Direct sums and free modules

Section 2.18 Direct sums and free modules

Specimen 36. Direct sums and products of modules.

Let \(R\) be a ring, and let \((M_i)_{i\in I}\) be an indexed family of \(R\)-modules for some nonempty index set \(I\text{.}\)

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The direct product of \((M_i)_{i\in }\) is the \(R\)-module whose underlying set is \(\prod_{i\in I}M_i\text{,}\) the set of all tuples \((m_i)_{i\in I}\text{,}\) where \(m_i\in M_i\) for all \(i\in I\text{,}\) and whose module addition and scalar multiplication are defined component-wise as follows:

\begin{align*} (m_i)_{i\in I}+ (n_i)_{i\in I} \amp = (m_i+n_i)_{i\in I}\\ r\cdot (m_i)_{i\in I} \amp = (rm_i)_{i\in I} \end{align*}

for all \(m=(m_i), n=(n_i)\in \prod_{i\in I}M_i\) and \(r\in R\text{.}\)

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The direct sum of \((M_i)_{i\in I}\text{,}\) denoted \(\bigoplus_{i\in I}M_i\) is the \(R\)-submodule of \(\prod_{i\in I}M_i\) consisting of all tuples \((m_i)_{i\in I}\) for which \(m_i=0\) for all but finitely many \(i\in I\text{:}\) i.e.,

\begin{align} \bigoplus_{i\in I} M_i \amp =\{(m_i)_{i\in I}\in \prod_{i\in I}M_i\mid m_i=0 \text{ for all but finitely many } i\in I\}\text{.}\tag{2.18.1} \end{align}

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For all \(j\in I\) we define the \(j\)-th projection map \(\pi_j\colon \prod_{i\in I}M_i\to M_j\) by \(\pi_j((m_i)_{i\in I})=m_j\text{,}\) and the \(j\)-th inclusion map \(\iota_j\colon M_j\to \bigoplus_{i\in I}M_i\) by \(\iota_j(m)=(m_i)_{i\in I}\text{,}\) where \(m_i=0\) for all \(i\neq j\) and \(m_j=m\text{.}\) The maps \(\pi_j\) and \(\iota_j\) are \(R\)-module homomorphisms for all \(j\in I\text{.}\)

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Proof that direct products and sums are modules.

From the component-wise definitions of addition and scalar multiplication on \(\prod_{i\in I}M_i\text{,}\) it follows in a straightforward manner that \(\prod_{i\in I}M_i\) satisfies the axioms of an \(R\)-module, and that \(\bigoplus_{i\in I}M_i\) is indeed a submodule of \(\prod_{i\in I}M_i\text{.}\) Indeed, we know already that component-wise addition satisfies the group axioms, making \(\prod_{i\in I}M_i\) an abelian group. Below we provide highly condensed one-line proofs remaining four axioms:

\begin{align*} (r+s)\cdot (m_i) \amp = ((r+s)m_i)=(rm_i+sm_i)=r\cdot (m_i)+s\cdot (m_i)\\ r\cdot ((m_i)+(n_i)) \amp = r\cdot (m_i+n_i)=(r(m_i+n_i))=(rm_i+rn_i)=r\cdot (m_i)+r\cdot (n_i) \\ r\cdot (s\cdot (m_i)) \amp =r\cdot (sm_i)_{i\in I}=(rs m_i)=(rs)\cdot (m_i)\\ 1\cdot (m_i) \amp =(1m_i)_{i\in I}=(m_i)\text{.} \end{align*}

We delegate the proofs of the remaining claims, including the fact that \(\pi_j\) and \(\iota_j\) are \(R\)-module homomorphisms, to the reader.

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Theorem 2.18.1. Mapping properties of direct sums and products.

Let \(R\) be a ring, and let \((M_i)_{i\in I}\) be an indexed family of \(R\)-modules for some nonempty index set \(I\text{.}\)

Mapping property of direct products.

