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Modern Algebra: Groups, Rings, Modules, Fields

Sam Macdonald

Section 13.2 Free Modules

Subsection Bases

“The basis of all good human behavior is kindness.”
―Eleanor Roosevelt

Definition 13.15. Basis and Free Module.

A subset \(A\) of an \(R\)-module \(M\) is a basis of \(M\text{,}\) if the set \(A\) generates \(M\) and is linearly independent. An \(R\)-module M is a free \(R\)-module if \(M\) admits at least one a basis.

Example 13.16. Free Modules.

  1. The zero module is free with \(\emptyset\) as its (only) basis. This holds since the empty set is vacuously linearly independent and it generates \(\{0\}\text{.}\)
  2. \(R\) is free since \(\{1\}\) is a basis for \(R\text{.}\) It generates and if \(r \cdot 1 = 0\) then \(r = 0\text{,}\) so it is linearly independent.
  3. More generally, \(R^n\) is free since \(e_1, \dots, e_n\) is a basis. This is called the standard basis of \(R^n\text{.}\) We’ve already seen that \(e_1, \dots, e_n\) generates \(R^n\) as an \(R\)-module. Suppose \(\sum r_i e_i = 0\text{.}\) Then \(\begin{bmatrix} r_1 \\ \vdots \\ r_n \end{bmatrix} = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix}\) and hence \(r_i = 0\) for all \(i\text{.}\)
  4. For any ring \(R\text{,}\) if \(I\) is a (two-sided) ideal such that \(I \ne 0\) and \(I \ne R\text{,}\) then \(M = R/I\) is not free. Since \(I \ne R\text{,}\) \(R/I\) is not the zero module and hence the empty set isn’t a basis. Let \(A\) be any non-empty subset. Then since \(I \ne 0\text{,}\) as shown above \(A\) is linearly dependent. We conclude that no subset of \(M\) is a basis.

Exercise 13.17.

Direct sums of free modules are free.
Bases of free modules are rarely unique.

Example 13.18. Bases are not Unique.

  1. If \(R\) is any ring, then any single unit forms a basis for \(R\) as a module over itself.
  2. For any ring \(R\) and any fixed element \(r \in R\text{,}\) the set
    \begin{equation*} \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} r \\ 1 \end{bmatrix} \right \} \end{equation*}
    forms a basis for the free \(R\)-module \(R^2\text{.}\)

Exercise 13.19. Every Module over a Field is Free.

Let \(R\) be a commutative ring with \(1 \neq 0\text{.}\) Show that if every \(R\)-module is free then \(R\) is a field.

Remark 13.20.

A key difference between free modules over rings that are not fields and vector spaces is that not every linearly independent subset of a free module can be extended to a basis. For example, \(\{3\}\) is a linearly independent subset of \(\Z\text{,}\) but it cannot be extended to a basis. Indeed, any set of the form \(\{3, n \}\) with \(n \ne 3\) is linearly dependent since \(n \cdot 3 + (-3) \cdot n = 0\text{.}\)
Likewise, over arbitrary rings, not every subset that generates a free module necessarily contains a basis. For instance, \(\{3,5\}\) generated \(\Z\) as a module over itself, but no subset of it is a basis.
(Note that \(\Z\) has precisely two bases as a module over itself: \(\{1\}\) and \(\{-1\}\text{.}\) )

Proof.

Let \(m\in M\text{.}\) As \(A\) is a basis of \(M\) it generates \(M\) as an \(R\)-module, and thus there exists \(r_1,\dots,\r_n\in R\) and \(a_1,\dots,a_n\in A\) such that \(m=r_1a_1+\dots+r_na_n\text{.}\)
Suppose there also exists \(s_1,\dots,\s_n\in R\) such that \(m=s_1a_1+\dots+s_na_n\text{.}\) Then
\begin{equation*} 0 = m-m = r_1a_1+\dots+r_na_n - s_1a_1+\dots+s_na_n = \sum_{i=1}^n a_i(r_i - s_i). \end{equation*}
Since \(A\) is a basis of \(M\) it is linearly independent, and thus \(r_i - s_i = 0\) for all \(i\text{.}\)

Proof.

Given a function \(j: B \to N\text{,}\) define \(\varphi: M\to N\) as follows: Given \(m \in M\text{,}\) by Lemma 13.21 \(m\) can be written uniquely as a finite sum \(m = \sum_{b \in B} r_b b\text{.}\) We set
\begin{equation*} \varphi(m) = \sum_{b \in B} r_b j(b). \end{equation*}
Note that \(\varphi\) is a well-defined function by the uniqueness of the equation \(m = \sum_{b \in B} r_b b\text{.}\)
We need to prove \(f\) is an \(R\)-module homomorphism. I’ll just show it preserves scaling — the proof for addition is similar. Given \(x \in R\) and \(m \in M\text{,}\) we have \(m = \sum_{b \in B} r_b b\) for some \(r_b \in R\text{,}\) and hence \(xm = \sum_b (xr_b) b\text{.}\) By definition of \(f\text{,}\)
\begin{equation*} \varphi(xm) = \sum_b (xr_b) j(b) = x \sum_b r_b j(b) = x \varphi(m). \end{equation*}
Finally, for any \(y \in B\) we have \(y = \sum_{b \in B} r_b b\) where \(r_b = 1\) if \(b = y\) and \(r_b = 0\) if \(b \ne y\text{.}\) So \(f(y) = j(y)\) by construction. This proves existence.
Let \(\phi:M\to N\) be another \(R\)-module homomorphism such that \(\phi(b)=\phi(b)\) for each \(b \in B\text{.}\) Given \(m \in M\) we have \(m = \sum_{b \in B} r_b b\) and hence
\begin{equation*} \phi(m) = \sum_b \phi(r_b (b))= \sum_b r_b \phi(b) =\sum_b r_b j(b) = \varphi(m). \end{equation*}
and hence \(\phi = \varphi\text{.}\)

