# Thursday November 7th

## Projective Modules

**Definition:**
A **projective** module $P$ over a ring $R$ is an $R\dash$module such that the following diagram commutes:

```{=latex}
\begin{center}
\begin{tikzcd}
&  & P \arrow[dd, "f"] \arrow[lldd, "\exists \phi", dashed] \\
&  &                                                        \\
M \arrow[rr, "g"] &  & N
\end{tikzcd}
\end{center}
```

i.e. for every surjective map $g:M \surjects N$ and every map $f: P \to N$ there exists a lift $\phi: P \to M$ such that that $g \circ \phi = f$.

**Theorem**:
Every free module is projective.

*Proof:*
Suppose $M \surjects N \to 0$ and $F \mapsvia{f} N$, so we have the following situation:

```{=latex}
\begin{center}
\begin{tikzcd}
&  & x \arrow[d, hook] \arrow[llddd, dotted]                &  &   \\
&  & F \arrow[dd, "f"] \arrow[lldd, "\exists \phi", dashed] &  &   \\
&  &                                                        &  &   \\
M \arrow[rr, "g", two heads] &  & N \arrow[rr, dotted]                                   &  & 0
\end{tikzcd}
\end{center}
```

For every $x\in X$, there exists an $m_x \in M$ such that $g(m_x) = f(i(x))$.
By freeness, there exists a $\phi: F \to M$ such that this diagram commutes.

$\qed$

**Corollary:**
Every $R\dash$module is the homomorphic image of a projective module.

*Proof:*
If $M$ is an $R\dash$module, then $F \surjects M$ where $F$ is free, but free modules are surjective.

$\qed$

**Theorem:**
Let $P$ be an $R\dash$module. Then TFAE:

a. $P$ is projective.

b. Every SES $0 \to M \to N \to P \to 0$ splits.

c. There exists a free module $F$ such that $F = P \oplus K$ for some other module $K$.

*Proof:*

$a \implies b$:

We set up the following situation, where $s$ is produced by the universal property:

```{=latex}
\begin{center}
\begin{tikzcd}
               &             &                        & P \arrow[d, "\id", two heads, hook] \arrow[ld, "\exists s"] & \\
0 \arrow[r]    & M \arrow[r] & N \arrow[r, two heads] & P \arrow[r]                                                 & 0
\end{tikzcd}
\end{center}
```

$\qed$

$b \implies c$:

Suppose we have $0 \to M \to N \to P \to 0$ a SES which splits, then $N \cong M \oplus P$ by a previous theorem.

$\qed$

$c\implies a$:

We have the following situation:

```{=latex}
\begin{center}
\begin{tikzcd}
&  & F = P \oplus K \arrow[dd, "\pi", bend right] \arrow[lldddd, "\exists h", dotted]               \\
&  & \\
&  & P \arrow[uu, "\iota", bend right] \arrow[dd, "f"] \arrow[lldd, "\phi = \iota \circ h", dotted] \\
&  & \\
M \arrow[rr, two heads]  &  & N
\end{tikzcd}
\end{center}
```

By the previous argument, there exists an $h: F\to M$ such that $g\circ h = f \circ \pi$.
Set $\phi = h\circ \iota$.

*Exercise*: Check that $g\circ \phi = f$.

$\qed$

**Theorem:**
$\bigoplus P_i$ is projective $\iff$ each $P_i$ is projective.

*Proof:*

$\implies$:
Suppose $\oplus P_i$ is projective.

Then there exists some $F = K \oplus \bigoplus P_i$ where $F$ is free.
But then $P_i$ is a direct summand of $F$, and is thus projective.

$\impliedby$:
Suppose each $P_i$ is projective.

Then there exists $F_i = P_i \oplus K_i$, so $F \definedas \bigoplus F_i = \bigoplus (P_i \oplus K_i) = \bigoplus P_i \oplus \bigoplus K_i$.
So $\bigoplus P_i$ is a direct summand of a free module, and thus projective.

$\qed$

> Note that a direct sum has *finitely many* nonzero terms.
Can use the fact that a direct sum of free modules is still free by taking a union of bases.

*Example of a projective module that is not free:*

Take $R = \ZZ_6$, which is not a PID and not a domain.
Then $\ZZ_6 = \ZZ_2 \oplus \ZZ_3$, and $\ZZ_2, \ZZ_3$ are projective $R\dash$modules.
By previous statements, we know these are torsion as $\ZZ\dash$modules, and thus not free.

