07 - Modules

Module

Let $R$ be a ring with identity $1_{R}$

Module over $R$ is abelian group $(M, +)$ with

Multiplication Action $a : R \times M \to M$ of $R$ on $M$ written as
$(r, m) \mapsto r . m$
which satisfies

$1_{R} . m = m for all m \in M$

$(r_{1} . r_{2}) . m = r_{1} . (r_{2} . m) for all r_{1}, r_{2} \in R, m \in M$

$(r_{1} + r_{2}) . m = r_{1} . m + r_{2} . m for all r_{1}, r_{2} \in R and m \in M$

$r . (m_{1} + m_{2}) = r . m_{1} + r . m_{2} for all r \in R and m_{1}, m_{2} \in M$

Modules as Ring Homomorphisms into Endomorphism Rings

Let $M$ be an abelian group and ring $R$

Set $End (M) = Hom (M, M)$ of group homomorphisms from $M$ to itself is naturally a ring
where addition is given pointwise and multiplication is given by composition

Giving $M$ the structure of an $R$ -module through action $R \times M \to M$ is equivalent to
Giving ring homomorphism $ϕ : R \to End (M)$

Proof

As $M$ is abelian
If $f_{1}, f_{2} \in End (M)$ then $f_{1} + f_{2}$ is a group homomorphism
As $f_{1} + f_{2} = f_{2} + f_{1}$ then addition in $End (M)$ is commutative
Composition of functions gives $End (M)$ a multiplication which distributes over addition
Hence $End (M)$ is a ring

If map $a : R x M$ denoted by $(r, m) \mapsto r . m$ then equation $ϕ (r) (m) = r . m$
defines $ϕ$ in terms of $ϕ$ and $a$ conversely
Action $m \mapsto r . m$ is a homomorphism of abelian group $M$
with compatibility of $ϕ$ with multiplication and addition

Submodule

Let $M$ be an $R$ -module

Let $N \subseteq M$ then

$N$ is a submodule if it is an abelian subgroup of $M$ and for all $r \in R$ and $n \in N$ then
$r . n \in N$

Submodule Generated by Subset

Let $X$ be an subset of an $R$ -module $M$ then

Submodule generated or spanned by $X$ is
$⟨ X ⟩_{R} = ⟨ X ⟩ = N \supseteq X ⋂ N$
where $N$ goes over submodules of $M$ containing $X$

Equivalent Form

$⟨ X ⟩ = R . X = {i = 1 \sum k r_{i} x_{i} : r_{i} \in R, x_{i} \in X}$

Sum of Submodules

Let $N_{1}, N_{2}$ be submodules then
$N_{1} + N_{2} = {m + n : m \in N_{1}, n \in N_{2}}$

Linearly Independent of a Set of a Module

Let $M$ be a module over $R$
Let set $S \subseteq M$ then

$S$ is linearly independent if
$r_{1} s_{1} + r_{2} s_{2} + \dots + r_{k} s_{k} = 0 where r_{i} \in R, s_{i} \in S for 1 \leq i \leq k$
then
$r_{1} = r_{2} = \dots = r_{k} = 0$

Basis of Module

Let $M$ be a module over $R$
Let set $S \subseteq M$ then

$S$ is a basis for $M$ if and only if it is

Linearly Independent

Spans $M$

Free Module

Let $M$ be a module over $R$

$M$ is a Free module if it has a basis

Freely Generated Module

Let $M$ be a module over $R$

$M$ is finitely generated if it is generated by some finite subset

Quotient Modules

Let $N$ be a submodule of $M$

Module $M / N$ is the quotient module of $M$ by $N$ If
For $r \in R$ and $m + N \in M / N$ then
$r . (m + N) = r . m + N$
Well-defined as if $m_{1} + N = m_{2} + N$ then $m_{1} - m_{2} \in N$ so $r . (m_{1} - m_{2}) \in N$ hence
$r . m_{1} + N = r . m_{2} + N$

Module Homomorphism

Let $M_{1}, M_{2}$ be $R$ -modules then

$ϕ : M_{1} \to M_{2}$ is a $R$ -module homomorphism if

$ϕ (m_{1} + m_{2}) = ϕ (m_{1}) + ϕ (m_{2})$ for all $m_{1}, m_{2} \in M_{1}$

$ϕ (r . m) = r . ϕ (m)$ for all $r \in R$ , $m \in M_{1}$

So $ϕ$ respects addition and multiplication by rings elements with

$ker (ϕ) = {m \in M_{1} : ϕ (m) = 0}$ is a submodules of $M_{1}$
$im (ϕ) = {ϕ (m) : m \in M_{1}}$ is a submodules of $M_{2}$

