module

Let $n=rs$ be a natural bigger than $1$, show that $r \mathbb{Z}_n$ is a projective $\mathbb{Z}_n$-module. Well, when $r$ and $s$ are coprimes we have that $\mathbb{Z}_n \simeq r \mathbb{Z}_n \oplus s\...
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A subring of $R$ and a projective left $R$-module.

This question is from an exercise which states that if $S$ is a subring of $R$, if $R$ is a free $S$-module and $M$ is left projective $R$-module, then $M$ is projective $S$-module. This can be proven ...
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38 views

Is a localization at a maximal ideal of a polynomial ring a perfect ring?

There are several equivalent definitions for a perfect ring $R$ (not necessarily a commutative ring), for example: Every flat left $R$-module is projective; see wikipedia. Also, there is the notion of ...
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Explicit construction of a finite projective resolution of a direct summand

Let $M,N$ be modules over a Commutative ring $R$. If $$0\to P_n \to P_{n-1}\to \cdots \to P_1 \to P_0 \to M\oplus N \to 0$$ is a finite projective resolution of $M \oplus N$, where $n\ge 1$, then how ...
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Let $K$ be a field and set $R= K[x]/(x^2)$ so that $R=K \dotplus K\bar{x}$ with $\bar{x}^2=0$. Show that $R \ncong \bar{x}R \oplus \bar{x}R$

Let $K$ be a field and set $R= K[x]/(x^2)$ so that $R=K \dotplus K\bar{x}$ with $\bar{x}^2=0$. In the textbook, there is a sentence : "$R \ncong \bar{x}R \oplus \bar{x}R$ since $\bar{x}$ ...
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Show that $\mathbb{Q}/\mathbb{Z}$ is an injective module and that it is not a projective module

So this is what I have for the injectivity: Knowing $\mathbb{Z}$ is a principal domain, $\mathbb{Q}/\mathbb{Z}$ is an injective module if and only if $\mathbb{Q}/\mathbb{Z}$ is divisible. Checking the ...
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Projective modules over reduced ring

I was trying to go through the following result: I couldn't get the first line of the proof? Why does assuming $A$ is reduced and has connected spectrum suffice?
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Graded version of Quillen and Suslin's theorem (graded projectives of graded-free and finite rank modules are graded-free)

I'm trying to solve the following exercice about graded $R$-modules, when $R=\mathbb{R}[x_1,\ldots,x_n]$ is a polynomial ring : Let $M$ be a graded-free $R$-module of finite rank. Show that any ...
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Fractional ideal of a Dedekind domain A is a projective module

It can be known that any fractional ideal $\alpha$ of A can be generated by two elements, and is that enough to construct a reverse of the quotient map from $A^2$ to $\alpha$? And by the way, I wonder ...
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$V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective.

Question : Let $\operatorname{Rad}(R)$ denote the Jacobson Radical of $R.$ Show that $V\cdot\operatorname{Rad}(R)\subseteq\operatorname{Rad}(V)$ where equality holds if $V$ is free or projective. Show ...
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50 views

Show that if $M$ is projective then $\operatorname{Hom}_R(M,.)$ is an exact functor. [duplicate]

Is there any reference that contains this proof: Show that if $M$ is projective then $\operatorname{Hom}_R(M,.)$ is an exact functor. Or any help in the proof will be appreciated. We defined ...
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Projective modules restricted to smooth curves

I want to prove a coherent sheaf $M$ on $X$ is locally free if and only if this is true for $M|_{X'}$ , for all smooth curves $X'$ mapping to $X$. I think the only if direction is obvious. For the if ...
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Understanding an example of a finitely generated projective module which is not free.

Here is the example I know: Consider the ring $R = \mathbb Z_2 \times \mathbb Z_2$ and the submodule $\mathbb Z_2 \times \{0\}.$ it is by construction a direct summand of $R$ but certainly not free....
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Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically?

Why every module $P$ is the quotient of a free module implies the existence of this exact sequence specifically $$0 \rightarrow \ker \varphi \rightarrow F \xrightarrow{\varphi} P \rightarrow 0? $$ ...
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Cycles in an acyclic complex of projectives are projectives.

Let $P_\bullet$ be a non-negative chain complex in an abelian category $\mathcal A$ such that $H_n(P_\bullet) \cong 0$ and each $P_n$ is projective. Then each $Z_{n} = \partial(P_{n+1})$ is projective....
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Show gl dim$R=\sup\{pd_RI\}$, where I runs through all right ideals of $R$, where $R$ is not Wedderburn

Question: Show gl dim$R=\sup\{pd_RI\}+1$, where I runs through all right ideals of $R$, where $R$ is not Wedderburn. Thoughts: I know that gl dim$R=1$ iff $R$ is a hereditary ring, but not Wedderburn. ...
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If $N$ is nilpotent ideal of $R$, then global dimension of $R$ is the supremum of the projective dimension of $V$ over $R$ such that $VN=0$

Question: If $N$ is nilpotent ideal of $R$, then global dimension of $R$ is the supremum of the projective dimension of $V$ over $R$ such that $VN=0$ My thoughts: I (hopefully) know pd$_RV$ is the ...
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proving $ \mathbb{Z}/12 \not\cong \mathbb{Z}/6 \oplus \mathbb{Z}/2$ as $\mathbb{Z}/12$ modules

I know this is false for $\mathbb{Z}$ modules, by the fundamental theorem of finite abelian groups. I don't know if this still holds true when they are taken as modules over $\mathbb{Z}/12$.
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Finitistic dimension conjeture for $A^{op} $ implies the strong Nakayama conjecture for A

I have some trouble with some detail in the proof of the following theorem. Assume that the Finitistic dimension conjecture is true for $ A^{op} $ that is $ sup\{ proj.dim(M) \vert M \in mod(A^{op}) ~...
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rank$(R\otimes_R S)$ is the composition of $f^∗$ and rank$(P)$.

