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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 $R^{\mathrm{opp}}$. 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 $R\to \mathrm{End}(M)$.

Example. If $I$ is an ideal of $R$, then $I$ is a left $R$-module. In particular, $R$ is a left $R$-module and $R$-submodule is an ideal of $R$.

Submodule

Definition

Let $M_i(i\in I)$ be $R$-submodule of $M$. Then ${\cap }_{i\in I}{M}_i$ is an $R$-submodule of $M$. Define

$$\sum _{i\in I}{M}_i=\left\{\sum _{i\in I}{m}_i:m_i\in M_i,|\{i\in I:m_i\ne 0\}|<\infty \right\}$$

and it is a $R$-submodule. For any subset $S\subseteq M$, define

$$⟨S⟩=\left\{\sum _{\text{mbox}finitely}{a}_i{s}_i:a_i\in R,s_i\in S\right\}$$

and it is also a $R$-submodule.

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.

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.

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 $\sigma ={\phi }_{M^{\prime }}^{\prime -1}$.

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.

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

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 $IM=0$. See here.
• 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} See ^xy300k. \end{proof}

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} See ^n0vjji. \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 $(x_i{)}_{i\in I}$ also denoted by ${\sum }_{i\in I}R{x}_i$. Then we have a morphism of $R$-modules

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

Then

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

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, $\mathrm{im}(\sigma )=L$ and $M=\ker \pi \oplus \mathrm{im}(\sigma )$.

\begin{proof} See ^03hqr6. \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 $M_2\ne 0$.

Let $P$ be a direct summand of a free $R$-module. Show whenever $X\stackrel{\pi }{\twoheadrightarrow }Y$ is an epimorphism and $\phi :P\to Y$ is a morphism, there exists unique $\tilde{\phi }:P\to X$ such that the diagram commutes.

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 is unique.

Existence of tensor product: See here.

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 $N$.

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 $g\in {\mathrm{Hom}}_R(M,{\mathrm{Hom}}_R(N,P))$.

Remark. The meaning of “canonical” is

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

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} See ^ahhdq2. \end{proof}

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 $C{\otimes }_{\mathbb{Q}}V{\simeq }_{\mathbb{Q}}\mathbb{C}\otimes (\mathbb{Q}\oplus \cdots \oplus \mathbb{Q})\simeq (\mathbb{C}{\otimes }_{\mathbb{Q}}\mathbb{Q}{)}^n\simeq {\mathbb{C}}^n$.

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 $M/IM\simeq (R/I){\otimes }_{\mathbb{R}}M$.

CounterExample. See here.

• 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.

Categories

Definition

A category $\mathcal{C}$ consists of

• a class of object;
• for each pair $(A,B)$, a set ${\mathrm{Hom}}_{\mathcal{C}}(A,B)$, whose elements are called morphisms from $A$ to $B$;
• for each triple of objects $(A,B,C)$ one has a map $\mathrm{Hom}(A,B)\times \mathrm{Hom}(B,C)\to \mathrm{Hom}(A,C),(f,g)\to g\circ f$.

It is assumed that the objects and morphisms satisfy the following conditions.

• If $(A,B)\ne (C,D)$, then $\mathrm{Hom}(A,B)$ and $\mathrm{Hom}(C,D)$ are disjoint.
• If $f\in \mathrm{Hom}(A,B)$ and $g\in \mathrm{Hom}(B,C)$, $h\in \mathrm{Hom}(C,D)$, then $(hg)f=h(gf)$.
• (Unit) For every object $A$, we have an element $1_A\in \mathrm{Hom}(A,A)$ such that $f{1}_A=f$ for any $f\in \mathrm{Hom}(A,B)$ and $1_{A}g=g$ for any $g\in \mathrm{Hom}(B,A)$.

Example. See here.