Submodule Correspondence lemma

Let $M$ be an $R$ -module
Let $N$ be a submodule of $M$

Let $q : M \to M / N$ be the quotient map

If $S$ is a submodule of $M$ then $q (S)$ is a submodule of $M / N$
If $T$ a submodule of $M / N$ then $q^{- 1} (T)$ is a submodule of $M$

Map $T \mapsto q^{- 1} (T)$ is an injective from submodules of $M / N$ to submodules of $M$ containing $N$
Thus $M / N$ correspond bijectively to submodules of $M$ which contain $N$

Proof

For $^{- 1} (T)$
If $m_{1}, m_{2} \in q^{- 1} (T)$ then
$q (m_{1}), q (m_{2}) \in T$
As $T$ is a submodule then
$q (m_{1}) + q (m_{2}) = q (m_{1} + m_{2}) \in T$
Hence $m_{1} + m_{2} \in q^{- 1} (T)$

If $r \in R$ then
$q (r . m_{1}) = r . q (m_{1}) \in T$
As $q (m_{1}) \in T$ and $T$ is a submodule so $r . m_{1} \in q^{- 1} (T)$
Hence $q^{- 1} (T)$ is a submodule of $M$

Similar process for $q (S)$

As $T$ is any subset of $M / N$ then $q (q^{- 1} (T)) = T$ as $q$ is surjective

Then $q^{- 1} (T)$ is a submodule in $M$ then map $S \mapsto q (S)$ is surjective in submodules in $M$ to submodules in $M / N$
Then $T \mapsto q^{- 1} (T)$ is an injective map

As $q (N) = {0} \subseteq T$ then for any submodule $T$ of $M / N$ then
$N \subseteq q^{- 1} (T)$
Hence image of map $T \mapsto q^{- 1} (T)$ consists of submodules of $M$ containing $N$

Suppose $S$ is an arbitrary submodule of $M$ and consider $q^{- 1} (q (S))$ then
$q^{- 1} q (S) = {m \in M : q (m) \in q (S)} = {m \in M : \exists s \in S such that m + N = s + N} = {m \in M : \exists s \in S such that m \in s + N} = S + N$
As $S$ contains $N$ then $S + N = S$ so $q^{- 1} (q (S)) = S$
Hence any submodule $S$ containing $N$ is preimage of submodule of $M / N$

Annihilator of a Element

Let $M$ be an $R$ -module
Let $m \in M$ then

Annihilator of $m$ is defined as
$Ann_{R} (m) = {r \in R : r . m = 0}$

Torsion Element

Let $M$ be an $R$ -module
Let $m \in M$ then

$m$ is a Torsion Element if
$Ann_{R} (m) \neq = {0}$

Torsion Module

Let $M$ be an $R$ -module

$M$ is a Torsion Module if for all $m \in M$ then
$m is torsion element$

Torsion-Free Module

Let $M$ be an $R$ -module

$M$ is Torsion Free if $M$ has no non-zero torsion elements

Cyclic Module

Let $M$ be an $R$ -module

$M$ is a Cyclic Module if it is geenrated by a single element

Torsion Submodule and Torsion-Free Quotient lemma

Let $M$ be an $R$ -module
Let $M^{tor} = {m \in M : Ann_{R} (m) \neq = {0}}$ then

$M^{tor}$ is a submodule of $M$ and Quotient Module $M / M^{tor}$ is a torsion-free module

Proof

Let $x, y \in M^{tor}$ then

Exists nonzero $s, t \in R$ such that
$s . x = t . y = 0$
As $s . t \in R ∖ {0}$ as $R$ is an integral domain hence
$(s . t) (x + y) = t . (s . x) + s . (t . y) = 0$
Thus if $r \in R$ then $s . (r . x) = r . (s . x) = 0$ hence
$M^{tor}$ is a submodule of $M$

Suppose $x + M^{tor}$ is a torsion element in $M / M^{tor}$ then

Exists nonzero $r \in R$ such that $r . (x + M^{tor}) = 0 + M^{tor}$
So $r . x \in M^{tor}$

So by definition, exists $s \in R$ such that $s . (r . x) = 0$
But $s . r \in R ∖ {0}$ as $R$ is an integral domain

As
$(s . r) . x = 0 for x \in M^{tor}$
Hence
$x + M^{tor} = 0 + M^{tor}$
So $M / M^{tor}$ is torsion free

