Annihilator of a Element

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

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

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Torsion Element

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

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

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Torsion Module

Let $M$ be an $R$ -module

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

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Torsion-Free Module

Let $M$ be an $R$ -module

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

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Cyclic Module

Let $M$ be an $R$ -module

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

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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

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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 $M / I M$

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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

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9.1 Homomorphisms between Free Modules

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
$(ψ + ϕ) (m) = ψ (m) + ϕ (m) and (r . ψ) (m) = r . (ψ (m))$

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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
$ϕ_{f} (v + tw) = i = 1 \sum n (r_{i} + t s_{i}) f (x_{i)} = i = 1 \sum n r_{i} f (x_{i}) + t i = 1 \sum n s_{i} f (x_{i}) = ϕ_{f} (v)$

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Matrix Form for Homomorphism of Free Modules corollary

Let $ϕ : F_{1} \to F_{2}$ be a homomorphism of free modules with bases
$X_{1} = {e_{1}, \dots, e_{m}} and X_{2} = {f_{1}, \dots, f_{n}}$
respectively

Then $ϕ$ is determined by matrix $A = (a_{ij}) \in Mat_{n, m} (R)$ given by
$ϕ (e_{i}) = j = 1 \sum n a_{ji} f_{j}$
Conversely given matrix $A$ , determines unique $R$ -homomorphism $ϕ_{A} : F_{1} \to F_{2}$

Proof

Follows from Homomorphisms from free modules are determined by a basis
As map $ϕ$ records values of $ϕ$ on $X_{1}$ then
Matrix $A$ records once $X_{1}$ and $X_{2}$ are known

Conversely define function $f : X_{1} \to N$ by
$ϕ (e_{i}) = j = 1 \sum n a_{ji} f_{j}$
which extends uniquely up to $R$ -module homomorphism $ϕ_{A} : F_{1} \to F_{2}$

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Matrix Representation of Composition of Homomorphisms lemma

Let $F_{1}, F_{2}, F_{3}$ be free modules with bases
$X_{1} X_{2} X_{3} = {e_{1}, e_{2}, \dots, e_{n}} = {f_{1}, f_{2}, \dots, e_{m}} = {g_{1}, g_{2}, \dots, g_{l}}$
respectively

Let $ϕ : F_{1} \to F_{2}$ with $ψ : F_{2} \to F_{3}$

Let matrix of $ϕ$ with respect to bases $X_{1}$ and $X_{2}$ is $A$
Let matrix of $ψ$ with respect to bases $X_{2}$ and $X_{3}$ is $B$ then

Matrix of Homomorphism $ϕ \circ ϕ$ with respect to bases $X_{1}$ and $X_{3}$ is
$B A$

Proof

$ψ \circ ϕ (e_{i}) = ψ (j = 1 \sum m a_{ji} f_{j}) = j = 1 \sum m a_{ji} ψ (f_{j}) = j = 1 \sum m a_{ji} (k = 1 \sum l b_{kj} g_{k}) = k = 1 \sum l (j = 1 \sum m b_{kj} a_{ji}) g_{k}$
Hence matrix of $ψ \circ ϕ$ has $(k, i)$ entry of
$j = 1 \sum m b_{kj} a_{ji} \equiv (B A)_{ki}$

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Change of Basis and the General Linear Group corollary

Let $F$ be a free module with basis
$X = {e_{1}, \dots, e_{n}}$
Set of Isomorphisms $ψ : F \to F$ corresponds under map to
$GL_{n} (R) = {A \in Mat_{n} (R) : \exists B \in Mat_{n} (R) such that A . B = B . A = I_{n}}$
is the group of units in noncommutative ring $Mat_{n} (R)$

Given two bases $X = {x_{1}, \dots, x_{n}}$ and $Y = {y_{1}, \dots, y_{n}}$ then
Exists unique isomorphism $ψ : F \to F$ such that
$ψ (x_{i}) = y_{i}$

Proof

Composition of Morphisms corresponds to matrix multiplication
So map sending homomorphism to corresponding $n \times n$ matrix is a ring map

If $X$ and $Y$ are bases then there is unique module homomorphism
$ψ : F \to F such that ψ (x_{i}) = y_{i}$
and unique module homomorphism
$ϕ : F \to F such that ϕ (y_{i}) = x_{i}$
As composition $ϕ \circ ψ$ satisfies $ϕ \circ ψ (x_{i}) = x_{i}$ then
$ϕ \circ ψ = Id$

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Change of Bases Matrices

Let $F_{1}, F_{2}$ be free modules with bases $X_{1}$ and $X_{2}$ respectively then
$_{X_{2}} [ϕ]_{X_{1}}$
denotes the matrix of homomorphism $ϕ$ with respect to bases $X_{1}$ and $X_{2}$

Let $Y_{1}$ and $Y_{2}$ be another pair of bases for $F_{1}$ and $F_{2}$ respectively
Let
$Q =_{Y_{1}} [id_{F_{1}}]_{X_{1}} and P =_{X_{2}} [id_{F_{2}}]_{Y_{2}}$
Then matrices $P$ and $Q$ are the change of bases matrices for pairs of bases $X_{2}, Y_{2}$ and $X_{1}, Y_{1}$

Note that if $F$ is a free module with two bases $X, Y$ then matrix $_{Y} [id_{F}]_{X}$ has columns given by $Y$ -coordinates of basis $X$

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Oxford Maths Notes

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09. Free, Torsion and Torsion-Free Elements

9.1 Homomorphisms between Free Modules

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