inj proj modules
Definition
All modules will be assumed to be $A$-modules where $A$ is a commutative ring with identity.
Given any surjective map $$M\overset{\pi}{\longrightarrow} N\to 0$$ we say that the module $P$ is projective if for any map $\phi$ there exists a map $\psi$ that extends it:
Dually, given any injective map $$0\to L\overset{i}{\longrightarrow}M$$ we say that the module $I$ is injective if for all there always exists an extension
Properties
- $P$ is a projective module iff $P$ is a direct summand of free module, i.e. there exists $n$ such that $$A^n=P\oplus Q.$$
- Any finitely generated projective module over a local ring is free.
Injective modules
- (Existence of injective modules) For all $M$ there exists an injective module $I$ such that $M\hookrightarrow I$.
- (Criterion for injectivity) Assume that for any ideal $I$, we have that the inclusion $I\hookrightarrow A$ and any homomorphism $j:I\to Q$ extends to a map $\tilde{j}:A\to Q$. Then $Q$ is an injective module.
- An abelian group $Q$ (which is a $\mathbb{Z}$-module) is injective iff $Q$ is divisible by elements of $\mathbb{Z}$, i.e. For all $q\in Q$ and all $N\in \mathbb{Z}$ there exists $\tilde{q}\in Q$ such that $\tilde{q}N=q$.
- Any $\mathbb{Z}$-module is embedded in an injective module, i.e. any abelian group is embedded in a divisible group.
- Let $\phi:A\to B$ be a homomorphism of rings, and $Q$ an injective $A$-module. Then $\tilde{Q}=\text{Hom}_A(B,Q)$ is an injective $B$-module, where $$\text{Hom}_A(B,Q)={\psi:B\to Q\mid \psi(ab)=a\cdot\psi(b)}$$
- Any $A$ module can be put in some injective module.
- Let $M$ be an $A$-module and $\bar{M}$ be the abelian group obtained by forgetting the module structure on $M$. We claim that $\bar{M}$ is injective, which by example two it suffices to show that it is divisible.
- Then $\text{Hom}_\mathbb{Z}(A,\bar{M})$ is an injective module by property 4
Examples
- (Proj) The free module $P=A^n$ is projective: it suffices to consider where to send the basis of $P$.
- (inj) divisible group $\mathbb{Z}/(n)$