Rank of Module lemma

Let $M$ be a finitely generated free $R$ -module then

Rank $rk (M)$ of $M$ is the size of a basis for $M$ and is uniquely determined

Proof

Let $X = {x_{1}, \dots, x_{n}}$ be a basis for $M$
Let $I$ be the maximal ideal of $R$

Let $I M$ be the submodule generated by set
$I M = {i . m : i \in I, m \in M}$
Let
$M_{I} = {i = 1 \sum n r_{i} x_{i} : r_{i} \in I}$
As $I$ is an ideal then $M_{I}$ is a submodule of $M$

As $X$ generates $M$ then
Any element in form $r . m$ where $r \in I$ and $m \in M$ means $I M \subseteq M_{I}$

As $I M$ contains $r_{i} x_{i}$ for $r_{i} \in I$ where $i \in {1, 2, \dots n}$
Then sums $\sum_{i = 1}^{n} r_{i} x_{i}$ are also in $I M$ so $M_{I} \subseteq I M$ hence
$M_{I} = I M$
where submodule $M_{I} = I M$ does not depend on choice of basis of $X$

Let $q : M \to M / I M$ be the quotient map
Quotient Module $M / I M$ is a module for $R$ as well as quotient field $k = R / I$ through
$(r + I) . q (m) = q (r . m)$
If $r - r^{'} \in I$ then
$r . m - r^{'} . m = (r - r^{'}) . m \in I M$
Hence $q (r^{'} . m) = q (r . m)$

Suppose $X$ is a basis for $M$ then $q (X)$ is a basis for $k$ -vector space $M / I M$
Assume $∣ X ∣ = dim_{k} (M / I M)$ where dimension is independent of $X$
As $X$ generates $M$ and $q$ is surjective then $q (X)$ generates $M / I M$
Suppose
$i = 1 \sum n c_{i} q_{i} (x_{i}) = 0 \in M / I M where c_{i} \in k$
For $r_{i} \in R$ for $c_{i} \in R / I$ then
$0 = i = 1 \sum n c_{i} q (x_{i}) = i = 1 \sum n q (r_{i} x_{i}) = q (i = 1 \sum n r_{i} x_{i})$
Then
$y = i = 1 \sum k r_{i} x_{i} \in ker (q) = I M$
As $I M = M_{I}$ then for $r_{i} \in I$ for each $i$ then $r_{i} \in I$ for each $i$ so
$c_{i} = 0$
Hence $\overline{X}$ is linearly independent and hence a $k$ -basis for $M / I M$

Submodules of Free Modules over a PID Are Free

Let $M$ be a finitely generated free module over $R$ a PID

Let $X = {e_{1}, \dots, e_{n}}$ be a basis

If $N$ is a submodule of $M$ then
$N$ is free and has rank at most $n$ elements

Proof

Proof by induction on $n = ∣ X ∣$

If $n = 1$ then
$ϕ : R \to M$
is homomorphism defined by $ϕ (r) = r e_{1}$
By First Isomorphism Theorem then $M ≅ R$

As submodules of $R$ is an ideal and $R$ is a PID then any submodules of $N$ of $R$ are cyclic
Hence
$N = R a for a \in R$
As $R$ is a integral domain then
${a} is a basis of N$
where $a \neq = 0$ so $N$ is free of rank $0$ or $1$

Suppose $n > 1$

Let $W = R e_{1} + R e_{2} + \dots + R e_{n - 1}$
Let $N_{1} = N \cap W$

As $N_{1}$ is a submodule of free module $W$ and $W$ has rank $n - 1$
Then by induction $N_{1}$ is free of rank $k \leq n - 1$

Let ${v_{1}, \dots, v_{k}}$ be a basis of $N_{1}$
By second isomorphism theorem then
$N / N_{1} ≅ (N + W) / W \subseteq M / W$
As $M / W$ then it has basis ${e_{n} + W}$ so $N / N_{1}$ is either zero or free of rank $1$
If $N = N_{1}$ then done

