Free Module

Free

An $R$ -module $M$ is called free if

M ≅ ⨁_{i \in I} R .

Rank

A free $R$ -module $M ≅ ⨁_{i = 1}^{n} R$ is said to be of rank $n$ .

The following theorem just the usefulness of this concept.

Theorem

A module is free if and only if it has a basis.

Intuitively, if this isomorphism exists, we can construct a basis of $M$ by taking a standard basis in $⨁_{i \in I} R$ and applying the isomorphism. On the other hand if there is a basis for $M$ , we construct an isomorphism by mapping each vector of the basis to the standard basis of $⨁_{i \in I} R$ . In particular, we set the coordinate vectors to match in their respective bases.

Lemma

$⨁_{i \in I} R$ has a basis.

This result is a simple technical result, where the basis can be constructed just like the standard basis in the vector space $F^{n}$ .

Proof

Define, for $j \in I$ , $e_{j} = (r_{i})_{i \in I}$ where $r_{j} = 1$ and $r_{k} = 0$ for all $k \neq j$ . It is then clear that for any $(r_{i})_{i \in I} \in ⨁_{i \in I} R$ we can write

(r_{i})_{i \in I} = \sum_{i \in J} r_{i} e_{i}

where $J = {i \in I : r_{i} \neq 0}$ . Note that the sum is well defined because from the definition of the direct sum, only finitely many $r_{i}$ will be non-zero.

This set is linearly independent because for any sum

\sum_{i \in J} r_{i} e_{i} = 0

over a finite $J \subseteq I$ , we have for each component

0 + 0 + \dots + 0 + r_{i} + 0 + \dots + 0 = 0

and thus each $r_{i} = 0$ .

Proof

First assume that $M ≅ ⨁_{i \in I} R$ is a free module. Then let $ψ : ⨁_{i \in I} R \to M$ be a module homomorphism.

If $m \in M$ we can write $ψ^{- 1} (m)$ in the basis ${e_{i} : i \in I}$ of $⨁_{i \in I} R$ as

ψ^{- 1} (m) = \sum_{i \in J} r_{i} e_{i}

for some finite subset $J$ of $I$ and arbitrary $r_{i} \in I$ .

As such, taking $ψ$ of both sides and using the fact that it is a homomorphism we have

m = (ψ \circ ψ^{- 1}) (m) = ψ (\sum_{i \in J} r_{i} e_{i}) = \sum_{i \in J} r_{i} ψ (e_{i})

and therefore $m$ is in the span of ${ψ (e_{i}) : i \in I}$ for arbitrary $m$ . This set is also linearly independent because

\sum_{i \in J} r_{i} ψ (e_{i}) = 0 ⟹ ψ^{- 1} (\sum_{i \in J} r_{i} ψ (e_{i})) = \sum_{i \in J} r_{i} e_{i} = 0

and so $r_{i} = 0$ for all $i \in I$ because ${e_{i} : i \in I}$ is a basis. Therefore ${ψ (e_{i}) : i \in I}$ defines a basis of $M$ .

For the converse, assume that $M$ has a basis ${m_{i} : i \in I}$ and therefore if $m \in M$ we have a finite set $J$ and coefficients $r_{i}$ such that

m = \sum_{i \in J} r_{i} m_{i} .

For convenience, we set $r_{i} = 0$ if $i \in I - J$ . Uniqueness of this expression allows us to define a map $ϕ : M \to ⨁_{i \in I} R$ by

ϕ (m) = ϕ (\sum_{i \in J} r_{i} m_{i}) = (r_{i})_{i \in I} .

This map is a ring homomorphism because

\begin{aligned} ϕ (s m) & = ϕ (\sum_{i \in J} s r_{i} m_{i}) \\ = (s r_{i})_{i \in I} \\ = s (r_{i})_{i \in I} \\ = ϕ (m) \end{aligned}

and

\begin{aligned} ϕ (m + n) & = ϕ (\sum_{i \in J} r_{i} m_{i} + \sum_{i \in J} s_{i} m_{i}) \\ = ϕ (\sum_{i \in J} (r_{i} + s_{i}) m_{i}) \\ = (r_{i} + s_{i})_{i \in I} \\ = (r_{i})_{i \in I} + (s_{i})_{i \in I} \\ = ϕ (m) + ϕ (n) \end{aligned}

where $J$ is constructed from the union of indices in which either $r_{i}$ and $s_{i}$ are non-zero.

Example

Every number ring is free. number rings are dedekind rings