Questions tagged [injective-module]

A module $I$ over a ring $R$ is injective if $\hom_{R}({-},I)$ is exact. The notion of injective modules is dual to the notion of a projective module. In homological algebra injective modules are used for computing right derived functors.

396 questions
23
votes
2 answers

A problem about an $R$-module that is both injective and projective.

Let $R$ be a domain that is not a field, and let $M$ be an $R$-module that is both injective and projective. Prove that $M= \left \{ 0 \right \}$. This is exercise 7.52 of Rotman's Advanced Modern Algebra. Using theorems before exercises, because…
12
votes
2 answers

Is there any relation of injective modules to free modules?

Projective modules are direct summands of free modules. As I perceive it, projections and injections are dual notions. Based on that, I was looking whether there is a relation of injective modules to free modules (similar to the natural relation of…
12
votes
1 answer

The relationship of free, divisible, projective, injective, and flat modules.

In general, we have that: free $\Rightarrow$ projective $\Rightarrow$ flat injective $\Rightarrow$ divisible ( ($\Rightarrow$) be ($\Leftrightarrow$) in PIDs) Simple Counter-examples: projective but not free: $\mathbb{Z}_2$ is $\mathbb{Z}_6$ …
12
votes
1 answer

Injective Modules Motivation & Intuition

A module $M$ over a commutative ring $R$ is called a 'injective module' if it satisfies certain universal property explaned here. Question: Is there any intuition how to think concretely about injective modules? Do them naturally arise as…
11
votes
0 answers

when are graded injective modules graded and injective?

Define a graded injective module over a graded ring $R$ to be an injective object in $GrMod-R$ (the category of right graded $R$-modules). From the little research I have done, a graded injective module is not necessarily injective. However, if it…
11
votes
1 answer

Divisible module which is not injective

On searching for some example of divisible module but not injective, I come across one in T.Y.Lam, Lectures on Modules and Rings. He considers the $\mathbb{Z}[x]$-module $M=\mathbb{Q}(x)/\mathbb{Z}[x]$, where $\mathbb{Q}(x)$ denotes the quotient…
user49685
  • 4,659
10
votes
2 answers

Injective resolutions of a complex

Let $\mathcal{A}$ be an abelian category, $M\in\mathcal{A}$. An injective resolution of $M$ is a quasi-isomorphism $M\longrightarrow I$, where $I$ is a complex of injective objects. This can be made more explicit: It is the same thing as an exact…
user114885
10
votes
1 answer

Let $R$ be a commutative domain with field of fractions $F$. Prove that $F$ is an injective $R$-module.

Let $R$ be a commutative integral domain with field of fractions $F$. Prove that $F$ is an injective $R$-module. I have tried to apply Baer's criterion: every $R$-module homomorphism from any ideal $I$ of $R$ can be extended to an $R$-module…
delueze
  • 545
9
votes
2 answers

On the existence of finitely generated injective modules (Bruns and Herzog, Exercise 3.1.23)

Suppose that $R$ is a local Noetherian ring. Suppose that there exists a non-zero finitely generated injective module $M$. How can I prove that $R$ is Artinian? It is easy if $R$ is Cohen-Macaulay, because we know that if there exists a nonzero…
9
votes
1 answer

$R/Ra$ is an injective module over itself

Let $R$ be a PID, $a\in R$ be a nonzero nonunit in $R$. Prove that $R/Ra$ is an injective module over itself. If $R$ is a PID, every $R$- divisible module is injective, but the question concerns with $R/Ra$-module, so I have no idea to solve this…
9
votes
1 answer

Construction of injective hulls without axiom of choice

Motivation: It is known that without the axiom of choice (AC), it is not provable that all categories of modules have enough injectives, let alone injective hulls. Still, there are examples of rings where one can explicitly write down 'candidate'…
Hanno
  • 20,120
  • 2
  • 36
  • 64
8
votes
1 answer

Integral domain with a finitely generated non-zero injective module is a field

Suppose that $R$ is a integral domain. Suppose that there exists a non-zero finitely generated injective module $M$. How can I prove that $R$ is field?
8
votes
1 answer

$Q$ is an injective module iff injections from $Q$ always split

Clarification of terminology: We say an injection $A \xrightarrow{i} B$ splits iff the induced short exact sequence $0 \to A \xrightarrow{i} B \to B/i(A) \to 0$ splits. Similarly, we say a surjection $B \xrightarrow{p} C$ splits iff $0 \to \ker p…
8
votes
1 answer

Prove that every $\mathbb{Z}/6\mathbb{Z}$-module is projective and injective. Find a $\mathbb{Z}/4\mathbb{Z}$-module that is neither.

I want to show that every $\mathbb{Z}/6\mathbb{Z}$-module is a direct sum of projective modules. As abelian group, $\mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/3\mathbb{Z}$, but is it the direct sum of $\mathbb{Z}/2\mathbb{Z}$…
8
votes
1 answer

Characterization of projective and injective modules

In Theorems 4.7 and 8.4 Hilton & Stammbach give two lists of 5 different characterizations of projective and injective modules, respectively. Even though I can follow the proofs they give, I'd like to get rid of the characterization involving free…
hjhjhj57
  • 4,211
1
2 3
26 27