Otherwise let $v_{k + 1}$ so that $v_{k + 1} + N_{1}$ is a basis of $N / N_{1}$

Claim ${v_{1}, \dots, v_{k + 1}}$ is a basis for $N$
If $m \in N$ then since ${v_{k + 1} + N_{1}}$ is a basis of $N / N_{1}$ then
Exists $r_{k + 1} \in R$ such that
$m + N_{1} = r_{k + 1} v_{k + 1} + N_{1}$
However
$m - r_{k + 1} v_{k + 1} \in N_{1}$
So as $N_{1}$ has basis ${v_{1}, \dots, v_{k}}$ then exists $r_{1}, \dots, r_{k} \in R$ such that
$m - r_{k + 1} v_{k + 1} = i = 1 \sum k r_{k} v_{k} = 0$
Hence
$m = i = 1 \sum k + 1 r_{i} v_{i}$
Suppose $\sum_{i = 1}^{k + 1} s_{i} v_{i} = 0$ then
$0 + N_{1} = i = 1 \sum k + 1 s_{i} v_{i} + N_{1} = s_{k + 1} v_{k + 1} + N_{1}$
Hence $s_{k + 1} = 0$ as ${v_{k + 1} + N_{1}}$ is a basis for $N / N_{1}$
But
$i = 1 \sum k s_{k} v_{k} = 0 ⟹ s_{i} = 0 for all 1 \leq i \leq k$
Hence
${v_{1}, \dots, v_{k + 1}}$
is a basis for $N$ and since $k + 1 \leq (n - 1) + 1 = n$ then done

Module Homomorphisms

Let $M, N$ be $R$ -modules

Let $Hom_{R} (M, N)$ denote the set of module homomorphisms from $M$ to $N$
with property if $ϕ, ψ \in Hom_{R} (M, N)$ then $ϕ + ψ$ is a module homomorphism where
$(ψ + ϕ) (m) = ψ (m) + ϕ (m) and (r . ψ) (m) = r . (ψ (m))$

Homomorphisms from Free Modules Are Determined by a Basis lemma

Let $M$ and $N$ be $R$ -module
Let $ϕ : M \to N$ be a $R$ -module homomorphism then

Let $X$ be a spanning set for $M$ , then $ϕ$ is uniquely determined by the restriction to $X$

Let $X$ be a basis for $M$
For any function $f : X \to N$ then
Exists unique $R$ -module homomorphism $ϕ_{f} : M \to N$

Proof

If $v \in M$ then as $X$ spans $M$ then there is $x_{1}, \dots, x_{n} \in X$ and $r_{1}, \dots, r_{n} \in R$ such that
$v = i = 1 \sum n r_{i} x_{i}$
As $ϕ$ is a $R$ -homomorphism then
$ϕ (v) = i = 1 \sum n r_{i} ϕ (x_{i})$
Hence $ϕ (v)$ is uniquely determined by
${ϕ (x_{i}) : 1 \leq i \leq n}$
If $X$ is a basis of $M$ then
For $f : X \to N$ then there is function $ϕ_{f} : M \to N$ by
$ϕ_{f} (v) = i = 1 \sum n r_{i} f (x_{i}) where v = i = 1 \sum n r_{i} x_{i}$
with $ϕ_{f}$ is $R$ -linear by uniqueness

If $v = \sum_{i = 1}^{n} r_{i} x_{i}$ and $w = \sum_{i = 1}^{n} s_{i} x_{i}$ and for $t \in R$ then
$v + tw = i = 1 \sum n (r_{i} + t s_{i}) x_{i}$
Thus
$ϕ_{f} (v + tw) = i = 1 \sum n (r_{i} + t s_{i}) f (x_{i)} = i = 1 \sum n r_{i} f (x_{i}) + t i = 1 \sum n s_{i} f (x_{i}) = ϕ_{f} (v)$

Finitely Generated Modules as Quotients of Free Modules

Let $M$ be a nonzero finitely generated module then
Exists $n \in N$ as surjective homomorphism $ϕ : R^{n} \to M$ with

$R^{n} / ker (ϕ) ≅ M$

Le $M$ and $ϕ$ be as of $(1)$ then there exists free module $R^{m}$ with $m \leq n$ and
injective homomorphism $ϕ : R^{m} \to R^{n}$ such that

$im (ψ) = ker (ϕ)$
with
$M ≅ R^{n} / im (ψ)$

Proof

Let ${m_{1}, m_{2}, \dots, m_{n}}$ be any finite subset of $M$
Let ${e_{1}, \cdot, e_{n}}$ be a basis for $R^{n}$

Extend map $e_{i} \mapsto m_{i}$ for $1 \leq i \leq n$ to homomorphism $ϕ : R^{n} \to M$
by Homomorphisms from free modules are determined by a basis

