# Friday January 31st

## Tensor and Hom

Let $M, N$ be $R\dash$modules, then we define

\begin{align*}
\hom_R(M, N) \definedas \theset{f: M\to N \suchthat f\text{ is an $R\dash$module map }}
.\end{align*}

Recall that $R\dash$module maps satisfy

- $f: (M, +) \to (N, +)$ a morphism of abelian groups
- For all $r\in R$, for all $m\in M$, $f(rm) = rf(m)$.

Note that $\hom_R$ is a commutative group, and is in fact an $R\dash$module with structure given by $(r\cdot f) \cdot m \mapsto r f(m) = f(rm)$.

> Note that the proof of this fact uses commutativity in a key way.

Facts:

\begin{align*}
\hom_R(R, N) &= N \\
\hom_R\qty{\bigoplus_{s\in S} R_s, N} &= N^S \\
\hom_R(M, R) &\definedas M\dual
.\end{align*}


Note:
Infinite dimensional vector spaces over fields are never isomorphic to its dual. 

**Exercise:**
Think about $M\dual$ and $(M\dual)\dual$.

Recall the map

\begin{align*}
\iota: M &\to (M\dual)\dual = \hom_R(\hom_R(M, R), R) \\
x &\mapsto ( \ell: M\to R \mapsto \ell(x) \in R )
.\end{align*}

**Exercise:**
If $R = k$ is a field, then show that $\iota$ is injective iff $\dim M$ is finite.

Is this always injective? No! Counterexample:
Take $R = \ZZ$ and $M = \ZZ/p\ZZ$, then $M\dual = \hom_\ZZ(\ZZ/p\ZZ, \ZZ) = 0$.

It can also fail to be surjective in the infinite dimensional case -- the space $M\dual$ is strictly larger than $M$.

Definition (Reflexive Modules)
: $M$ is *reflexive* if $\iota: M \mapsvia{\sim} (M\dual)\dual$ is an isomorphism.

**Exercise:**
Show the following:

- If $M$ is free and finitely generated, then $M$ is reflexive.
- If $R=k$ is a field, then $M$ is reflexive iff $M$ is finitely generated.
- There exists a ring $R$ and a reflexive $R\dash$module $M$ that is not finitely generated.

## Free and Torsion Modules

Let $R$ be a domain, and for all $a\in R^\bullet$ the map $[a]:R\to R$ is injective, and $[a] \in \hom_R(R, R) = R$.

Definition (Torsion Submodule)
: $$M[\tors] \definedas \theset{ m\in M \suchthat \ann(m) \neq (0) } \leq M$$ is the **torsion submodule** of $M$.

Definition (Torsion and Torsionfree Modules)
: $M$ is **torsion** iff $M = M[\tors]$, and $M$ is **torsionfree** iff $M[\tors] = (0)$.

Exercise
: Show that if $0 \to A \to B \to C \to 0$, then

	- Show that if $B$ is torsion then $A, C$ are torsion.
	- If $A, C$ are torsion, must $B$ be torsion?
	- Show that if $B$ is torsion-free then $A$ is torsion-free but $C$ need not be torsion-free.
	- If $A, C$ are torsion-free, must $B$ be torsion-free?

> Note: $0 \to \ZZ/2 \to \ZZ/4 \to \ZZ/2 \to 0$ is an extension that isn't a semidirect product!

Fact
: Free modules are torsion-free.

Note that we need to be in a domain to even talk about torsion.

Proposition (Torsionfree implies submodule of f.g. free)
: Let $R$ be a domain and $M$ an $R\dash$module.
	Then

	a. $M / M[\tors]$ is torsion-free.
	b. If $M$ is finitely generated, then $M$ is torsion free iff $M$ is isomorphic to a submodule of a finitely-generated free module.

Proposition (Implication Chain)
: Free $\implies$ projective $\implies$ flat $\implies_{R\text{ a domain }}$ torsion-free.

Proof (of (a))
: 	Let $x\in M/M[\tors]$ such that $\exists r\in R^\bullet$ such that $rx = 0$.
		Lift $x$ to $\tilde x\in M$, then $r\tilde x \in M[\tors]$.
		Then $\exists r' \in R^\bullet$ such that $$0 = r'(r\tilde x) = (r' r)\tilde x \definedas r_2\tilde x$$ for some $r_2 \neq 0$.
		But then $\tilde x\in M[\tors]$, and so $x = 0$ in $M/M[\tors]$.

Proof (of (b))
: Let $M = \generators{x_1, \cdots, x_r}$ with $r\geq 1$ and $x_i \neq 0$.
	After reordering, there exists some $s$ with $1\leq s \leq r$ such that $x_1, \cdots, x_s$ are $R\dash$linearly independent, and for all $i > s$, $\theset{x_j}_{j \leq s} \union x_i$ is linearly *dependent*.
	
	Then define $F\definedas \generators{x_1, \cdots, x_s}$; this is a finitely generated free module.
	If $r=s$, we done.

	Otherwise, $r< s$, then $\forall i > r$ there exists an $a_i \in R^\bullet$ such that $a_i x_i \in F$.
	So we can take $a \definedas a_{s+1} \cdots a_r \neq 0$; then $aM \subset F$.
	Since $M$ is torsion-free, the multiplication maps are injective, so $[a]: M \mapsvia{\cong} M \subset F$, so $M \injects F$ embeds $M$ into a free module.

Does this work with $M$ not finitely generated? No, we can't take an infinite product for $a$.
Is every torsion-free module a submodule of a free module? No.

Remark
: This fails without finite generation, see Theorem 3.56 on ordinal filtration.
  If $R$ is a PID and $F$ is a free $R\dash$module and $M \leq F$ as an $R\dash$submodule, then $M$ is free.

Thus if $R$ is a PID, "subfree" $\iff$ free.
Does torsion-free imply free?
No, take $R=\ZZ$ and $M = (\QQ, +)$, this is not finitely generated and torsion-free but not a free $\ZZ\dash$module.

Definition (Divisible and Uniquely Divisible Modules)
: For $R$ a domain, $M$ is *divisible* if $\forall a\in M^\bullet$ iff $$[a]: M \surjects M$$ is a surjection.
	$M$ is *uniquely divisible* if for all $a\in M^\bullet$, $[a]: M \mapsvia{\cong} M$ is an isomorphism, i.e. $M$ is torsion-free and divisible.

Exercise
: Show that $(\QQ, +)$ is a uniquely divisible $\ZZ\dash$module.

Exercise
: Let $R$ be a domain with fraction field $K$, with $R\neq K$.
	Show that a nonzero free $R\dash$module is not divisible but $(K, +)$ is a divisible torsion-free $R\dash$module.
	Thus $(K, +)$ is a torsion-free module $R\dash$module that is not free.

Remark
: Finitely generated torsion free modules are embedded in free modules.
  Note that in the spectrum of properties earlier (projective, free, etc), the two extremes are equal for f.g. PIDs.

Exercise
: Let $R$ be a Noetherian domain which is not a PID.
	Then an ideal $I\normal R$ with $I$ f.g., not principal, and a torsion-free $R\dash$module.
	Show that since $I$ is not principal, $I$ is not free as an $R\dash$module.

So ideals can't contain linearly independent elements, so they have to be free of rank 1 and thus principal.
Thus finitely generated torsion-free is strictly *weaker* then free in this setting.