Projective, Injective and Flat Modules

Definition

Let $M$ be an $R$-module.

• $M$ is called projective $R$-module if ${\mathrm{Hom}}_R(M,\cdot )$ is exact.
• $M$ is called injective $R$-module if ${\mathrm{Hom}}_R(\cdot ,M)$ is exact.
• $M$ is called flat $R$-module if $M{\otimes }_R\cdot$ is exact.

Theorem

The following statements about a left $R$-module $P$ are equivalent.

• Whenever there is an $R$-epimorphism $M\stackrel{g}{\to }N$, and an $R$-homomorphism $P\stackrel{\gamma }{\to }N$, there exists an $R$-homomorphism $\overline{\gamma }:P\to M$ such that $\gamma =g\overline{\gamma }$. (lifting property)
• For each epimorphism $f:{}_{R}M\to {}_{R}N$, the map ${\mathrm{Hom}}_R(P,M)\to {\mathrm{Hom}}_R(P,N)$ is an epimorphism.
• For each bimodule structure ${}_R{P}_S$, the functor ${\mathrm{Hom}}_R(P,\cdot ):R\text{mbox}-module\to S\text{mbox}-module$ is exact.
• For every exact sequence $0\to M\stackrel{f^{\prime }}{\to }M\stackrel{g}{\to }{M}^{\prime \prime }\to 0$ in $R$-module, the sequence $0\to {\mathrm{Hom}}_R(P,M^{\prime })\stackrel{{\mathrm{Hom}}_R(P,f)}{\to }{\mathrm{Hom}}_R(P,M)\stackrel{{\mathrm{Hom}}_R(P,g)}{\to }{\mathrm{Hom}}_R(P,M^{\prime \prime })\to 0$ is exact.

In fact, $P$ is a projective $R$-module.

\begin{proof} i)→ii) Assume that i) holds, and $f:M\to N$ is an epimorphism. For any $\gamma \in {\mathrm{Hom}}_R(P,N)$, there exists $\overline{\gamma }:P\to M$ such that $f\circ \overline{\gamma }=\gamma$. Thus, the map

$${\mathrm{Hom}}_R(P,M)\to {\mathrm{Hom}}_R(P,N),\overline{\gamma }\mapsto f\circ \overline{\gamma }$$

is surjective and so ii) holds.

ii)→i) Assume that ii) holds, then for any $g:M\to N$, the induced map $\phi :{\mathrm{Hom}}_R(P,M)\to {\mathrm{Hom}}_R(P,N),\tau \mapsto g\circ \tau$ is surjective. For any $\gamma \in {\mathrm{Hom}}_R(P,N)$, there exists $\overline{\gamma }\in \mathrm{Hom}(P,M)$ such that $g\circ \overline{\gamma }=\gamma$ by $\phi$ surjective. Thus $\overline{\gamma }$ is what we desired and we proved i).

ii)←>iii) Recall that the functor $F={\mathrm{Hom}}_R(P,\cdot )$ is always left-exact, so it remains to show that for every epimorphism $g:M\to M^{\prime \prime }$, the induced map ${\mathrm{Hom}}_R(P,g):{\mathrm{Hom}}_R(P,M)\to {\mathrm{Hom}}_R(P,M^{\prime \prime })$ is also an epimorphism. It is exactly ii). Another direction is similar.

Finally, iv) is equivalent to iii). Now we finish the proof. \end{proof}

Theorem

The following statements about a (left) $R$-module $P$ are equivalent:

• $P$ is projective.
• Every epimorphism $M\stackrel{f}{\to }P$ splits, that is, there exists $R$-homomorphism $g:P\to M$ such that $fg=1_P$.
• $P$ is isomorphic to a direct summand of a free left $R$-module.

\begin{proof} i) → ii) Take $\gamma ={\mathrm{id}}_P:P\to P$ and apply ^9f21f9 i).

ii)→iii) Note that every module is a quotient of free modules, then by ^qtklfv we finish the proof.

iii)→i) Assume that $P\oplus Q=F$ is a free $R$-module. Note that the free module $F={\oplus }_{i\in I}R$ is projective as the functor

$$M\mapsto {\mathrm{Hom}}_R(F,M)=\prod _{i\in I}{\mathrm{Hom}}_R(R,M)=\prod _{i\in I}M$$

is exact. Then ${\mathrm{Hom}}_R(F,\cdot )={\mathrm{Hom}}_R(P,\cdot )\times {\mathrm{Hom}}_R(Q,\cdot )$ as functors, hence both $P$ and $Q$ are projective. \end{proof}

Corollary

A finitely generated projective $R$-module is a direct summand of $R^n$ for some $n\in {\mathbb{N}}_{+}$.

\begin{proof} It is a direct corollary of ^3rkmd6, iii). \end{proof}

Schanuel's lemma

Let $R$ be a ring with identity. If $0\to K\to P\to M\to 0$ and $0\to K^{\prime }\to P^{\prime }\to M\to 0$ are short exact sequences of $R$-modules and $P$ and $P^{\prime }$ are projective, then $K\oplus P^{\prime }$ is isomorphic to $K^{\prime }\oplus P$.

\begin{proof} Define the following submodule of $P\oplus P^{\prime }$, where $\varphi :P\to M$ and ${\varphi }^{\prime }:P^{\prime }\to M$

$$X=\left\{(p,q)\in P\oplus P^{\prime }:\varphi (p)={\varphi }^{\prime }(q)\right\}.$$

The map $\pi :X\to P$, where $\pi$ is defined as the projection of the first coordinate of $X$ into $P$, is surjective. Since ${\varphi }^{\prime }$ is surjective, for any $p\in P$, one may find a $q\in P^{\prime }$ such that $\varphi (p)={\varphi }^{\prime }(q)$. This gives $(p,q)\in X$ with $\pi (p,q)=p$. Now the kernel of the map $\pi$ is

$$\begin{array}{rl}\ker \pi & =\{(0,q):(0,q)\in X\}\\ & =\left\{(0,q):{\varphi }^{\prime }(q)=0\right\}\\ & \cong \ker {\varphi }^{\prime }\cong K^{\prime }.\end{array}$$

We may conclude that there is a short exact sequence

$$0\to K^{\prime }\to X\to P\to 0.$$

Since $P$ is projective and this sequence splits, so $X\cong K^{\prime }\oplus P$ by ^3rkmd6. Similarly, we can write another map ${\pi }^{\prime }:X\to P^{\prime }$, and the same argument as above shows that there is another short exact sequence

$$0\to K\to X\to P^{\prime }\to 0$$

and so $X\cong P^{\prime }\oplus K$. Combining the two equivalences for $X$ gives the desired result. \end{proof}

Remark. This argument is very similar to the proof of Goursat’s lemma.