# Friday February 7th

## Projective Modules

For $X$ a topological space and $\pi:E\to X$ a real vector bundle on $X$.
Then $$\Gamma(E, X) = \theset{\sigma: X\to E \suchthat \pi \circ \sigma = \id_X}$$ is naturally a module over the ring $C(X, \RR)$ of continuous real-valued functions.
For $p\in X$, the fibers $\sigma(p) \in \pi\inv(p)$ are vector spaces, and we can consider $f(p)\sigma(p)$ for any $f\in C(X, \RR)$.
For trivial bundles $\RR^n \cross X \mapsvia{\pi} X$ with a global section

\begin{align*}
\sigma: X &\to \RR^n \cross X \\
p &\mapsto (\tilde\sigma(p), p)
.\end{align*}

Then $\tilde\sigma: X \to \RR^n$, or equivalently a collection of $n$ continuous functions $\tilde\sigma_j \to \RR$.
Thus $\Gamma(X, E) \cong C(X, \RR)^n$.

Theorem (Swan)
:   Suppose $X$ is compact. 
    Then

    a. $\Gamma(X, E)$ is a finitely generated projective $C(X, \RR)\dash$module, i.e. $\pi$ is a direct summand of a trivial vector bundle on $X$, and

    b. There is an equivalence of categories between vector bundles on $X$ and finitely generated projective $C(X, \RR)\dash$modules.

Example
: Let $X$ be the two points space $\theset{1, 2}$.
  Take a 0-dimensional vector space over $1$ and a 1-dimensional vector space over $2$.

  ![Image](figures/2020-02-07-11:30.png)\

Remark
: Such cheap examples exist on $X$ iff $X$ is disconnected.

Definition (Splitting an Exact Sequence)
: Recall that it $0 \to A \to B \mapsvia{f} C \to 0$ is exact, then a **splitting** is a map $\sigma: C\to B$ such that $f\circ \sigma = \id_C$.
  Then $B = A \oplus \sigma(C) \cong A \oplus C$.

Exercise
: Take $R=\ZZ$ and find a SES such that $B\cong_\ZZ A \oplus B$ but the sequence is *not* split.

Definition (Projective Module)
: A module $P$ is **projective** iff $0 \to M \to N \to P \to 0$ is split.

Exercise
: Show that free implies projective.

> Hint: Lift basis and use universal property.

Theorem (Projective is Direct Summand of Free)
: If $P$ is projective, then there exists a $K$ such that $P\oplus K$ is free.

Idea: summands can be *both* a submodule and a quotient module.

Proof
: Choose a free $F$ and an $R\dash$module surjection $q:F \surjects P$ with $K = \ker q$ to obtain $0 \to K \to F \to P \to 0$.
  Since $P$ is projective, this sequence splits and thus $F \cong K \oplus P$ is free.

*Comment:*
If $P$ is finitely generated, then we can take $K$ (and hence $F$) to be finitely generated module. 
A quotient of a finitely-generated module is also finitely generated, and $F \cong K \oplus P$.

Theorem (Lifting Property of Projectives)
:   If there exists a $K$ such that $P\oplus K$ is free, then $P$ satisfies this lifting property:

    \begin{center}
    \begin{tikzcd}
    &  & P \arrow[dd, "f"] \arrow[lldd, "\exists \tilde f", dotted] &  &   \\
    &  &                                                            &  &   \\
    M \arrow[rr] &  & N \arrow[rr]   &  & 0
    \end{tikzcd}
    \end{center}

Proof
:   Choose $K$ such that $P \oplus K$ is free, and let $\theset{f_i}_{i\in I}$ be a basis for $F$.
    Then write $F = P \oplus K$ and $f_i = p_i + k_i$ where $p_i \in P, k_i \in K$.
    Then we can construct a unique $g: F\to M$  by sending $f_i$ to $m_i$:

    \begin{center}
    \begin{tikzcd}
    &  & \{f_i\} \arrow[d, hook], &  &   \\
    &  & F = P\oplus K \arrow[dd, "\pi"] \arrow[lldddd, "\exists! g"', dotted] &  &   \\
    &  &                                                                       &  &   \\
    &  & P \arrow[dd, "f"] \arrow[uu, "{\iota(p) = (p, 0)}"', bend right=60]   &  &   \\
    &  &                                                                       &  &   \\
    M \arrow[rr, "q"] &  & N \arrow[rr]                                                          &  & 0 \\
    \{m_i\}           &  & \{n_i\}                                                               &  &  
    \end{tikzcd}
    \end{center}

    Then $q\circ g\circ \iota = (q\circ g) \circ \iota = (f\circ \pi) \circ \iota = f \circ (\pi \circ \iota) = f$ since $\iota$ is a section.

> Todo: Revisit!

This $P$ is projective iff

- Every length 2 resolution of $P$ splits.
- $P$ is a direct summand of a free module.
- $P$ satisfies this lifting property.

If $P$ satisfies this lifting property, we have:

\begin{center}
\begin{tikzcd}
&  &              &  &              &  & P \arrow[lldd, "\exists \sigma"'] &  &   \\
&  &              &  &              &  &                                   &  &   \\
0 \arrow[rr] &  & M \arrow[rr] &  & N \arrow[rr] &  & P \arrow[uu, "\id_P"'] \arrow[rr] &  & 0
\end{tikzcd}
\end{center}

Exercise
: Show free implies projective in as many ways as you can (using any of these properties).

Remark
: An easy consequence of the lifting property implies that the functor $M \mapsto \hom_R(P, M)$ is covariant and exact.

> Natural question: for any new property of modules, is there a class of rings for which this coincides with known properties?

*Question:*
How different is projective from free?

Free $\implies$ projective $\implies$ subfree $\implies_{R \text{ a domain }}$ torsion-free.

For $R$ a PID and $M$ finitely generated, these are all equivalent (hence the diminished importance of projectivity when studying the structure theorem).
Recall (Theorem 3.56) that if $R$ is PID, then subfree $\implies$ free and projective $\iff$ free, but $(\QQ, +)$ is torsion-free but not free.

> Recall $\spec(R_1 \cross R_2) = \spec R_1 \disjoint \spec R_2$

Example (Projective does not imply free):*
:   Let $R_1, R_2$ be rings and consider $R = R_1 \cross R_2$.
    Then recall that $I\normal R$ implies $I = I_1 \cross I_2$ for $I_i \normal R_i$.
    Take $M_1 \definedas R_1 \cross 0 \normal R$, and $M_2 \definedas 0 \cross R_2 \normal R$.

    Then $M_1 \oplus M_2 = M_1 + M_2 = R$, so both $R_i$ are projective.
    They are not free though, since $\ann M_1 = M_2$ and vice-versa.

Example
: Let $R = \CC \cross \CC$, so $\spec R = \theset{1, 2}$, then $M_1 = \CC\cross 0 \to \spec R$, and we can construct "cheap" bundles in analogy to the disconnected topological case.

*Next question:*
What is an example of a nonfree projective module over a domain.