Modules

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 29 May 2011

Modules

Let $R$ be a ring with identity $1\in R$.

• A left $R$-module is a set $M$ with functions

  $\begin{array}{ccc}M\times M& \to & M\\ (m_1,m_2)& ⟼& m_1+m_2\end{array}$     and     $\begin{array}{ccc}R\times M& \to & M\\ (r,m)& ⟼& rm\end{array}$,

  addition and $R$-action, such that

  (a)   If $m_1,m_2,m_3\in M$ then $(m_1+m_2)+m_3=(m_1+(m_2+m_3)$,

  (b)   If $m_1,m_2\in M$ then $m_1+m_2=m_2+m_1$,

  (c)   There exists a zero, $0\in M$, such that if $m\in M$ then $0+v=v+0=v$,

  (d)   If $m\in M$ then there exists an additive inverse to $m$, $-m\in M$, such that $(-m)+m=m+(-m)=0$,

  (e)   If $r_1,r_2\in R$ and $m\in M$ then $r_1\left(r_{2}m\right)=\left(r_1{r}_2\right)m$,

  (f)   If $m\in M$ then $1m=m$,

  (g)   If $r\in R$ and $m_1,m_2\in M$ then $r(m_1+m_2)=r{m}_1+r{m}_2$,

  (h)   If $r_1,r_2\in R$ and $m\in M$ then $(r_1+r_2)m=r_{1}m+r_{2}m$,

$R$-modules are the analogues of group actions except for rings.

Note that conditions (a), (b), (c) and (d) in the definition of a left $R$-module imply that every left $R$-module is an abelian group under addition.

HW: Show, using Ex. 2.2.5, Part I, that the element $0\in M$ is unique.
HW: Show, using Ex. 2.2.5, Part I, that if $m\in M$ then the element $-m\in M$ is unique.
HW: Show that if $M$ is a left $R$-module and $m\in M$ then $0\cdot m=0$.

Important examples of modules are:

(a)   If $R$ is a ring then $R$ is a left $R$-module.

(b)   All abelian groups are $\mathbb{Z}$-modules.

(c)   If $R$ is a field then the $R$-modules are vector spaces.

$R$-module homomorphisms are for comparing $R$-modules.

Let $R$ be a ring and let $M$ and $N$ be $R$-modules.

• An $R$-module homomorphism from $M$ to $N$ is a function $f:M\to N$ such that

  (a)   If $m_1,m_2\in M$ then $f(m_1+m_2)=f\left(m_1\right)+f\left(m_2\right)$,

  (b)   If $r\in R$ and $m\in M$ then $f\left(rm\right)=rf\left(m\right)$.

• A $R$-module isomorphism is a bijective $R$-module homomorphism.
• Two left $R$-modules $M$ and $N$ are isomorphic, $M\simeq N$, if there exists a vector space isomorphism $f:M\to N$ between them.

Note that condition (a) in the definition of an $R$-module homomorphism implies that $f$ is a group homomorphism.
HW: Show that if $M$ and $N$ are left $R$-modules and if $f:M\to N$ is an $R$-module homomorphism then $f\left(0_M\right)=0_N$, where $0_M$ and $0_N$ are the zeros in $M$ and $N$, respectively.
HW: Let $f:M\to N$ be an $R$-module homomorphism. Show that if $m\in M$ then $f(-m)=-f\left(m\right)$.

• A submodule of an $R$-module $M$ is a subset $M\subseteq M$ such that

  (a)   If $n_1,n_2\in N$ then $n_1+n_2\in N$,

  (b)   $0\in N$,

  (c)   If $n\in N$ then $-n\in N$,

  (d)   If $n\in N$ and $r\in R$ then $rn\in N$.

• The zero $R$-module, $\left(0\right)$, is the set containing only $0$ with operations $0+0=0$ and $r\cdot 0=0$, for $r\in R$.

• Let $M$ be a left $R$-module and let $S$ be a subset of $M$. The submodule generated by $S$ is the submodule $\left(S\right)$ of $M$ such that

  (a)   $S\subseteq \left(S\right)$,

  (b)   If $T$ is a submodule of $M$ and $S\subseteq T$ then $\left(S\right)\subseteq T$.

The submodule $\left(S\right)$ is the smallest submodule of $M$ containing $S$. Think of $\left(S\right)$ as gotten by adding to $S$ exactly those elements of $V$ that are needed to make a submodule.

Cosets

• A subgroup of a left $R$-module $M$ is a subset $N\subseteq M$ such that

  (a)   If $n_1,n_2\in N$ then $n_1+n_2\in N$,

  (b)   $0\in N$,

  (c)   If $n\in N$ then $-n\in N$,

Let $M$ be a left $R$-module and let $N$ be a subgroup of $M$. We will use the subgroup $N$ to divide up the module $M$.

• A coset of $N$ in $M$ is a set $m+N=\{m+n|n\in N\}$, where $m\in M$.
• $M/N$ (pronounced "$M$ mod $N$") is the set of cosets of $N$ in $M$.

Let $M$ be a left $R$-module and let $N$ be a subgroup of $M$. Then the cosets of $N$ in $M$ partition $M$.

Notice that the proofs of Proposition (mdptn) and Proposition (gpptn) are essentially the same.
HW: Write a very short proof of Proposition (mdptn) by using (gpptn).