This is Weibel's K-book exercise 2.12 If $f : R →S $ is a homomorphism of commutative rings, there is a continuous map $f^*: Spec(S) → Spec(R)$ sending $p$ to $f^{-1}(p).$ If $P$ is a finitely ...
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If $F$ is free, then the functor $M \mapsto \operatorname{Hom}_A(F,M)$ is exact.

This is from Serge Lang's Algebra 3rd Edition (see the picture down below if you feel confused). My problem is, how to prove that if $F$ is free, then $M \mapsto \operatorname{Hom}_A(F,M)$ is exact. ...
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62 views

Property of transpose Tr in category of modules (Auslander-Reiten theory)

I am reading the book Elements of the Representation Theory of Associative Algebras: Volume 1 , and I have a question on its proof of main properties of transpose $Tr$, appearing in Auslander-Reiten ...
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If M is a finitely generated projective module then the Homomorphism ring is also projective

Ok so the problem is the following: $\newcommand{\Hom}{\operatorname{Hom}}$ Given $M$ a finitely generated and projective left $R$-module, prove that $\Hom_{R}(M,R)$ seen as a right $R$-module is ...
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Confusion in the equivalent statement of projective module

The following statements are equivalent If $P$ is an $R-$module, and if it sastisfies the following equivalent properties then it is called a projective module. $1$- Every exact sequence of the form $...
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Is the projective dimension of modules with finite projective dimension bounded?

I know that a Noetherian local ring $R$ has finite global dimension if and only if $R$ is regular in which case $\mathrm{gldim}{R} = \dim{R}$. Therefore, for regular rings, every module has projective ...
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Finiteness and extension of a ring of scalars

Let $A\subset B$ be rings and $P$ a projective left $A$-module. Write $\rho:A\rightarrow B$ for the canonical injection. Suppose $\rho^*(P)=\rho_*(B)\otimes_A P$ is finitely generated. The author says,...
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Question on injective envelopes

Let $A$ be an Artin algebra and let $\text{mod}(A)$ denote the category of finitely generated left $A$-modules. Let $S$ be a simple module in $\text{mod}(A)$ and let $\iota_S: S \rightarrow I(S)$ be ...
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An Exercise on projective covers and injective envelopes of simple modules over an Artin algebra

Let $A$ be an artin algebra and let $\text{mod}(A)$ denote the category of finitely generated left $A$-modules. Let $S$ be a simple module in $\text{mod}(A)$. Let $\iota_S: S \rightarrow I(S)$ and $\...
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Is a bounded complex of projective $R$-modules of finite type quasi-isomorphic to a bounded complex of free $R$-modules of finite type?

The question is the one in the title. Let $(P,d)$ be a bounded complex of projective $R$-modules. My attempt is to consider, since projective is summand of free, for every integer $i$ a free module $F^...
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56 views

Two equivalent definitions of projective module $P$ (one involving right inverse), how do we make sense of these “replacements” in the proof?

I'm going through Rosenberg's Algebraic K-theory and its applications. The following is a screenshot from the first few pages. The definition of projective left $R$-module $P$ is: for any surjective ...
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77 views

Proof that $\text{Hom}(E_1,F_1)\otimes\text{Hom}(E_2,F_2)$ and $\text{Hom}(E_1\otimes E_2,F_1\otimes F_2)$ are isomorphic

Let $C$ be a commutative ring and $E_1,E_2,F_1,F_2$ $C$-modules. Suppose $E_1$ and $F_1$ are projective and finitely generated. I want to show that the canonical mapping $$\lambda:\text{Hom}(E1,F1)\...
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Why does the short exact sequence for projective module split?

I want to understand the definition of surjective module in terms of splitting sequence. The definition says for a projective $R$-module $P$, the following short exact sequence $$0 \to A \xrightarrow{...
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K Theory: Book Recommendations

Good people! So I've been hoping to get into K Theory for a while now, and the book that I have been trying to use (and failing) has been Charles Weibel's book by that very title. The book itself isn'...
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Trivial Group Representation is Projective implies Semisimple

Let $G$ be a group and $\mathbb k$ a field. Then, let $\mathfrak 1$ be the trivial $\mathbb k G$-module. According to Lorenz's A Tour of Representation Theory, he states "it turns out $\mathfrak ...
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Condition for a module to be finitely generated projective

Let $A$ be a ring and $E$ a left $A$-module. Write $\theta$ for the canonical homomorphism from $E^*\otimes_AE$ into $\text{End}_A(E)$ such that $$\theta(x^*\otimes y)(x)=\langle x,x^*\rangle y$$ for ...
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83 views

Projective, semi-simple, non-singular modules [closed]

Can you give an example of a projective and semi-simple non-singular module?
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30 views

Computing the number of projective-injective modules

Let $A$ be a finite dimensional $\mathbb{K}$-algebra, where $\mathbb{K}$ is an algebraically closed field. How does one compute (homologically) the number of projective-injective $A$-modules? Maybe ...
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48 views

Understanding the Proof of Maschkes's Theorem

I am currently reading "A Course in Ring Theory" by Passman, and I know other questions about this proof have been asked here before, but I really want to understand the proof in the book. ...

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