Let \(\pi_i\) be the \(i\)-th projection map from \(\prod_{i\in I}M_i\text{.}\) For any \(R\)-module \(N\) and tuple of \(R\)-module homomorphisms \((\phi_i)_{i\in I}\text{,}\) where \(\phi_i\colon N\rightarrow M_i\text{,}\) the map \(\phi=\prod_{i\in I}\phi_i\) defined as

\begin{align} \phi (n) \amp =(\phi_i(n))_{i\in I}\tag{2.18.2} \end{align}

is the unique \(R\)-module homomorphism \(\phi\colon N\rightarrow \prod_{i\in I}M_i\) satisfying \(\phi\circ \pi_i=\phi_i\) for all \(i\in I\text{.}\)

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It follows that the map

\begin{align*} \Phi\colon \Hom_R(N, \prod_{i\in I}M_i) \amp\rightarrow \prod_{i\in I}\Hom(N, M_i) \\ \phi \amp \longmapsto (\pi_i\circ\phi)_{i\in I} \end{align*}

is an isomorphism of \(R\)-modules with inverse

\begin{align*} (\phi_i)_{i\in I} \amp \longmapsto \prod_{i\in I}\phi_i\text{.} \end{align*}

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Mapping property of direct sums.

Let \(\iota_i\) be the \(i\)-th inclusion map from \(M_i\) to \(\bigoplus_{i\in I}M_i\text{.}\) For any \(R\)-module \(N\) and tuple of \(R\)-module homomorphisms \((\phi_i)_{i\in I}\text{,}\) where \(\phi_i\colon M_i\rightarrow N\text{,}\) the map \(\phi=\bigoplus_{i\in I}\phi_i\) defined as

\begin{align} \phi ((m_i)_{i\in I}) \amp = \sum_{i\in I}\phi_i(m_i)\tag{2.18.3} \end{align}

is the unique \(R\)-module homomorphism \(\phi\colon \bigoplus_{i\in I}M_i\rightarrow N\) satisfying \(\phi\circ \iota_i=\phi_i\) for all \(i\in I\text{.}\)

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It follows that the map

\begin{align*} \Phi'\colon \Hom_R(\bigoplus_{i\in I}M_i, N) \amp\rightarrow \prod_{i\in I}\Hom(M_i, N) \\ \phi \amp \longmapsto (\phi\circ \iota_i)_{i\in I} \end{align*}

is an isomorphism of \(R\)-modules with inverse

\begin{align*} (\phi_i)_{i\in I} \amp \longmapsto \sum_{i\in I}\phi_i\text{.} \end{align*}

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Proof.

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Example 2.18.2. \(R[x]\) and \(R[[x]]\) as \(R\)-modules.

Let \(R\) be a nonzero commutative ring. Prove that as \(R\)-modules, we have \(R[x]\cong\bigoplus_{i=1}^\infty R\) and \(R[[x]]\cong \prod_{i=1}^\infty R\text{.}\)

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Solution.

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Specimen 37. Free module.

Let \(R\) be a ring, and let \(I\) be a nonempty set. The module \(\bigoplus_{i\in I} R\) is called the free \(R\)-module module on the set \(I\text{.}\) An \(R\)-module \(M\) is free if \(M\cong_R \bigoplus_{i\in I} R\) for some nonempty set \(I\text{.}\)

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In analogy to the case of vector spaces, we will denote by \(e_i\) the element of \(\bigoplus_{i\in I}R\) whose \(i\)-th component is \(1\in R\) and whose other components are \(0\in R\text{,}\) and call \(e_i\) the \(i\)-th standard basis element of \(\bigoplus_{i\in I}R\text{.}\)

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Theorem 2.18.3. Mapping property of free modules.

Let \(R\) be a ring, let \(I\) be a nonempty set, and let \(e_i\) be the \(i\)-th standard basis element of \(\bigoplus_{i\in I}R\text{.}\) For any \(R\)-module \(N\) and any tuple \((n_i)_{i\in I}\in \bigoplus_{i\in I}N \text{,}\) the \(R\)-module homomorphism \(\phi\colon \bigoplus_{i\in I}R\rightarrow N\) defined as \(\phi((r_i)_{i\in I})=\sum_{i\in I}r_in_i\) is the unique homomorphism satisfying \(\phi(e_i)=n_i\) for all \(i\in I\text{.}\)

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As a result, the map

\begin{align*} \Phi\colon \Hom_R(\bigoplus_{i\in I}R, N) \amp\rightarrow \prod_{i\in I}N \\ \phi \amp \longmapsto (\phi(e_i))_{i\in I} \end{align*}

is an isomorphism of \(R\)-modules.

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Proof.