## Endomorphisms as Matrices

> See section 7.1 in Hungerford

Let $M_{m, n}(\RR)$ denote $m\times n$ matrices with coefficients in $R$.
This is an $R\dash R$ bimodule, and since $R$ is not necessarily a commutative ring, these two module actions may not be equivalent.

If $m=n$, then $M_{n,n}(R)$ is a ring under the usual notions of matrix addition and multiplication.

**Theorem:**
Let $V, W$ be vector spaces where $\dim V = m$ and $\dim W = n$.
Let $\hom_k(V, W)$ be the set of linear transformations between them.

Then $\hom_k(V, W) \cong M_{m, n}(k)$ as $k\dash$vector spaces.

*Proof:*
Choose bases of $V, W$.
Then consider

\begin{align*}
T: V \to W \\
v_1 \mapsto \sum_{i=1}^n a_{1, i} ~w_i \\
v_2 \mapsto \sum_{i=1}^n a_{2, i} ~w_i \\
\vdots
\end{align*}


This produces a map
\begin{align*}
f: \hom_k(V, W) &\to M_{m, n}(k) \\
T &\mapsto (a_{i, j})
,\end{align*}

which is a matrix.

> *Exercise: Check that this is bijective.*

$\qed$

**Theorem:**
Let $M, N$ be free left $R\dash$modules of rank $m, n$ respectively.
Then $\hom_R(M, N) \cong M_{m, n}(R)$ as $R\dash R$ bimodules.

*Notation:*
Suppose $M, N$ are free $R\dash$modules, then denote $\beta_m, \beta_n$ be fixed respective bases.
We then write $[T]_{\beta_m, \beta_n} \definedas (a_{i, j})$ to be its *matrix representation*.

**Theorem**:
Let $R$ be a ring and let $V, W, Z$ be three free left $R\dash$modules with bases $\beta_v, \beta_w, \beta_z$ respectively.
If $T: V \to W, S: W\to Z$ are $R\dash$module homomorphisms, then $S \circ T: V \to Z$ exists and
$$
[S \circ T]_{\beta_v, \beta_z} = [T]_{\beta_v, \beta_w} [S]_{\beta_w, \beta_z}
$$

*Proof:*
Exercise.

Show that
$$
(S \circ T)(v_i) = \sum_j^t \sum_k^m a_{ik} b_{kj} z_j
.$$

$\qed$

## Matrices and Opposite Rings

Suppose $\Gamma: \hom_R(V, V) \to M_n(R)$ and $V$ is a free left $R\dash$module.
By the theorem, we have $\Gamma(T \circ S) = \Gamma(S) \Gamma(T)$.
We say that $\Gamma$ is an **anti-homomorphism**.

To address this mixup, given a ring $R$ we can define $R^{op}$ which has the same underlying set of $R$ but with the modified multiplication
$$
x \cdot y \definedas yx \in R
.$$

If $R$ is commutative, then $R \cong R^{op}$.

$\qed$

**Theorem**:
Let $R$ be a unital ring and $V$ an $R\dash$module.

Then $\hom_R(V, V) \cong M_n(R^{op})$ as rings.

*Proof*:
Since $\Gamma(S \circ T) = \Gamma(T) \Gamma(S)$, define a map
\begin{align*}
\Theta: M_{n, n}(R) &\to M_{n, n}(R^{op}) \\
A &\mapsto A^t
.\end{align*}

Then
$$
\Theta(AB) = (AB)^t = B^t A^t = \Theta(B) \Theta(A)
,$$
so $\Theta$ is an anti-isomorphism.

Thus $\Theta\circ \Gamma$ is an anti-anti-homomorphism, i.e. a usual homomorphism.

$\qed$

**Definition:**
A matrix $A$ is **invertible** iff there exists a $B$ such that $AB = BA = \id_n$.

**Proposition:**
Let $R$ be a unital ring and $V, W$ free $R\dash$modules with $\dim V = n, \dim W = m$.
Then

1. $T \in \hom_R(V, W)$ is an isomorphisms iff $[T]_{\beta_v, \beta_w}$ is invertible.

2. $[T\inv]_{\beta_v, \beta_w} = [T]_{\beta_v, \beta_w}\inv$.

**Definition:**
We'll say that two matrices $A, B$ are **equivalent** iff there exist $P, Q$ invertible such that $PAQ = B$.