Module Constructions and Universal Properties

Preliminary

Recall the definition of left $R$-module.

• $1_{R}a=a$.

Link to original

Similarly we can define right $R$-module, where $(r,a)\to ar$. A left module can be converted to a right module and vice versa, by considering $R^{\mathrm{op}}$. In particular, if $R$ is commutative, then a left module is also a right module.

For any $r\in R$, define ${\phi }_r:M\to M,m\mapsto rm$. Then ${\phi }_r$ is a group homomorphism and ${\phi }_r\in \mathrm{End}(M)$. It deduces that

$$R\to \mathrm{End}(M),r\mapsto {\phi }_r$$

is a ring homomorphism. It is a motivation of the following theorem.

Theorem

Let $M$ be an abelian group. Then $M$ is an $R$-module iff there is a ring homomorphism $R\to \mathrm{End}(M)$.

Facts.

• A submodule of finitely generated module may not finitely generated. Similarly, recall that subring of Noetherian ring is not Noetherian.
  • A ring $R$ is a $R$-module itself. Its submodules are just its ideals. Let us consider $R=\mathbb{R}\left[x_1,x_2,\dots ,x_n,\dots \right]$ and $I=⟨x_1,x_2,\dots ,x_n,\dots ,⟩$. Note that $R$ is finitely generated by $1_{\mathbb{R}}$ and $I$ is not finitely generated.
• If $R$ is a Noetherian ring, then all submodule of a finitely generated module is finitely generated.

Definition

Let $\phi :M\to M^{\prime }$ be a $R$-homomorphism. Define $\mathrm{coker}(\phi )=M^{\prime }/\mathrm{im}(\phi )$ and $\mathrm{coim}(\phi )=M/\ker (\phi )$. Furthermore, there exists a unique isomorphism $\overline{\phi }:\mathrm{coim}\phi \to \mathrm{im}\phi$ such that the following diagram commutes.

$$\text{require}AMScd\begin{array}{ccc}M& \stackrel{\phi }{\to }& M^{\prime }\\ {\pi }_{\ker \phi }\downarrow & & \uparrow \\ \mathrm{coim}(\phi )& \stackrel{\overline{\phi }}{\to }& \mathrm{im}(\phi )\end{array}$$

Ideal Action on Modules and R/I-Module Structures

Definition

Let $M$ be an $R$-module, and let $I$ be an ideal of $R$. Denote by $IM$ the submodule of $M$ generated by $\{am:a\in I,m\in M\}$.

Proposition

Let $M$ be a $R$-module.

• $R\to \mathrm{End}(M)$ factorizes through the natural surjection $R\to R/I$ iff $IM=0$.
• The module $M/IM$ is naturally endowed with a structure of $R/I$-module.
• Let $\phi :M\to M^{\prime }$ be an $R$-homomorphism, then $\phi (IM)\subseteq I{M}^{\prime }$ and $\phi$ induces a morphism of $R/I$-modules $M/IM\to M^{\prime }/I{M}^{\prime }$.

\begin{proof} i) Note that $R\to \mathrm{End}(M)$ factorizes through the natural surjection $R\to R/I$ means there exists ${\rho }_I:R/I\to \mathrm{End}(M)$ such that the diagram

commutes. By the universal property of ring homomorphism, ${\rho }_I$ exists iff $I\subseteq \ker \rho$ iff $IM=0$.

ii) We define $R/I$ acting on $M/IM$ as $(r+I)\cdot (m+IM)=rm+IM$, then it is trivial to check $M/IM$ is a $R/I$-module.

iii) For any $x\in IM$, $x$ can be written as a finite sum of $x={\sum }_j{a}_j{m}_j$. Then

$$\phi (x)=\sum \phi (a_j{m}_j)=\sum a_j\phi (m_j)\in I{M}^{\prime }$$

and so $\phi (IM)\subseteq I{M}^{\prime }$. Define $\overline{\phi }:M/IM\to M^{\prime }/I{M}^{\prime }$ naturally, and it is easy to check it is a morphism between $R/I$-modules. \end{proof}

Weird Examples

• There is a module having no linearly independent elements.
• There is a module, some of whose submodules have no complement.
• There is a finitely generated module, one of its submodule is not finitely generated.
• There is a module with linearly independent subset $S$ such that any element is not a linear combination of other elements of $S$.
• There is a module: a minimal generated set is not basis, and a maximal linearly independent set not basis.
• There is a free module with non-free submodules and non-free quotient modules.
• There is an $R$-module $M$: there exists $r\in R$, $m\in M$ with $r,m\ne 0$ such that $rm=0$.
• There is a free module such that
  • linearly independent subset is not contained in any basis;
  • a generated subset does not contain any basis.

