LEC ALG4 3

Projective modules

Proposition 386.1(Proposition).

Let $\alpha :M\to N$ be a surjective $R$-module homomorphism, and let $\phi :F\to N$ be a $R$-module homomorphism with $F$ free. Then, there exists an $R$-module homomorphism $\tilde{\phi }:F\to M$ such that the following diagram commutes.

Proof.

Let $\{e_i:i\in I\}\subseteq F$ be a basis of $F$. For all $i\in I$, let $x_i\in M$ be such that $\alpha (x_i)=\phi (e_i)$ (possible since $\alpha$ is surjective). Extend the map $e_i\mapsto x_i$ to an $R$-module homomorphism $\tilde{\phi }:F\to M$. It is now clear that $\alpha \circ \tilde{\phi }=\phi$.□

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Modules with the property described in Prp 1 are called projective modules.

Definition 386.2(Projective modules).

A $R$-module $P$ is said to be projective if for all surjective $\alpha :M\twoheadrightarrow N$ and $R$-module homomorphisms $\phi :P\to N$ there exists an $R$-module homomorphism $\tilde{\phi }:P\to M$ such that the following diagram commutes:

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Clearly, free modules are projective.

Warning

Prp 1 and Def 2 are not describing universal properties! As is evident by the proof of Prp 1, $\tilde{\phi }$ does not have to be unique.

Corollary 386.3(Corollary).

Let $\alpha :M\twoheadrightarrow P$ be a surjective $R$-module homomorphism with $P$ projective. Then there exists an $R$-module homomorphism $\beta :P\to M$ such that $\alpha \circ \beta ={\text{id}}_P$.

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Proposition 386.4(Proposition).

Let $P$ be a projective module. Let $\alpha :M\twoheadrightarrow P$ be a $R$-module homomorphism, and $\beta$ be supplied by Def 2 such that $\alpha \circ \beta ={\text{id}}_P$. Then,

1. $\alpha$ is surjective and $\beta$ is injective.

2. $M=\ker \alpha \oplus \text{im}\beta$.

Proof.

$(1)$ is just set theory.

$(2)$ Let $x\in M$. Write $y=\beta \alpha (x)$. Since $\alpha (x-y)=0$, we have $x-y\in \ker \alpha$. So, $x$ can be expressed as $x=(x-y)+y$. Next, let $x\in \ker \alpha \cap \text{im}\beta$ and let $y\in P$ be such that $x=\beta (y)$. Then, $0=\alpha (x)=\alpha \beta (y)=y$. Thus, $x=0$. We are done by Thm 99.1.□

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Proposition 386.5(Proposition).

1. If $P$ is projective, there exists $Q$ such that $P\oplus Q\cong F$ for free $F$.
2. If $P\oplus P^{\prime }$ is projective, then $P$ and $P^{\prime }$ are projective.

Thus, $P$ is projective iff it is a direct summand of a free module.

Proof.

$(1)$ follows from Prp 373.13 and Prp 4.

For $(2)$, Let $\phi :P\to N$ and $\alpha :M\twoheadrightarrow N$ be given as in Def 2. The trivial map from $P^{\prime }$ to $N$ together with $\phi$ determines a map ${\phi }^{\prime }:P\oplus P^{\prime }\to N$ such that $\phi ={\phi }^{\prime }\circ i_P$. Since $P\oplus P^{\prime }$ is projective, there exists $\tilde{\phi }:P\oplus P^{\prime }\to M$ such that ${\phi }^{\prime }=\alpha \circ \tilde{\phi }$. Thus, $\alpha \circ (\tilde{\phi }\circ i_P)=(\alpha \circ \tilde{\phi })\circ i_P={\phi }^{\prime }\circ i_P=\phi$. Thus, $P$ is projective.

A symmetric argument shows that $P^{\prime }$ is projective.□

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Rank of a free module, à la Kummini

Here’s how Kummini proved Thm 373.9⇒.

We first prove the invariance of basis cardinality for vector spaces.

Proposition 386.6(Proposition).

Let $V$ be a $k$-vector space with bases $X$ and $Y$. Then $|X|=|Y|$.

Proof.

We have seen the case when $|X|$ and $|Y|$ are finite in Thm 74.2. The case of one being finite and the other being infinite is ruled out by Lem 74.5. So, assume $|X|$ and $|Y|$ are infinite.

For all $x\in X$, there exists a finite subset $Y_x$ of $Y$ such that $x\in ⟨Y_x⟩$. Therefore, ${\bigcup }_{x\in X}{Y}_x$ generates $V$. Since no proper subset of $Y$ can generate $V$, we have

$$Y=\bigcup _{x\in X}{Y}_x.$$

Since each $Y_x$ is finite, basic set theory says $|Y|⩽|X|$. Reverse the argument to obtain $|X|=|Y|$.□

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Proposition 386.7(Proposition).

Let $F$ be a free $R$-module with bases $X$ and $Y$. Then $|X|=|Y|$.

Proof.