The result follows from the mapping property of direct sums applied to the case where \(M_i=R\) for all \(i\in I\) and the fact that \(\Hom_R(R,N)\cong N\) via the isomorphism \(\psi\mapsto \psi(1)\text{.}\)

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Definition 2.18.4. Finitely generated module.

An \(R\)-module \(M\) is finitely generated if there are elements \(m_1,m_2,\dots, m_n\in M\) satisfying \(M=(m_1,m_2,\dots, m_n)\text{.}\)

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Example 2.18.5. Finitely generated modules.

Let \(R\) be a ring. Prove that a nonzero \(R\)-module \(M\) is finitely generated if and only if we have \(M\cong_R R^n/M'\) for some positive integer \(n\) and submodule \(M'\subseteq R^n\text{.}\)

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Solution.

Assume \(M=(m_1,m_2,\dots, m_n)\text{.}\) By Theorem 2.18.3, there is an \(R\)-module homomorphism \(\phi\colon R^n\rightarrow M\) defined as \(\phi((r_1,r_2,\dots, r_n))=\sum_{i=1}^n r_im_i\text{.}\) By definition of \((m_1,m_2,\dots, m_n)\text{,}\) it is clear that \(\phi\) is surjective. The first isomorphism theorem for modules now implies that \(M\cong R^n/M'\text{,}\) where \(M'=\ker\phi\text{.}\)

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Definition 2.18.6. Linear independence and span.

Let \(M\) be an \(R\)-module, and let \((m_i)_{i\in I}\) be a tuple of elements of \(M\) indexed by a nonempty set \(I\text{.}\)

\((m_i)_{i\in I}\) is linearly independent if for all \((r_i)_{i\in I}\in \bigoplus_{i\in I}R\text{,}\) if \(\sum_{i\in I}r_im_i=0\text{,}\) then \(r_i=0\) for all \(i\in I\text{.}\)
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\((m_i)_{i\in I}\) spans \(M\) (or is a spanning set for \(M\)) if \(M=(\{m_i\mid i\in I\})\text{.}\)
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\((m_i)_{i\in I}\) is a basis of \(M\) if \((m_i)_{i\in I}\) is both linearly independent and spans \(M\text{.}\)
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Corollary 2.18.7. Free modules and bases.

Let \(R\) be a nonzero ring.

Let \(I\) be a nonempty set. The tuple \((e_i)_{i\in I}\) is a basis of \(\bigoplus_{i\in I}R\text{.}\)
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An \(R\)-module \(M\) is free if and only if it has a basis.
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Let \(M\) be a free \(R\)-module with basis \((m_i)_{i\in I}\text{.}\) Given any \(R\)-module \(N\text{,}\) the map

\begin{align*} \Hom_R(M,N) \amp \rightarrow N^I\\ \phi \amp \mapsto (\phi(m_i))_{i\in I} \end{align*}

is an isomorphism of \(R\)-modules.

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In plain English, given any tuple \((n_i)_{i\in I}\) of elements of \(N\text{,}\) there is a unique \(R\)-module homomorphism \(\phi\in \Hom_R(M,N)\) satisfying \(\phi(m_i)=n_i\text{;}\) and in particular, an \(R\)-module homomorphism \(\phi\in \Hom_R(M,N)\) is uniquely determined by its values \(\phi(m_i)\) at the basis elements \(m_i\text{.}\)
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Proof.

Given \((r_i)_{i\in I}\in \bigoplus_{i\in I}R\text{,}\) we have \((r_i)=\sum_{i\in I}r_ie_i\text{.}\) It follows immediately that \((e_i)_{i\in I}\) spans \(\bigoplus_{i\in I}R\text{.}\) Furthermore, for all \((r_i)\in \bigoplus_{i\in I}R\text{,}\) we have

\begin{align*} \sum_{i\in I}r_ie_i=0 \amp \iff (r_i)_{i\in I}=0 \\ \amp \iff r_i=0 \text{ for all } i\in I\text{.} \end{align*}

Thus \((e_i)_{i\in I}\) is linearly independent, and we conclude that \((e_i)_{i\in I}\) is a basis of \(\bigoplus_{i\in I}R\text{.}\)

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If \(M\) is free, then there is an isomorphism \(\phi\colon \bigoplus_{i\in I}R\rightarrow M\text{,}\) from whence it follows easily that \((\phi(e_i))_{i\in I}\) is a basis for \(M\text{.}\)
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Conversely, assume that \((m_i)_{i\in I}\) is a basis for \(M\text{.}\) From the mapping property of free modules, the map

\begin{align*} \phi\colon \bigoplus_{i\in I}R \amp \rightarrow M\\ (r_i)_{i\in I} \amp \longmapsto \sum_{i\in I}r_im_i \end{align*}

is an \(R\)-module homomorphism. Since \((m_i)\) spans \(M\text{,}\) this map is surjective. Since \((m_i)\) is linearly independent, the kernel of this map is \(\{0\}\text{.}\) Thus \(\phi\) is an isomorphism, and we conclude that \(M\) is free.

