Proof (of Lemma 11.1.3)

Let \(m \in M\). Then \[ 0_R m = (0_R + 0_R) m = 0_Rm + 0_Rm. \] Since \(M\) is an abelian group, the element \(0_Rm\) has an additive inverse, \(-0_Rm\), so adding it on both sides we see that \[ 0_M = 0_Rm. \] Moreover, \[ m + (-1_R)m = 1_R m + (-1_R)m = (1_R - 1_R)m = 0_R m = 0_M, \] so \((-1_R)m = -m\).

Proof (of Lemma 11.1.5)

First, we show 1). If \(M\) is a \(\mathbb{Z}\)-module, then \((M,+)\) is an abelian group by definition of module. Conversely, if \((M,+)\) is an abelian group then there is a unique \(\mathbb{Z}\)-module structure on \(M\) given by the formulas below. The uniqueness of the \(\mathbb{Z}\) action follows from the identities below in which the right hand side is determined only by the abelian group structure of \(M\). The various identities follow from the axioms of a module: \[ \begin{cases} i \cdot m = (\underbrace{1 + \cdots + 1}_i) \cdot m = \underbrace{1 \cdot m + \cdots +1 \cdot m}_i = \underbrace{ m + \cdots + m}_i & \text{ if } i>0\\ 0 \cdot m = 0_M & \\ i \cdot m = - (-i) \cdot m = - (\underbrace{m + \cdots + m}_{-i}) & \text{ if } i<0. \end{cases} \] We leave it as an exercise to check that this \(\mathbb{Z}\)-action really satisfies the module axioms.

Now we show 2). If \(M\) is a \(\mathbb{Z}/n\) module, then \((M,+)\) is an abelian group by definition, and \(nm= \underbrace{ m + \cdots + m}_n=\underbrace{[1]_n \cdot m + \cdots +[1]_n \cdot m}_n=[0]_nm=0_M\).

Conversely, there is a unique \(\mathbb{Z}/n\)-module structure on \(M\) given by the formulas below, which are analogous to the ones above: \[ \begin{cases} [i]_n \cdot m = (\underbrace{[1]_n+ \cdots + [1]_n}_i) \cdot m = \underbrace{[1]_n \cdot m + \cdots +[1]_n \cdot m}_i= \underbrace{ m + \cdots + m}_i & \text{ if } i>0\\ 0 \cdot m = 0_M &\\ [i]_n \cdot m = - (-[i]_n) \cdot m = - (\underbrace{m + \cdots + m}_{-i}) & \text{ if } i<0. \end{cases} \] These formulas are well-defined, meaning they are independent of the choice of representative for \([i]_n\), because of the assumption that \(nm=0_M\). Again checking that this \(\mathbb{Z}/n\)-action really satisfies the module axioms is left as an exercise.

Proof (of Lemma 11.2.4)

Exercise.

Proof (of Lemma 11.2.9)

Let \(r,s \in R\) and \(m,n \in M\). One checks that the axioms in the definition of a module hold for the given action using properties of ring homomorphisms. For example: \[ (r+s)m=\phi(r + s)m= (\phi(r)+\phi(s))m=\phi(r)m + \phi(s)m=rm+sm. \] The remaining properties are left as an exercise.

Proof (of Lemma 11.3.11)

There are many things to check, including:

• The addition and the \(R\)-action are both well-defined: given \(f,g\in \mathrm{Hom}_R(M,N)\) and \(r\in R\), we always have \(f+g, rf\in \mathrm{Hom}_R(M,N)\).
• The axioms of an \(R\)-module are satisfied for \(\mathrm{Hom}_R(M,N)\).

We leave the details as exercises.

Proof (of Lemma 11.3.12)

Let \(f\!:M\to \mathrm{Hom}_R(R,M)\) be given for each \(m\in M\) by \(f(m)=\phi_m\) where \(\phi_m\) is the map defined by \(\phi_m(r)=rm\) for all \(r\in R\). Now we have many things to check:

• \(f\) is well-defined, meaning that for any \(m\in M\), its image \(f(m) = \phi_m\) is an element of \(\mathrm{Hom}_R(R,M)\), since \[ \phi_m(r_1+r_2)=(r_1+r_2)m=r_1m+r_2m=\phi_m(r_1)+\phi_m(r_2) \] \[ \phi_m(r_1r_2)=(r_1r_2)m=r_1(r_2m)=r_1\phi_m(r_2) \] for all \(r_1,r_2\in R\).
• \(f\) is an \(R\)-module homomorphism, since \[ \phi_{m_1+m_2}(r)=r(m_1+m_2)=rm_1+rm_2=\phi_{m_1}(r)+\phi_{m_2}(r) \] \[ \phi_{r'm}(r)=r(r'm)=(rr')m=r'(rm)=r'\phi_{m}(r) \]
• \(f\) is injective, since \(\phi_m=\phi_{m'}\) implies in particular that \(\phi_m(1_R)=\phi_{m'}(1_R)\), which by definition of \(\phi_{-}\) means that \(m=m'\).
• \(f\) is surjective, since for \(\psi \in \mathrm{Hom}_R(R,M)\) we have \(\psi(r)=\psi(r1_R)=r\psi(1_R)\) for all \(r\in R\), so \(\psi=\phi_{\psi(1_R)}\).

This shows that \(f\) is an \(R\)-module isomorphism.

Proof (of Lemma 11.3.15)

Among the many things to check here, we will only check the well-definedness of the \(R\)-action on \(M\), and leave the others as exercises. To check well-definedness, consider \(m+N=m'+N\). Then \(m-m'\in N\), so \(r(m-m')\in N\) by the definition of submodule. This gives that \(rm-rm'\in N\), hence \(rm+N=rm'+N\).

Proof (of Theorem 11.3.18)

By 817, we already know that \(\overline{f}\) is a well-defined homomorphism of groups under \(+\) and that it is the unique one such that \(\overline{f} \circ q = f\). It remains only to show \(\overline{f}\) is an \(R\)-linear map: \[ \overline{f}(r (m +N)) = \overline{f} (rm + N) = f(rm) = r f(m) = r \overline{f}(m + N). \] where the third equation uses that \(f\) preserves scaling.

Proof (of Theorem 11.3.19)

If we forget the multiplication by scalars in \(R\), by the First Isomorphism Theorem for Groups, we know that there is an isomorphism of abelian groups under \(+\), given by \[ \begin{aligned} M/\mathrm{ker}(h) &\longrightarrow & \mathrm{im}(h) \\ m+\mathrm{ker}(h) &\longmapsto& h(m). \end{aligned} \] It remains only to show this map preserves multiplication by scalars. And indeed: \[ \begin{aligned} \overline{h}(r(m+\ker(h))) & = \overline{h}(rm+\ker(h)) & \textrm{by definition of the \(R\)-action on } M/\ker(h)\\ & = h(rm) & \textrm{by definition of } \overline{h} \\ & = rh(m) & \textrm{ since \(h\) is an \(R\)-module homomorphism} \\ & = r \overline{h}(m+ \ker h) & \textrm{by definition of } h. \end{aligned} \]

Proof (of Theorem 11.3.20)

We know that \(A+B\) and \(A \cap B\) are submodules of \(M\). By the Diamond Isomorphism Theorem for Groups, there is an isomorphism of abelian groups \[\begin{aligned} A/(A \cap B) &\cong (A+B)/B \\ a + (A\cap B) &\mapsto a + B\end{aligned} \] It remains only to show \(h\) preserves multiplication by scalars: \[ h(r(a+(A \cap B))) = h(ra + A \cap B) = ra + B = r(a +B) = rh(a + (A \cap B)). \]

Proof (of Theorem 11.3.21)

From 817, we know that \(B/A\) is a subgroup of \(M/A\) under \(+\). Given \(r \in R\) and \(b +A \in B/A\) we have \(r(b+A) = rb + A\) which belongs to \(B/A\) since \(rb \in B\). This proves \(B/A\) is a submodule of \(M/A\). By the Cancelling Isomorphism Theorem for Groups, there is an isomorphism of abelian groups \[\begin{aligned} (M/A)/(B/A) &\cong M/B \\ (m+A) + B/A &\mapsto m + B \end{aligned} \] and it remains only to show this map is \(R\)-linear: \[ \begin{aligned} h(r((m+A) + B/A)) = & h(r(m+A) + B/A) = h((rm + A) + B/A) \\ & = rm + B = r(m +B)\\ & = r h((m+A) + B/A). \end{aligned} \]

Proof (of Theorem 11.3.22)