Remark 13.23.

The uniqueness only uses that \(B\) generates \(M\) as an \(R\)-module.

Remark 13.24.

Another way of stating Theorem 13.22 is that there is a bijection of sets
\begin{equation*} \Hom_R(M,N) \leftrightarrow\Fun(B, N) \end{equation*}
given by sending a homomorphisms \(f: M \to N\) to its restriction \(f|_B:B \to N\) to \(B\text{.}\) (Here, \(\Hom_R(M,N)\) is the set of all \(R\)-module homomorphisms from \(M\) to \(N\) and \(\Fun(B, N)\) is the set of all functions from \(B\) to \(N\text{.}\))
What Theorem 13.22 tells us is that to create a map between \(M\) and any other module, all we must do is specify where the basis elements are sent.

Proof.

Given a module \(M\text{,}\) first take a generating set for \(M\text{,}\) say \(\Gamma=\left\{m_{i}\right\}_{i \in \Lambda}\text{.}\) Notice that a generating set always exists: for example, we can take \(\Gamma=M\text{,}\) though of course that is a bit of an overkill, since it’s quite likely that some elements can be obtained from linear combinations of others. Next, we construct a free module on the set \(\Lambda\text{;}\) more precisely, we take a free module on as many generators as generators for \(M\) that we picked. Now the map
\begin{equation*} \begin{aligned} & \bigoplus_{i \in \Lambda} R \stackrel{\pi}{\longrightarrow} M \\ & \left(r_{i}\right) \longmapsto \sum_{i \in \Lambda} r_{i} m_{i} . \end{aligned} \end{equation*}
Notice this map actually makes sense: the tuples \(\left(r_{i}\right)\) have only finitely many nonzero entries, and thus \(\sum_{i \in \Lambda} r_{i} m_{i}\) is a (finite) linear combination of our chosen generators. Moreover, since we chose the \(m_{i}\) to be generators for \(M\text{,}\) this map \(\pi\) is surjective. It is also easy to check that it is an \(R\)-module homomorphism: in fact, this is the \(R\)-module homomorphism we would get from the UMP for Free \(R\)-Modules by setting \(e_{i} \mapsto m_{i}\text{.}\)
By the First Isomorphism Theorem for Modules,
\begin{equation*} M \cong \bigoplus_{i \in \Lambda} R / \operatorname{ker} \pi \end{equation*}

Subsection Rank

Definition 13.26. Module Rank.

Let \(R\) be a non-zero commutative ring and let \(M\) be a free \(R\)-module. The cardinality of any basis of \(M\) is called the rank of \(M\text{.}\)
But what about if two bases have different ranks? As it turns out, any two bases of the same free module will have the same cardinality, as we will prove in Theorem 13.33.
 1 
Mostly. The rest is proved in Theorem 13.49.

Example 13.27. Rank of \(R^n\).

The \(R\)-module \(R^n\) has rank \(n\text{.}\) Thus \(R\) has rank \(1\) as a module over itself, as this is the case when \(n=1\text{.}\)
Bases need not be finite.

Example 13.28. Module with Infinite Basis.

Let \(R\) be any ring and \(M = R[x]\) (which, recall, is an \(R\)-module due to the evident ring map \(R \to R[x]\)). Then the countably infinite set \(\{1,x,x^2,\dots, x^n,\dots\}\) is a basis. The fact that this set is a basis is essentially part of the definition of \(R[x]\text{.}\) Lemma 13.21 says that every polynomial is uniquely expressible as an \(R\)-linear combination of (a finite subset of) \(1,x,x^2,\dots\text{.}\)

Proof.