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It is easy to see that the map \(\phi\mapsto (\phi(m_i))_{i\in I}\) is an \(R\)-module homomorphism from \(\Hom_R(M,N)\) to \(N^I\text{.}\) To show that it is an isomorphism, we construct an inverse map. Given any tuple \(n=(n_i)_{i\in I}\in N^I\text{,}\) we define \(\phi_n\colon M\rightarrow N\text{,}\) as follows: given any \(m\in M\) we can write

\begin{align} m \amp =\sum_{i\in I} a_i m_i\tag{2.18.4} \end{align}

for some \((a_i)_{i\in I}\in \bigoplus_{i\in I}R\text{,}\) and we define

\begin{align*} \phi_n(m) \amp =\sum_{i\in I}a_i n_i\text{.} \end{align*}

As usual, once we know that \(\phi_n\) is well defined, it follows very easily that \(\phi_n\in \Hom_R(M,N)\) and that the two maps

\begin{align*} \Hom_R(M,N)\amp \rightarrow N^I \amp N^I\amp\rightarrow \Hom_R(M,N) \\ \phi \amp \mapsto (\phi(m_i))_{i\in I} \amp n=(n_i)_{i\in I}\amp \mapsto \phi_n \end{align*}

are inverses of one another.

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What exactly are the obstacles to \(\phi_n\) being well defined? Firstly, it is important to note that the sum \(\sum_{i\in I}a_in_i\) is in fact a finite sum, since \((a_i)_{i\in I}\in \bigoplus_{i\in I}R\text{.}\) Secondly, there is the question of whether the definition of \(\phi_n\) depends on how we choose the coefficients \((a_i)_{i\in I}\text{.}\) The issue here is easily resolved: since \((m_i)_{i\in I}\) is a basis, for any \(m\in M\text{,}\) there is a unique \((a_i)_{i\in I}\in \bigoplus_{i\in I}R\) satisfying \(m=\sum_{i\in I}a_im_i\text{.}\) This is a consequence of the linear independence property of \((m_i)_{i\in I}\text{:}\)

\begin{align*} \sum_{i\in I}a_im_i=\sum_{i\in I}b_im_i\amp \implies \sum_{i\in I}a_im_i-\sum_{i\in I}b_im_i=0\\ \amp \implies \sum_{i\in I}(a_i-b_i)m_i=0\\ \amp \implies a_i-b_i=0 \text{ for all } i\in I\amp ((m_i)_{i\in I} \text{ lin. ind.})\\ \amp \implies a_i=b_i \text{ for all } i\in I\text{.} \end{align*}

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Corollary 2.18.8. Isomorphic free modules.

Let \(R\) be a ring, and let \(I\) and \(J\) be nonempty sets.

If \(\abs{I}=\abs{J}\text{,}\) then \(\bigoplus_{i\in I} R\cong_R\bigoplus_{j\in J}R\text{.}\)
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If \(R\) is commutative, then \(\bigoplus_{i\in I}R\cong_R \bigoplus_{j\in J}R\) if and only if \(\abs{I}=\abs{J}\text{.}\)
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Proof.

