§ Projective modules in terms of universal property

§ (1): Universal property / Defn

• $P$ is projective iff for every epimorphism $e: E \to B$, and every morphism $f: P \to B$, there exists a lift $\tilde{f}: P \to E$.

     e
   E ->> B
   ^   ^
  f~\  | f
     \ |
       P

§ Thm: every free module is projective

• Let $P$ be a free module. Suppose we have an epimorphism $e: M \to N$ and a morphism $f: P \to N$. We must create $\tilde f: M \to N$
• Let $P$ have basis $\{ p_i \}$. A morphism from a free module is determined by the action on the basis. Thus, we simply need to define $\tilde f(p_i)$.
• For each $f(p_i): N$, there is a pre-image $m_i \in M$ such that $e(m_i): N = f(p_i): N$.
• Thus, define $\tilde{f}(p_i) = m_i$. This choice is not canonical since there could be many such $m_i$.
• Regardless, we have succeeded in showing that every free module is projective by lifting $f: P \to N$ to a map $\tilde f: M \to N$.

§ (1 => 2): Projective as splitting of exact sequences

• $P$ is projective iff every exact sequence $0 \to N \to M \xrightarrow{\pi} P \to 0$ splits.
• That is, we have a section $s: P \to M$ such that $\pi \circ s = id_P$.

• PROOF (1 => 2): Suppose $P$ solves the lifting problem. We wish to show that this implies that exact sequence splits.
• Take the exact sequence:

            pi
0 -> N -> M -> P -> 0
               ^
               | idP
               P

• This lifts into a map $P \to M$ such that the composition is the identity:

            pi
0 -> N -> M -> P -> 0
          ^   ^
       idP~\  | idP
            \ |
             P

• This gives us the section s = idP~ such that pi . s = idP from the commutativity of the above diagram.

§ (2 => 3): Projective as direct summand of free module

• $P$ is projective iff it is the direct summand of a free module. So there is a another module $N$ such that $P \oplus N \equiv R^n$.
• We can always pick a surjective epi $\pi: F \to P$, where $F$ is the free module over all elements of $P$.
• We get our ses $0 \to ker(\pi) \to F \to P \to 0$. We know this splits because as shown above, projective splits exact sequences where $P$ is the surjective image.
• Since the sequence splits, the middle term $F$ is a direct sum of the other two terms. Thus $F \simeq \ker \pi \oplus P$.

§ Splitting lemma

• If an exact sequence splits, then middle term is direct sum of outer terms.

§ (3 => 1): Direct summand of free module implies lifting

• Let's start with the diagram:

  e
E ->>B
     ^
    f|
     P

• We know that $P$ is the direct summand of a free module, so we can write a P(+)Q which is free:

  e
E ->>B
     ^
    f|
     P <<-- P(+)Q
         pi

• We create a new arrow f~ = f . pi which has type f~: P(+)Q -> B. Since this is a map from a free module into B, it can be lited to E. The diagram with f~ looks as follows:

  e
E ->>B <--
     ^    \f~
    f|     \
     P <<-- P(+)Q
         pi

• After lifting f~ to E as g~, we have a map g~: P(+)Q -> E.

--------g~--------
|                |
v e              |
E ->>B <--       g~
     ^    \f~    |
    f|     \     |
     P <<-- P(+)Q
         pi

• From this, I create the map g: P -> E given by g(p) = g~((p, 0)). Thus, we win!

§ Non example of projective module

• Z/pZ is not projective.
• We have the exact sequence 0 -> Z -(xp)-> Z -> Z/kZ -> 0 of multiplication by p.
• This sequence does not split, because Z (middle) is not a direct summand of Z (left) and Z/kZ (right), because direct summands are submodules of the larger module. But Z/pZ cannot be a submodule of Z because Z/pZis torsion while Z is torsion free.

§ Example of module that is projective but not free

• Let $R \equiv F_2 \times F_2$ be a ring.
• The module $P \equiv F_2 \times \{0\}$ is projective but not free.
• It's projective because it along with the other module $Q \equiv \{0\} \times F_2$ is isomorphic to $R$. ( $P \oplus Q = R$).
• It's not free because any $R^n$ will have $4^n$ elements, while $P$ has only two element.
• Geometrically, we have two points, one for each $F_2$. The module $P$ is a vector bundle that only takes values over one of the points. Since the bundle different dimensions over the two points (1 versus 0), it is projective but not free.
• It is projective since it's like a vector bundle. It's not free because it doesn't have constant dimension.

§ References

• video