Let \(A\) be a basis of \(M\) and \(B\) be a basis of \(N\) such that \(|A|=|B|\text{.}\) Thus there exists a bijection \(j:M\to N\) with inverse \(j\inv:N\to M\text{.}\) We invoke the UMP for Free \(R\)-Modules to build \(R\)-module homomorphisms \(\varphi:M\to N\) and \(\phi:N\to M\) such that \(\varphi = j(a)\) for all \(a\in A\) and \(\phi = j\inv(b)\) for all \(b\in B\text{.}\)
We’ll show that \(\varphi\) and \(\phi\) are mutual inverses. For this note that \(\phi\circ \varphi:N\to N\) is an \(R\)-module homomorphism and \((\phi\circ \varphi)(b)=\phi(j(b))=j^{-1}(j(b))=b\) for every \(b\in B\text{.}\) Since the identity map \(\id_N\) is also an \(R\)-module homomorphism such that \(id_N(b)=b\) for every \(b\in B\text{,}\) by the uniqueness clause in the UMP for Free \(R\)-Modules, we have \(\phi\circ \varphi=\id_n\text{.}\) Similarly \(\varphi\circ \phi=\id_M\text{.}\)

Example 13.30. \(M\cong R^n\) as \(R\)-modules.

If \(M\) is a free \(R\)-module that has a basis of cardinality \(n\text{,}\) then \(M \cong R^n\) as \(R\)-modules. This is because \(R^n\) is a free \(R\) module of rank \(n\text{,}\) and thus \(M \cong R^n\) by Corollary 13.29.

Remark 13.31.

Beware that the cardinality of a basis of a free modules is not an isomorphism invariant in general! There exist rings \(S\) such that \(S^n\) and \(S^m\) are isomorphic \(S\)-modules for all positive integers \(n\) and \(m\text{.}\)

Proof.

https://eloisagrifo.github.io/Teaching/818/PSet3solutions.pdf

Proof.

Let \(R\) be a non-zero commutative ring and let \(M\) be a free \(R\)-module with two bases, \(A\) and \(B\text{,}\) with ranks \(m\) and \(n\text{,}\) respectively. Since \(A\) and \(B\) are finite, by Example 13.30 we have \(A\cong R^m\) and \(B\cong R^n\text{.}\)
I will prove this statement by taking it as already known that it holds in the special case when \(R\) is a field. (We will prove it for fields later in Theorem 13.49.)
Since \(R\) is not the zero ring, it contains at least one maximal ideal \(\fm\) by Theorem 9.40. Recall that \(R/\fm\) is a field by Theorem 9.36
Given an isomorphism \(f: R^n \xra{\cong} R^m\) of \(R\)-modules, by Lemma [provisional cross-reference: cite] we have an induced homomorphism of \(R/I\)-module \(\ov{f}: R^n/IR^n \to R^m /IR^m\text{.}\) Likewise, the inverse map \(f^{-1}: R^m \xra{\cong} R^n\) induces a map \(\ov{f^{-1}}: R^m/IR^m \to R^n /IR^n\text{.}\) Also by that Lemma [provisional cross-reference: cite] we have \(\ov{f} \circ \ov{f^{-1}} = \ov{f \circ f^{-1}} = \ov{\id} = \id\) and similarly \(\ov{f^{-1}} \circ \ov{f}\) is the identity. That is, we have an isomorphism \(R^n/IR^n \cong R^m /IR^m\) of \(R/I\)-modules.
Next, I claim that there is an isomorphism
\begin{equation*} R^n/IR^n \cong (R/I)^n \end{equation*}
of \(R/I\)-modules. Define \(R^n \to (R/I)^n\) in the evident way (modding out by \(I\) entry-wise). It is a surjective map of \(R\)-modules with kernel \(I R^n\) and thus, by the First Isomorphism Theorem for Modules, it induces an isomorphism
\begin{equation*} g: R^n/I R^n \xra{\cong} (R/I)^n \end{equation*}
given by \(g((r_1, \dots, r_n)^\tau + IR^n) = (r_1 + I, \dots, r_n + I)^\tau\) (where \(\tau\) denotes taking the transpose). Now, what we have said so far only shows that \(g\) is isomorphism of \(R\)-modules, but it is easy to see that \(g\) is in fact \(R/I\)-linear (I’ll leave that to you) and thus it is an isomorphism of \(R/I\)-modules.
Putting the results proven so far together, we conclude that \((R/I)^n\) and \((R/I)^m\) are isomorphic as \(R/I\)-modules. Since \(R/I\) is a field and since we are assuming the result holds for fields, we deduce that \(n = m\text{.}\)

Remark 13.34.

Checking the definition carefully, we see that both the empty set and the set \(\{0\}\) form bases for the zero module over the \(0\) ring. This gives an example of a module with two bases of different cardinalities, justifying why the suppostion that \(1\ne 0\) was used in Theorem 13.33.

Exercise 13.35. Bases of ideals in commutative rings.

  1. Assume \(R\) is a non-zero, commutative ring and \(I\) is a non-zero ideal. Prove \(I\) is free as an \(R\)-module if and only if \(I = R \cdot a\) for a non-zerodivisor \(a\text{.}\) (Recall that an element \(a\) is a non-zerodivisor in \(R\) provided \(a \ne 0\) and \(xa = 0\) implies \(x = 0\) for all \(x \in R\text{.}\) )
  2. Let \(F\) be a field and \(R = F[x,y]\text{.}\) Let \(I = R \cdot \{x, y\}\text{,}\) the ideal consisting of all polynomials with \(0\) constant term. Prove \(I\) is not free as an \(R\)-module.
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