Projective Module

一些很简单的projective module的内容, 比较有意思的是最后一个定理. 这个这个定理给出了local ring上的finite generated projective module一定是free的. 这是Kaplanski定理的一个特殊情况, 在这里的证明使用了Nakayama引理数维数.

~~(头图快不够了, 估计以后要用美少女来代替了 (ㄒoㄒ)/)~~

Projective Module

Definition 1. An $R$ -module $P$ is projective if for any exact sequence $A\xrightarrow{p}B\to 0$ and $f:P\to B$ , there exists a homomorphism $g:P\to A$ s.t. the following diagram commutes:

The homomorphism $p:A\to B$ will induce a homomorphism from $\mathrm{Hom}(P,B)$ to $\mathrm{Hom}(P,A)$ , i.e. $g$ is a lifting of $f$ .

First, we have: free module $F$ is projective. Take a basis $\{f_{i}\}$ of $F$ , from surjectivity of $p$ , $\exists a_{i}\in A$ s.t. $p(a_{i})=f(f_{i})$ . Then assign $f_{i}$ to $a_{i}$ and generate $g$ from it.

Theorem 1. The functor $\mathrm{Hom}(P,-)$ is exact if and only if $P$ projective.

Proof

Proof. For the exact sequence $0\to A\to B\to C\to 0$ , we have the sequence $0\to \mathrm{Hom}(P,A)\to \mathrm{Hom}(P,B)\to \mathrm{Hom}(P,C)$ exact. $B\to C\to 0$ exact, so $\mathrm{Hom}(P,B)\to \mathrm{Hom}(P,C)$ is surjective if and only if $P$ projective. ◻

The projective module is a direct summand of a free module, i.e. $P$ is projective if and only if $\exists M$ such that $P\oplus M$ is free module. We first show the theorem:

Theorem 2. An $R$ -module $P$ is projective if and only if the exact sequence $0\to A\to B\to P\to 0$ splits.

Proof

Proof. Consider the exact sequence $$0\to A\xrightarrow{f}B\xrightarrow{g}P\to 0$$

If $P$ projective, take the exact sequence $B\to P\to 0$ and $id_{P}:P\to P$ , it induces $h:P\to B$ with $g\circ h=id_{P}$ . So the sequence splits.

Conversely, suppose every exact sequence $0\to A\to B\to P\to 0$ splits. Consider any diagram

take the free presentation $0\to \mathrm{Ker}\pi\to F\xrightarrow{\pi} P\to 0$ of $P$ , by assumption

0\to \mathrm{Ker}\pi\to F\xrightarrow{\pi}P\to 0

splits. So $\exists h:P\to F$ s.t. $h\circ \pi=id_{P}$ . Since $F$ is projective, $\exists g:F\to B$ s.t. $f\circ \pi=p\circ g$

Thus $f=f\circ \pi\circ h=p\circ g\circ h$ , $P$ is projective. ◻

Theorem 3. $P$ is projective if and only if it’s a direct summand of a free $R$ -module $F$ .

Proof

Proof. If $P$ is projective, the projective presentation $0\to \mathrm{Ker}\pi\to F\xrightarrow{\pi}P\to 0$ splits, so $F=P\oplus \mathrm{Ker}\pi$ .

Conversely, using similar diagram chasing skill for

◻

A direct corollary is

Theorem 4. Direct summands and sums of projective modules are projective modules.

Proof

Proof. Omitted. ◻

We have shown that for projective module $P$ and any module $M$ , $\mathrm{Ext}_{R}^{1}(P,M)=0$ in the Ext functor and Tor functor

Theorem 5. If $\mathrm{Ext}_{R}^{1}(P,M)=0$ for every $R$ -module $M$ , then $P$ is projective.

Proof

Proof. For a free(projective) resolution

\cdots\to F_{1}\to F_{0}\to P\to 0

Take $M=\mathrm{Ker}(F_{0}\to P)$ , then $\mathrm{Ext}_{R}^{1}(P,M)=0$ . Take an element $\xi \in \mathrm{Ext}_{R}^{1}(P,M)$ given by the quotient of $\pi:F_{1}\to M$ . Since $\xi$ is zero, there is a map $s:F_{0}\to M$ s.t. $\pi=s\circ d_{1}$ , $d_{1}:F_{1}\to F_{0}$ . Thus $F_{0}=\mathrm{Ker}s\oplus \mathrm{Ker}(F_{0}\to P)=P\oplus \mathrm{Ker}(F_{0}\to P)$ . ◻

For projective module, it is not necessary to be free. For example, for the ring $\mathbb{Z} /\mathbb{6Z}$ , the ideal $(2)$ is projective since $(2)\oplus (3)\cong \mathbb{Z} /6\mathbb{Z}$ . Actually, any infinitely generated projective module over any Dedekind domain. We proof some weaker result:

Theorem 6. Any finite generated projective module $M$ of a local ring $R$ is free.

Proof

Proof. Take a minimal generating basis $m_{1},\dots m_{n}$ of $M$ , here minimal means the minimal number of generators. Let $F=\bigoplus\limits_{i=1}^{n}Re_{i}$ , and $\varphi:F\to M$ with $\varphi: \sum r_{i}e_{i}\mapsto \sum r_{i}m_{i}$ . Take $K=\mathrm{Ker}\varphi$ . If $\sum r_{i}m_{i}=0\in M$ , pass it to the quotient $M /\mathfrak{m}M$ , $\sum r_{i}\bar{m_{i}}=0\in M /\mathfrak{m}M$ (Recall $R /\mathfrak{m}\otimes_{R} M\cong M /\mathfrak{m}M$ !). So $r_{i}\in \mathfrak{m}$ . Hence $K\subset \mathfrak{m}F$ . Since $M$ is projective, the sequence

0\to K\to F\to M\to 0

splits. Thus $F\cong M\oplus K$ . Since $F$ and $M$ have same number of genenrators, $\dim_{R /\mathfrak{m}}(R/\mathfrak{m}\otimes_{R}F)=\dim_{R /\mathfrak{m}}(R /\mathfrak{m}\otimes_{R}M)\Rightarrow F /\mathfrak{m}F\cong M /\mathfrak{m}M$ .
We have

R /\mathfrak{m}\otimes_{R} F\cong F /\mathfrak{m}F\cong R /\mathfrak{m}\otimes_{R}(M\oplus K)\cong (R /\mathfrak{m}\otimes_{R}M)\oplus (R /\mathfrak{m}\otimes_{R}K)\cong M /\mathfrak{m}M\oplus K/\mathfrak{m}K

Thus $K /\mathfrak{m}K=0$ . Since $K$ is finite generated, by Nakayama lemma, $K=0$ . Thus $F=M$ . ◻

A more stronger result can be seen in Projective modules over a local ring.