Let \(M=\bigoplus_{i\in I}R\) and \(M'=\bigoplus_{j\in J}R\text{.}\) Assume \(\abs{I}=\abs{J}\text{.}\) By definition, there is a bijection of sets \(f\colon I\rightarrow J\text{.}\) Let \(e_i\) be the \(i\)-th standard basis element of \(M=\bigoplus_{i\in I} R\) for all \(i\in I\text{,}\) and let \(f_j\) be the \(j\)-th standard basis of \(M'=\bigoplus_{j\in J}R\) for all \(j\in J\text{.}\) The mapping property of free modules implies there are unique \(R\)-module homomorphisms \(\phi\colon M\rightarrow M' \) and \(\psi\colon M'\rightarrow M\) satisfying \(\phi(e_i)=f_{\alpha(i)}\) for all \(i\in I\) and \(\psi(f_j)=e_{\alpha^{-1}(j)}\) for all \(j\in J\text{.}\) It follows that the homomorphisms \(\psi\circ\phi\in \End_R(\bigoplus_{i\in I}R)\) and \(\phi\circ\psi\in \End_R(\bigoplus_{j\in J}R)\) satisfy

\begin{align*} \psi\circ\phi(e_i) \amp = \psi(f_{\alpha(i)}) = e_{\alpha^{-1}(\alpha(i))}=e_i\\ \phi\circ\psi(f_j) \amp = \phi(e_{\alpha^{-1}(j)})=f_{\alpha(\alpha^{-1}(j))}=f_j \end{align*}

for all \(i\in I\) and \(j\in J\text{.}\) Since \(\id_{M}(e_i)=e_i\) for all \(i\) and \(\id_{M'}(f_j)=f_j\) for all \(j\in J\text{,}\) the uniqueness claim of the mapping property of free modules implies that

\begin{align*} \psi\circ\phi \amp = \id_M\\ \phi\circ\psi \amp = \id_{M'}\text{.} \end{align*}

Thus \(\psi=\phi^{-1}\text{,}\) and we conclude that \(\phi\) is a module isomorphism.

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This is left as a homework exercise. There we assume the result is true when \(R=F\) is a field. This is a standard result in linear algebra when \(\abs{I}\) is finite; when \(\abs{I}\) is infinite, we must use Zorn’s lemma (and the result is not so standard).
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Example 2.18.9. Homomorphisms between free modules.

Let \(R\) be a ring, and let \(m\) and \(n\) be positive integers.

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Prove that \(\Hom_R(R^m,R^n)\cong_R R^{mn}\text{.}\)

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Solution.

From Theorem 2.18.3, we have \(\Hom_R(R^m, R^n)\cong \prod_{i=1}^m R^n\text{.}\) We claim that \(\prod_{i=1}^m R^n\cong R^{mn}\text{.}\) For each \(1\leq i\leq m\text{,}\) let \(\pi_i\) be the projection map from \(\prod_{i=1}^m R^n\) onto the \(i\)-th copy of \(\R^n\text{.}\) For each \(1\leq j\leq n\text{,}\) let \(p_j\colon R^n\rightarrow R\) be the \(j\)-th projection map from \(R^n\) onto the \(j\)-th copy of \(R\text{.}\) Let \(I=\{1,2,\dots, m\}\times \{1,2,\dots, n\}\text{.}\) From the mapping property of \(R^{I}\text{,}\) the tuple \((\phi_{i,j})_{(i,j)\in I}\text{,}\) where \(\phi_{i,j}=p_j\circ \pi_i\text{,}\) give rise to a ring homomorphism \(\phi\colon \prod_{i=1}^m R^n\rightarrow R^I\text{,}\) defined as

\begin{align*} \phi(((r_{i,j})_{j=1}^n)_{i=1}^m) \amp= (\phi_{i,j}(((r_{i,j})_{j=1}^n)_{i=1}^m))_{(i,j)\in I} \\ \amp = (p_j(\pi_i(((r_{i,j})_{j=1}^n)_{i=1}^m)))_{(i,j)\in I}\\ \amp = (r_{i,j})_{(i,j)\in I}\text{.} \end{align*}

The map \(\phi\) is clearly surjective. To see that \(\ker\phi=\{0\}\text{,}\) observe that we have

\begin{align*} \phi((((r_{i,j})_{j=1}^n)_{i=1}^m))=\{0\} \amp \iff (r_{i,j})_{(i,j)\in I}=\{0\} \\ \amp \iff r_{i,j}=0 \text{ for all } i\in \{1,2,\dots, m\}, j\in \{1,2,\dots, n\}\\ \amp \iff (r_{i,j})_{j=1}^{n}=0 \text{ for all } i\in \{1,2,\dots, m\}\\ \amp \iff (((r_{i,j})_{j=1}^n)_{i=1}^m)=0 \text{.} \end{align*}

Lastly, since \(\abs{I}=mn\text{,}\) we conclude that \(R^I\cong R^{mn}\) by Theorem 2.18.3.

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