Quotient Module: Universal Property

Proposition

Let $M$ be an $R$-module, and let $N$ be an submodule of $M$. Let $L$ be an $R$-module and $\phi :M\to L$ a morphism such that $N\subseteq \ker (\phi )$. Then there exists a unique morphism ${\phi }_N:M/N\to L$ such that the following diagram commutes.

Moreover, $\ker ({\phi }_N)=\ker (\phi )/N$.

Remark. Modules $M$ and $N$ are defined as above. The pair $(M/N,{\pi }_N)$ is unique in the following sense.

• If $(M^{\prime },{\pi }^{\prime })$ is a pair such that

  • $M^{\prime }$ is an $R$-module
  • ${\pi }^{\prime }:M\twoheadrightarrow M^{\prime }$ is a surjective morphism satisfying $\ker {\pi }^{\prime }\supseteq N$ and the property of $(M/N,{\pi }_N)$ described as above proposition, that is, for any morphism ${\phi }^{\prime }:M\to L^{\prime }$ with $N\subseteq \ker \phi$, there exists a unique morphism ${\phi }_{M^{\prime }}^{\prime }:M^{\prime }\to L$ such that the diagram

  

  commutes,

• then $M^{\prime }\simeq M/N$.

• The proof is as follows: take $L^{\prime }=M/N$, then there exists a unique isomorphism $\sigma :M/N\stackrel{\sim }{\to }{M}^{\prime }$ such that the following diagram commutes, where $\sigma ={\phi }_{M^{\prime }}^{\prime -1}$.

Free Module

Theorem

Let $R$ be a commutative ring with $1$, and let $M$ be a finitely generated free $R$-module. Then any basis of $R$ has the same cardinality.

\begin{proof} First we claim that any basis of $M$ is a finite set. Suppose that $\{x_i:i\in I\}$ is a basis and $|I|=\infty$. Since $M$ is finitely generated, there exists $y_1,\cdots ,y_n$ such that $M=⟨y_1,\cdots ,y_n⟩$. Note that $y_i$ can be expressed as a linear combination of $\{x_i:i\in I_0\}$ with $|I_0|<\infty$. There exists $j\in I\setminus I_0$ such that $x_j$ is a linear combination of $y_i$ and so is a linear combination of $\{x_i:i\in I_0\}$, which is a contradiction.

Let $J$ be a maximal ideal of $R$. Consider $JM$ and $M/JM$, then $M/JM$ is an $R/J$-module. Since $F:=R/J$ is a field, $M/JM$ is a vector space over $F$. Let $x_1,\cdots ,x_r$ be a basis of $M$. It suffices to show $\overline{x_1},\cdots ,\overline{x_r}$ is a basis of $M/JM$. It is easy to show they generate $M/JM$, so it remains to show they are linearly independent. Suppose that $\overline{a_1}\overline{x_1}+\cdots +\overline{a_r}\overline{x_r}=0$, then $a_1{x}_1+\cdots +a_r{x}_r\in JM=J{x}_1+\cdots +J{x}_r$. There are $a_1^{\prime },\cdots ,a_r^{\prime }$ such that

$$a_1{x}_1+\cdots +a_r{x}_r=a_1^{\prime }{x}_1+\cdots +a_r^{\prime }{x}_r$$

with $a_i^{\prime }\in J$. Since $x_1,\cdots ,x_r$ is a basis of $M$, then $a_i\in J$ and so $\overline{a_i}=0$. Hence, $\overline{x_1},\cdots ,\overline{x_r}$ are linearly independent. \end{proof}

Definition

Let $I$ be a set. Write $R^I={\prod }_{i\in I}R$ and $R^{(I)}={\oplus }_{i\in I}R$. Let $M$ be a $R$-module. Take $x_i$, $i\in I$ be a family of elements of $M$. The submodule generated by $(x_i{)}_{i\in I}$ is denoted by ${\sum }_{i\in I}R{x}_i$, and we have a morphism of $R$-modules

$$\phi :R^{(I)}\to M,(r_i{)}_{i\in I}\mapsto \sum _{i\in I}{r}_i{x}_i.$$

Then

• $\phi$ injective iff $x_i,i\in I$ are linearly independent;
• $\phi$ surjective iff $x_i$, $i\in I$ generate $M$;
• $\phi$ isomorphism iff $M$ is free with a basis $\{x_i{\}}_{i\in I}$.