${m_{1}, m_{2}, \dots, m_{n}}$ is a generating set of $M$ is equivalent to map $ϕ$ being surjective
So by surjectivity of $ϕ$ and First Isomorphism Theorem then
$R^{n} / ker (ϕ) ≅ M$
As $R$ is a PID then submodule $ker (ϕ)$ is a free submodule of rank $m \leq n$
Let ${x_{1}, \dots, x_{m}}$ be a basis of of $ker (ϕ)$

Define $ψ$ by sending standard basis of $R^{m}$ to ${x_{1}, \dots, x_{m}}$
Hence $ϕ$ is injective and has image $ker (ϕ)$

Presentation of a Module

Let $M$ be a finitely generated $R$ -module
Pair of maps $ϕ, ψ$ such that $ψ : R^{m} \to R^{n}$
$im (ϕ) = M and im (ψ) = ker (ϕ)$
where $ϕ, ψ$ come from Finitely generated modules as quotients of free modules

Then $ϕ, ψ$ are called the presentation of the finitely generated modules $M$

Resolution of a Module

Let $M$ be a finitely generated $R$ -module
Let $ϕ, ψ$ be the presentation of $M$ then

If $ψ$ is injective then
$ϕ, ψ$ are called the resolution of the finitely generated modules $M$

Torsion-Free Modules over PID corollary

Let $M$ be a finitely generated torsion-free module over $R$ where $R$ is a PID then

$M$ is free

Generally if $M$ is a finitely generated $R$ -module then
Rank $s$ of free part of $M$ given in structure theorem is
$rk (M / M^{tor})$
Hence it is unique

Proof

By Structure Theorem for Finitely Generated Modules over a Euclidean Domain then
$M$ is isomorphic to module to module in form
$R^{s} \oplus (i = 1 ⨁ r R / c_{i} R)$
Let $F = R^{s}$ and $N = ⨁_{i = 1}^{r} R / c_{i} R$ so $M = F \oplus N$

If $a \in R / c_{i} R$ then as $c_{i} ∣ c_{k}$ then $c_{k} (a) = 0$
If $m \in N$ then say $m = (a_{1}, \dots, a_{k})$ then
$a_{i} \in R / c_{i} R ⟹ c_{k} (a_{1}, \dots, a_{m}) = (c_{k} a_{1}, \dots, c_{k} a_{k}) = (0, \dots, 0)$
Hence $N$ is torsion

Otherwise suppose $m = (f, n)$ where $f \in F$ and $n \in N$ so $r (f, n)) = (r . f, r . n) = (0, 0)$
Thus
$f = 0$
as a free module is torsion free

Hence $M^{tor} = N$

Thus $M$ is torsion free if an only if $M = F$ is free
By Second Isomorphism Theorem then
$F ≅ M / M^{tor}$
Hence $s = rk (F) = rk (M / M^{tor})$

Structure Theorem for Finitely Generated Abelian Groups corollary

Let $A$ be a finitely generated abelian group then

Exists integer $r \in Z_{\geq 0}$ and integers $c_{1}, c_{2}, \dots, c_{k} \in Z$ greater than $1$ such that
$c_{1} ∣ c_{2} ∣ \dots ∣ c_{k} and A ≅ Z^{r} \oplus (Z / c_{1} Z) \oplus \dots \oplus (Z / c_{k} Z)$
where $s, c_{1}, \dots, c_{k}$ are uniquely determined

Proof

Same as Structure theorem for finitely generated abelian groups
but insist $c_{i} > 0$ as unit group $Z^{\times} = {\pm 1}$

Chinese Remainder Theorem for PIDs

Let $d_{1}, \dots, d_{k}$ be a set of pairwise coprime elements of PID such that
$hcf {d_{i}, d_{j}} = 1 for i \neq = j$
Then
$R / (d_{1} d_{2} \dots d_{k}) R = i = 1 ⨁ k R / d_{i} R$

Proof

As $d_{i}$ s are pairwise coprime then for $i < k$ define
$c_{i} = d_{i + 1} \dots d_{k}$
Hence for each $i$ then $hcf {d_{i}, c_{i}} = 1$
Thus for $1 \leq i \leq k - 1$ then
$d_{i} R + c_{i} R = R and d_{i} R \cap c_{i} R = d_{i} c_{i} R$
Hence by Chinese Remainder Theorem and induction on $k$ then
$R / (d_{1} \dots d_{k}) R = R / (d_{1} c_{1} R) ≅ R / d_{1} R \oplus R / c_{1} R ≅ i = 1 ⨁ k R / d_{i} R$
where use induction on factor $R / c_{1} R$ as $c_{1} = d_{2} \dots d_{k}$

Oxford Maths Notes

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07 - Modules

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