Quotient modules $\leftrightarrow$ Submodules

Let $M$ be a left $R$-module and let $N$ be a subgroup of $M$. We can try to make the set $M/N$ of cosets of $N$ in $M$ into an $R$-module by defining an addition operation and an action of $R$. This doesn't work with just any subgroup of $N$, the subgroup must be a submodule.

Let $N$ be a subgroup of a left $R$-module $M$. Then $N$ is a submodule of $M$ if and only if $M/N$ with operations given by $$(m_1+N)+(m_2+N)=(m_1+m_2)+N\text{and}r(m+N)=rm+N$$ is a left $R$-module.

Notice that the proofs of Proposition 2.2.5 and Proposition 1.1.8 are essentially the same.
HW: Write a shorter proof of Proposition 2.2.5 by using Proposition 1.1.8.

• The quotient module $M/N$ is the left $R$-module of cosets of a submodule $N$ of an $R$-module vector $M$ with operations given by $(m_1+N)+(m_2+N)=(m_1+m_2)+N$ and $r(m+N)=rm+N$.

We have made $M/N$ into a left $R$-module when $N$ is a submodule of $V$.

HW: Show that if $N=M$ then $M/N\simeq \left(0\right)$.

Kernel and image of a homomorphism

• The kernel of an $R$-module homomorphism $f:M\to N$ is the set $$\ker f=\{m\in M|f\left(m\right)=0_N\},$$ where $0_N$ is the zero element of $N$.
• The image of an $R$-module homomorphism $f:M\to N$ is the set $$\mathrm{im}f=\left\{f\right(m\left)\right|m\in M\}.$$

Let $f:M\to N$ be a linear transformation. Then

(a)   $\ker f$ is a submodule $M$.

(b)   $\mathrm{im}f$ is a submodule of $N$.

Let $f:M\to N$ be a linear transformation. Let $0_M$ be the zero element of $M$. Then

(a)   $\ker f=\left\{0_M\right\}$ if and only if $f$ is injective.

(b)   $\mathrm{im}f=N$ if and only if $f$ is surjective.

Notice that the proof of Proposition (mdinjsur)(b) does not use the fact that $f:M\to N$ is a homomorphism, only the fact that $f:M\to N$ is a function.

(a)   Let $f:M\to N$ be an $R$-module homomorphism and let $K=\ker f$. Define $$\begin{array}{cccc}\hat{f}:& M/\ker f& ⟶& N\\ & m+K& ⟼& f\left(m\right)\end{array}$$ Then $\hat{f}$ is a well defined injective $R$-module homomorphism.

(b)   Let $f:M\to N$ be an $R$-module homomorphism and define $$\begin{array}{cccc}f\prime :& M& ⟶& \mathrm{im}f\\ & m& ⟼& f\left(m\right)\end{array}$$ Then $f\prime$ is a well defined surjective $R$-module homomorphism.

(c)   If $f:M\to N$ is an $R$-module homomorphism then $$M/\ker f\simeq \mathrm{im}f,$$ where the isomorphism is an $R$-module isomorphism.

Direct sums

Suppose $M$ and $N$ are $R$-modules. The idea is to make $M\times N$ into an $R$-module.

• The direct sum $V\oplus W$ of two left $R$-modules $M$ and $N$ is the set $M\times N$ with operations given by $$(m_1,n_1)+(m_2,n_2)=(m_1+m_2,n_1+n_2)\text{and}r(m,n)=(rm,rn),$$ for $m,m_1,m_2\in M$, $n,n_1,n_2\in N$ and $r\in R$. The operations in $M\oplus N$ are componentwise.
• More generally, given left $R$-modules $M_1,M_2,\dots ,M_{\mathrm{s}}$, the direct sum $M_1\oplus M_2\oplus \cdots \oplus M_{\mathrm{s}}$ is the set $M_1\times M_2\times \cdots \times M_{\mathrm{s}}$ with the operations given by $$(m_1,\dots ,m_i,\dots ,m_s)+(m_1,\dots ,n_i,\dots ,n_s)=(m_1+n_1,\dots ,m_i+n_i,\dots ,m_s+n_s)$$ $$\text{and}r(m_1,\dots ,m_i,\dots ,m_s)=(r{m}_1,\dots ,r{m}_i,\dots ,r{m}_s),$$ where $m_i,n_i\in M_i$, $r\in R$, and $m_i+n_i$ and $r{m}_i$ are given by the operations in $M_i$.

HW: Show that these are good definitions, i.e. that, as defined above, $M\oplus N$ and $M_1\oplus M_2\oplus \cdots \oplus M_{\mathrm{n}}$ are left $R$-modules with zeros given by $(0_M,0_N$ and $(0_{M_1},\dots ,0_{M_s}$, respectively. ($0_{M_i}$ denotes the zero element in the left $R$-module $M_i$.)

Notes and References

These notes are written to highlight the analogy between groups and group actions, rings and modules, and fields and vector spaces.

References

[Ram] A. Ram, Notes in abstract algebra, University of Wisconsin, Madison 1993-1994.

[Bou] N. Bourbaki, Algèbre, Chapitre 9: Formes sesquilinéaires et formes quadratiques, Actualités Sci. Ind. no. 1272 Hermann, Paris, 1959, 211 pp. MR0107661.

[Ru] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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