Lemma

Let $L$ be a free $R$-module with basis $(x_i{)}_{i\in I}$. For any $R$-module $M$, the map ${\mathrm{Hom}}_R(L,M)\to M^I,\varphi \mapsto (\varphi (x_i){)}_{i\in I}$ is a bijection.

Proposition

Let $L$ be a free $R$-module.

• Let $\pi :X\twoheadrightarrow Y$ be an epimorphism of $R$-modules. For any $R$-morphism $\phi :L\to Y$, there exists a morphism $\tilde{\phi }:L\to X$ such that the diagram commutes.
• In particular, for any $R$-module $M$, any epimorphism $\pi :M\twoheadrightarrow L$ has a section, that is, there is a morphism $\sigma :L\to M$ such that $\pi \sigma ={\mathrm{id}}_L$. In that case, $\sigma$ is injective, $\mathrm{im}(\sigma )\simeq L$ and $M=\ker \pi \oplus \mathrm{im}(\sigma )$.

\begin{proof} i) Let $\{e_i{\}}_{i\in I}$ be a basis of $L$. For $i\in I$, let $x_i$ be an element of $X$ with $\pi (x_i)=\phi (e_i)$. Define $\tilde{\phi }:L\to X,e_i\to x_i$. Then $\tilde{\phi }$ is what we desire.

ii) is easy to prove by i). \end{proof}

Definition

Let $M$ be a module. If $M=M_1\oplus M_2$, then $M_1$ is called a direct summand of $M$.

• We say nonzero $M$ is simple if $M$ has only two submodules $\{0\}$ and $M$.
• We say $M$ is indecomposable if $M$ cannot express as $M=M_1\oplus M_2$ with $M_1\ne 0$ and $M_2\ne 0$.

Tensor Product

The tensor product $M{\otimes }_{R}N$ of two modules $M$ and $N$ “linearizes the bilinear map”.

Definition

Let $M$ and $N$ be $R$-module. The tensor product of $M$ and $N$ is pair $(M{\otimes }_{R}N,t_{M,N})$ where $M{\otimes }_{R}N$ is an $R$-module and $t_{M,N}:M\times N\to M{\otimes }_{R}N$ is a bilinear map, usually denoted $(m,n)\mapsto m{\otimes }_{R}n$ which satisfies the following universal properties: whenever $X$ is a $R$-module and $f:M\times N\to X$ is a bilinear map, there exists a unique $R$-morphism $\tilde{f}:M{\otimes }_{R}N\to X$ such that the following diagram commutes

Remark. Up to isomorphism, tensor product exists and it is unique. See ^aa5afc.

Basic Property

Proposition

Fix $m\in M$, then the map $N\to m{\otimes }_{R}N,n\mapsto m{\otimes }_{R}n$ is a morphism of $R$-modules. Its image $m{\otimes }_{R}N=\{m{\otimes }_{R}n:n\in N\}$ is a submodule of $M{\otimes }_{R}N$, is isomorphic to a quotient of $N$.

Remark. Recall Strange Things about Tensor Product. Take $M=\mathbb{Z}/2\mathbb{Z}$ and $N=\mathbb{Z}$, then $m\otimes N=1\otimes \mathbb{Z}\simeq \mathbb{Z}/2\mathbb{Z}$.

Proposition

We have unique isomorphisms

• $(M_1\otimes M_2)\otimes M_3\to M_1\otimes (M_2\otimes M_3)$ and $M_1\otimes M_2\to M_2\otimes M_1$;
• $M_1\otimes (M_2\oplus M_3)\to (M_1\otimes M_2)\oplus (M_1\otimes M_3)$;
• $M\otimes ({\oplus }_{i\in I}{M}_i)\to {\oplus }_{i\in I}(M{\otimes }_R{M}_i)$;
• $R\otimes M\to M,\lambda \otimes m\mapsto \lambda m$ and $R^{(I)}\otimes M\to M^{(I)},({\lambda }_i{)}_{i\in I}\otimes m\mapsto ({\lambda }_{i}m{)}_{i\in I}$.