From 817, we know there is a bijection between the set of subgroups of \(M\) and that contain \(N\) and subgroups of the quotient group \(M/N\), given by the same map \(\Psi\). We just need to prove that these maps send submodules to submodules. If \(K\) is a submodule of \(M\) containing \(N\), then by the Cancelling Isomorphism Theorem we know that \(K/N\) is a submodule of \(M/N\). If \(T\) is a submodule of \(M/N\), then \(\pi^{-1}(T)\) is an abelian group, by 817. For \(r \in R\) and \(m \in \pi^{-1}(T)\), we have \(\pi(m) \in T\), and hence \(\pi(rm) = r\pi(m) \in T\) too, since \(T\) is a submodule. This proves \(\pi^{-1}(T)\) is a submodule.

Proof (of Lemma 11.4.6)

The proof of (2) will be a problem set question. To show (1), note that if \(M=RA\) then \(M/N=R\bar{A}\), where \(\bar{A}=\{a+N \mid a\in A\}\).

Proof (of Lemma 11.4.11)

Suppose that if \(m\neq 0\) and \(A_1,A_2\) are finite subsets of \(A\) such that \[ m=\sum_{a\in A_1}r_aa=\sum_{b\in A_2}s_bb \] for some \(r_a, s_b \in R\). Then \[ \sum_{a\in A_1\cap A_2} (r_a-s_a)a+\sum_{a\in A_1\setminus A_2} r_aa-\sum_{a \in A_2\setminus A_1} s_aa=0. \] Since \(A\) is a linearly independent set, we conclude that \(r_a=s_a\) for \(a\in A_1\cap A_2\), \(r_a=0_R\) for \(a\in A_1\setminus A_2\), and \(s_a=0_R\) for \(a \in A_2\setminus A_1\). Set \[ B := \{a \in A_1\cap A_2 \mid r_a \neq 0_R\}. \] Then \[ m = \displaystyle\sum_{a\in B}r_aa \] is the unique way of writing \(m\) as a linear combination of elements of \(A\) with nonzero coefficients.

Proof (of Theorem 11.4.12)

We have two things to prove: existence and uniqueness.

Existence: Any \(0\neq m\in M\) can be written uniquely as \[ m=r_1b_1+\dots+r_nb_n \] with \(b_i\in B\) distinct and \(0 \neq r_i \in R\). Define \(h\!: M \to N\) by \[ \begin{cases} h(r_1b_1+\dots+r_nb_n) = r_1j(b_1) + \cdots +r_nj(b_n) & \text{ if } r_1b_1 + \cdots + r_nb_n \neq 0\\ h(0_M)=0_N \end{cases} \] One can check that this satisfies the conditions to be an \(R\)-module homomorphism (exercise!).

Uniqueness: Let \(h:M\to N\) be an \(R\)-module homomorphism such that \(h(b_i)=j(b_i)\). Then in particular \(h\!:(M,+)\to (N,+)\) is a group homomorphism and therefore \(h(0_M)=0_N\) by properties of group homomorphisms. Furthermore, if \(m=r_1b_1+\dots+r_nb_n\) then \[ h(m)=h(r_1b_1+\dots+r_nb_n)=r_1h(b_1)+\dots+r_nh(b_n)=r_1j(b_1)+\dots+r_nj(b_n) \] by the definition of homomorphism, and because \(h(b_i)=j(b_i)\).

Proof (of Corollary 11.4.13)

Let \(g:M\to N\) and \(h:N\to M\) be the module homomorphisms induced by the bijection \(j:A\to B\) and its inverse \(j^{-1}:B\to A\), which exist by the UMP for free module. We will show that \(h\) and \(g\) are inverse homomorphisms. First, note that \(g \circ h:N\to N\) is an \(R\)-module homomorphism and \((g \circ h)(b) = g(j^{-1}(b))=j(j^{-1}(b))=b\) for every \(b\in B\). Since the identity map \(\mathrm{id}_N\) is an \(R\)-module homomorphism and \(id_N(b)=b\) for every \(b\in B\), by the uniqueness in the UMP for free modules we have \(g \circ h = \mathrm{id}_n\). Similarly, one shows that \(h \circ g = \mathrm{id}_M\).

Proof (of Theorem 11.4.16)

You will show this in the next problem set (at least in the case where \(M\) has a finite basis).