Free Module

Free

An \(R\)-module \(M\) is called free if

\[ M \cong \bigoplus_{i \in I} R.\]

Rank

A free \(R\)-module \(M \cong \bigoplus_{i = 1}^n R\) 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 \(\bigoplus_{i \in I} R\) 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 \(\bigoplus_{i \in I} R\). In particular, we set the coordinate vectors to match in their respective bases.

Lemma

\(\bigoplus_{i \in I} R\) 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 \(\mathbb{F}^n\).

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 \((r_i)_{i \in I} \in \bigoplus_{i \in I} R\) 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 \(M \cong \bigoplus_{i \in I} R\) is a free module. Then let \(\psi : \bigoplus_{i \in I} R \to M\) be a module homomorphism.

If \(m \in M\) we can write \(\psi^{-1}(m)\) in the basis \(\{e_i : i \in I\}\) of \(\bigoplus_{i \in I} R\) as

\[ \psi^{-1}(m) = \sum_{i \in J} r_i e_i\]

for some finite subset \(J\) of \(I\) and arbitrary \(r_i \in I\).

As such, taking \(\psi\) of both sides and using the fact that it is a homomorphism we have

\[ m = (\psi \circ \psi^{-1})(m) = \psi\left(\sum_{i \in J} r_i e_i\right) = \sum_{i \in J} r_i \psi(e_i)\]

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

\[ \sum_{i \in J} r_i \psi(e_i) = 0 \implies \psi^{-1}\left(\sum_{i \in J} r_i \psi(e_i)\right) = \sum_{i \in J} r_i e_i = 0\]

and so \(r_i = 0\) for all \(i \in I\) because \(\{e_i : i \in I\}\) is a basis. Therefore \(\{\psi(e_i) : i \in I \}\) 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 \(\phi : M \to \bigoplus_{i \in I} R\) by

\[ \phi(m) = \phi\left(\sum_{i \in J} r_i m_i\right) = (r_i)_{i \in I}.\]

This map is a ring homomorphism because

\[\begin{align*} \phi(sm) &= \phi\left(\sum_{i \in J} sr_i m_i\right) \\ &= (sr_i)_{i \in I} \\ &= s(r_i)_{i \in I} \\ &= \phi(m) \\ \end{align*}\]

and

\[\begin{align*} \phi(m + n) &= \phi\left(\sum_{i \in J} r_i m_i + \sum_{i \in J} s_i m_i\right) \\ &= \phi\left(\sum_{i \in J} (r_i + s_i) m_i\right) \\ &= (r_i + s_i)_{i \in I} \\ &= (r_i)_{i \in I} + (s_i)_{i \in I} \\ &= \phi(m) + \phi(n) \\ \end{align*}\]

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