# Wednesday January 29th

Coming up: the modules $\bigoplus \ZZ, \hom_R(M, N), M \tensor_R N$, as well as various properties:

- Torsion
- Torsionfree
- Free
- Projective
- Flat
- Injective
- Divisible

We have a series of implication
$$
\text{free} \implies \text{projective} \implies \text{flat} \implies \text{torsionfree}
$$

## Universal Mapping Properties

Definition (Direct Product of Modules)
:   For a collection $\theset{M_i}$ of modules, the **direct product** is characterized by

    \begin{center}
    \begin{tikzcd}
    &  &                                                     &  & M_j \\
    N \arrow[rrrru, "\varphi_j", bend left] \arrow[rrrrd, "\varphi_k"', bend right] \arrow[rr, "\exists!", dotted] &  & \prod M_i \arrow[rrd, "\pi_k"] \arrow[rru, "\pi_j"] &  &     \\
    &  &                                                     &  & M_k
    \end{tikzcd}
    \end{center}

    Here we define the canonical projection by $\pi_j(m_1, \cdots, m_j, \cdots) = m_j$.

Fact
: $$\hom_R(N, \prod M_i) = \prod \hom_R(N, M_i)$$

Definition (Direct Sum of Modules)
:   For a collection $\theset{M_i}$ of modules, the **direct sum** is characterized by

    \begin{center}
    \begin{tikzcd}
    M_j \arrow[rrd, "\iota_i", hook] \arrow[rrrrd, "\varphi_j", bend left] &  &                                      &  &   \\
    &  & \bigoplus M_i \arrow[rr, "\exists!"] &  & N \\
    M_k \arrow[rru, "\iota_k"] \arrow[rrrru, "\varphi_k"', bend right]     &  &                                      &  &  
    \end{tikzcd}
    \end{center}

    Here we define the canonical *injection* by $\iota_j(m) = (0, 0, \cdots, m, 0, \cdots)$.

In this case, we can define $\phi(m_1, m_2, \cdots, m_i, \cdots) = \sum \phi_i(m_i)$, which makes sense because cofinitely many of the terms in this sum are zero.

Fact
: $$\hom_R\qty{\bigoplus_{s\in S} R, N} = \prod_{s\in S} \hom_R(R, N) = N^S$$

Fact
: $\hom_R(R, N) \cong N$ via the map $f\in\hom(R, N) \mapsto f(1)$.


## Free Modules

Definition (Spanning, Linear Independence, and Basis)
:   For $M$ an $R\dash$module and $S\subset M$,

    1. $S$ *spans* $M$ if $\generators{S} = M$, where $\generators{S}$ is the set of all finite linear combinations of elements in $S$.

    2. $S$ is $R\dash$linearly independent iff $\sum r_i m_i = 0 \implies r_i = 0$ for all $i$.

    3. $S$ is a *basis* for $M$ iff $S$ is a spanning $R\dash$linearly independent subset of $M$.

    If $M$ admits a basis, $M$ is said to be *free*.

Theorem (Free Modules are Quotients)
:   \hfill
  
    a. If $S = \theset{s_i}$ is a basis for $M$, then there is a surjection
    \begin{align*}
      \bigoplus_{s\in S} R &\to M \\
      r_i &\mapsfrom \sum r_i s_i
    .\end{align*}
    
    b. For any set $S$, the module $\bigoplus_{s\in S} R$ has a canonical basis
    $$
    \vector{e}_s = (0, 0, \cdots, 0, 1, 0, \cdots, 0)
    $$

    c. If $\phi: \bigoplus _{s\in S} R \to M$ is an isomorphism, then $\theset{\phi(\vector e_s)}_{s\in S}$ is a basis for $M$.

Fact
: Let $F$ be a free $R\dash$module, then $\ann(F) = R$ if $F = (0)$ and 0 otherwise.
  Moreover,

  - $\ann(\bigoplus M_i) = \intersect \ann(M_i)$
  - $\ann(R) = \theset{0}$

Proposition (Characterization of Freeness in terms of Rings)
:   For a ring $R\neq 0$, TFAE:
  
    a. Every $R\dash$module is free
    b. $R$ is a field

Proof 
:   $a \implies b$: 
    If $R$ is not a field, then $0 < I \normal R$ is proper, and since $\ann(R/I) = I$, we have $0 < \ann(R/I) < R$.
    So $\ann(R/I)$ is proper, and $R/I$ is thus not a field.

    The reverse implication is linear algebra.
    Every vector space has a basis by AOC (note that this is equivalent to Zorn's Lemma).

Fact
: Every $R\dash$module $N$ is the quotient of a free module.

This follows by taking the generating set $S = N$, then $\bigoplus_{n\in N}R \surjects N$ using a previous fact.

Fact
: $N$ is quotient of a finitely generated free module iff $N$ is finitely generated.

Exercise
: Show that for $0 \to A \to B \to C \to 0$ a SES of $R\dash$modules,

  a. If $A, C$ are finitely generated, then so is $B$.

  b. If $B$ is finitely generated, then so is $C$.

Example
: It is possible for $B$ to be finitely generated with $A < B$ and $A$ not finitely generated.
  Let $R$ be non-Noetherian.
  Equivalently, there exists $I\normal R$ that is not finitely generated.
  So take $B = R$ and $A = I$.
  For example, take $M = C([0, 1], R)$ the module of continuous functionals, which is non-Noetherian.

Examples of non-Noetherian rings:

1. $\theset{R_i}$ where each $R_i$ is infinite and ???; then $\prod R_i$ is non-Noetherian.

2. For $k$ a field, $T = \theset{t_n \suchthat 1\leq n < \infty}$, take $R = k[T]$.
   Then $I = \generators{T}$ is not finitely-generated.

Fact
: If $R$ is a Noetherian ring, then every finitely generated $R\dash$module is a Noetherian module.

Example
: Take $R, M = \ZZ$, which are free modules, and $S = \theset{2}$.
  Note that $S$ is $R\dash$linearly independent in $M$, but can not be extended to a basis, and $\generators{S} = 2\ZZ \neq \ZZ$.
  Similarly, $S' = \theset{2, 3}$ can not be reduced to a basis, while $\generators{S'} = \ZZ$.

Question: can $M$ have basis sets of different cardinalities?
Answer: sometimes, commutative rings have the *invariant basis property*.