Let $I$ be an ideal of $R$. Define $IF$ to be the submodule of $F$ generated by $\{rm:r\in I,m\in F\}$:

$$\begin{array}{r}IF:=\left\{\sum _{i=1}^n{r}_i{m}_i:n\in \mathbb{N},r_i\in I,m_i\in F\right\}.\end{array}$$

Observe that $F/IF$ is an $R/I$-module (this is not true in general; see Rmk 8): If $r_1+I=r_2+I$,

$$\begin{array}{r}(r_1+I)(m+IF)=(r_{1}m+IF)=(r_{2}m+IF)\stackrel{!}{=}(r_2+I)(m+IF).\end{array}$$

For $r\in R$ and $m\in F$, we may denote $r+I$ by $\overline{r}$ and $m+IF$ by $\overline{m}$.

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We now claim that $\overline{X}:=\{\overline{x}:x\in X\}$ is a basis of $F/IF$ as an $R/I$-module.

Firstly, if $Z\subseteq F$ generates $F$ as an $R$-module, then $\{\overline{z}:z\in Z\}$ generates $F/IF$ as an $R$-module, and hence as an $R/I$-module: For $m+IF\in F/IF$, with a little abuse of notation we can write

$$\begin{array}{rlr}m& =\sum _{i=1}^n{r}_i{z}_i& z_i\in Z\\ ⟹m+IF& =\sum _{i=1}^n{r}_i(z_i+IF)& \\ ⟹m+IF& =\sum _{i=1}^n(r_i+I)(z_i+IF).& \end{array}$$

Thus, $\overline{X}$ generates $F/IF$ as an $R/I$-module.

It remains to show that $\overline{X}$ is linearly independent. Fix $n\in \mathbb{N}$. Let $r_i\in R$ for $1⩽i⩽n$, $x_i\in X$ for $1⩽i⩽n$, such that

$$\sum _{i=1}^n{\overline{r}}_i{\overline{x}}_i=0.$$

Equivalently, ${\sum }_{i=1}^n{r}_i{x}_i\in IF$. Thus, ${\sum }_{i=1}^n{r}_i{x}_i$ can be expressed in the form ${\sum }_{i=1}^{k^{\prime }}{s}_i^{\prime }{m}_i$ with $s_i^{\prime }\in I$ and $m_i\in F$. Replacing each $m_i$ with its expression in the basis $X$, we get

$$\begin{array}{rlr}\sum _{i=1}^n{r}_i{x}_i=\sum _{i=1}^k{s}_i{\tilde{x}}_i.& & {\tilde{x}}_i\in X\end{array}$$

Since $X$ is a basis, we have $n=k$, and (WLOG) ${\tilde{x}}_i=x_i$ and $s_i=r_i$. Thus, $r_i\in I$, and hence ${\overline{r}}_i=0$ for all $i$¹.

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Using Thm 111.1, let $I$ be a maximal ideal of $R$. This makes $R/I$ a field and $F/IF$ an $R/I$-vector space. We have shown $\overline{X}$ and (similarly defined) $\overline{Y}$ are bases of $F/IF$ over $R/I$. The final result follows from Prp 6.□

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Remark 386.8(Remark).

Every $R/I$ module is naturally an $R$-module: $(a,x)\mapsto \overline{a}x$. In general, if $\phi :R\to S$ is a ring map and $M$ is an $S$-module, we can give $M$ an $R$-module structure by $(a,x)\mapsto \phi (a)x$. This is the unique module structure for $M$ as an $R$ module which is compatible with $\phi$.

The converse is generally false. For example, $R$ considered an an $R$-module cannot have a compatible $R/I$-module structure for any nonzero ideal $I$. If $y\in I$ is nonzero, then $(y,1_R)$ evaluates to $y$ under the $R$-module structure, while the action of $y+I=0_{R/I}\in R/I$, forced to be the trivial action in any $R/I$-module structure on $R$, yields $(y+I,1_R)=0_R$.

However, if $M$ is an $R$-module such that for all $x\in M$ and for all $a\in I$, $ax=0$², then the prescription

$$\begin{array}{rl}& R/I\times M\to M\\ & (\overline{a},x)\mapsto ax\end{array}$$

makes $M$ a compatible $R/I$-module.

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Let $M_1$ and $M_2$ be $S$-modules. Let $\phi :R\to S$. When can we say $M_1{\otimes }_R{M}_2=M_1{\otimes }_S{M}_2$?

Footnotes

1. Notice the claim this is implicitly proving: no element of $X$ lies in $IF$. If this were true, say $x^{\prime }=im\in X$ for some $i\in I$ and $m\in F$, then $x^{\prime }=im=i\left({\sum }_{i=1}^n{r}_i{x}_i\right)$ for some $x_1,\dots ,x_n\in X$. On the RHS, $i$ divides every coefficient. Thus, moving $x^{\prime }$ to the RHS yields a trivial linear combination. ↩

2. i.e, $I\subseteq {\text{Ann}}_R(M)$ - as is the case in the proof of Prp 7 ↩