Theorem

Let $M,N,P$ be $R$-module, we have a canonical isomorphism $\varphi :={\varphi }_{M,N,P}:{\mathrm{Hom}}_R(M{\otimes }_{R}N,P)\to {\mathrm{Hom}}_R(M,{\mathrm{Hom}}_R(N,P))$, where $\varphi (f)(m):N\to P,n\mapsto f(m\otimes n)$, and ${\varphi }^{-1}(g):(m,n)\mapsto g(m)n$ for any $g\in {\mathrm{Hom}}_R(M,{\mathrm{Hom}}_R(N,P))$.

\begin{proof} See ^60fa98. \end{proof}

Remark. The meaning of “canonical”: for any morphism $f:M^{\prime }\to M$, the following diagram commutes.

Multilinear Map

Definition

Define

$$L^n(M_1,\cdots ,M_n;N):=\{f\mid f:M_1\times \cdots \times M_n\to N\text{mbox}isan-linearmap\}.$$

In particular, $L^2(M_1\times M_2;N)\simeq L^1(M_1\otimes M_2,N)\simeq L^1(M_1,L^1(M_2,N))$.

Relation between Dual Modules and Homomorphisms

Proposition

Let $M,N$ be finite generated free $R$-module. Define $M^{\vee }={\mathrm{Hom}}_R(M,R)=L(M,R)$, then $M^{\vee }\otimes N\simeq {\mathrm{Hom}}_R(M,N)=L(M,N)$.

\begin{proof} Define $\phi :M^{\vee }\times N\to {\mathrm{Hom}}_R(M,N),(f,n)\mapsto \phi (f,n):m\mapsto f(m)n$. Since $\phi$ is bilinear, it induces a morphism $M^{\vee }\otimes N\to {\mathrm{Hom}}_R(M,N)$. On the other hand, let $v_1,\cdots ,v_n$ be a basis of $M$, and let $v_1^{\vee },\cdots ,v_n^{\vee }$ be a dual basis of $M^{\vee }$. Define $\psi :{\mathrm{Hom}}_R(M,N)\to M^{\vee }\otimes N,g\mapsto \sum v_i^{\vee }\otimes g(v_i)$. It is easy to verify $\phi \circ \psi$ and $\psi \circ \phi$ are identity. \end{proof}

Extension of Scalars

Proposition

• Assume that $R$ is a subring of a ring $R^{\prime }$. Let $M$ be an $R$-module. Then the bilinear map $R^{\prime }\times (R^{\prime }\otimes M)\to R^{\prime }\otimes M,({\lambda }_1^{\prime },{\lambda }_2^{\prime }\otimes m)\mapsto {\lambda }_1^{\prime }{\lambda }_2^{\prime }\otimes m$ defines a structure of $R^{\prime }$-module on the $R$-module $R^{\prime }\otimes M$.
• More generally, let $f:R\to T$ be a ring homomorphism. View $T$ as an $R$-module defined by $R\times T\to T,(\lambda ,\mu )\mapsto f(\lambda )\mu$. The bilinear map $T\times (T{\otimes }_{R}M)\to T{\otimes }_{R}M,({\mu }_1,{\mu }_2\otimes m)\mapsto {\mu }_1{\mu }_2\otimes m$ defines a $T$-module structure on the $R$-module $T{\otimes }_{R}M$.

Example. Let $V$ be an $n$-dimensional $\mathbb{Q}$-space, i.e., $V\simeq {\mathbb{Q}}^n$, then $\mathbb{C}{\otimes }_{\mathbb{Q}}V{\simeq }_{\mathbb{Q}}\mathbb{C}\otimes (\mathbb{Q}\oplus \cdots \oplus \mathbb{Q})\simeq (\mathbb{C}{\otimes }_{\mathbb{Q}}\mathbb{Q}{)}^n\simeq {\mathbb{C}}^n$.

Lemma

Let $I$ be an ideal of $R$ and let $M$ be an $R$-module. The morphism of $R$-module $M\to (R/I){\otimes }_{\mathbb{R}}M,\mapsto 1_{R/I}\otimes m$ induces an isomorphism $M/IM\simeq (R/I){\otimes }_{\mathbb{R}}M$.