11. Presentations and Canonical Form for Modules

Finitely Generated Modules as Quotients of Free Modules

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

$$R^n/\ker (\varphi )\cong M$$

2. Le $M$ and $\varphi$ be as of $(1)$ then there exists free module $R^m$ with $m\le n$ and
   injective homomorphism $\varphi :R^m\to R^n$ such that

$$\mathrm{im}(\psi )=\ker (\varphi )$$

with

$$M\cong R^n/\mathrm{im}(\psi )$$

Proof

Let $\{m_1,m_2,\cdots ,m_n\}$ 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 $1\le i\le n$ to homomorphism $\varphi :R^n\to M$
by Homomorphisms from free modules are determined by a basis

$\{m_1,m_2,\cdots ,m_n\}$ is a generating set of $M$ is equivalent to map $\varphi$ being surjective
So by surjectivity of $\varphi$ and First Isomorphism Theorem then

$$R^n/\ker (\varphi )\cong M$$

As $R$ is a PID then submodule $\ker (\varphi )$ is a free submodule of rank $m\le n$
Let $\{x_1,\cdots ,x_m\}$ be a basis of of $\ker (\varphi )$

Define $\psi$ by sending standard basis of $R^m$ to $\{x_1,\cdots ,x_m\}$
Hence $\varphi$ is injective and has image $\ker (\varphi )$

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Presentation of a Module

Let $M$ be a finitely generated $R$-module
Pair of maps $\varphi ,\psi$ such that $\psi :R^m\to R^n$

$$\mathrm{im}(\varphi )=M\text{ and }\mathrm{im}(\psi )=\ker (\varphi )$$

where $\varphi ,\psi$ come from Finitely generated modules as quotients of free modules

Then $\varphi ,\psi$ are called the presentation of the finitely generated modules $M$

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Resolution of a Module

Let $M$ be a finitely generated $R$-module
Let $\varphi ,\psi$ be the presentation of $M$ then

If $\psi$ is injective then
$\varphi ,\psi$ are called the resolution of the finitely generated modules $M$

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09 - Structure Theorem for Finitely Generated Modules over a Euclidean Domain

Structure Theorem for Finitely Generated Modules over a Euclidean Domain

Let $M$ be a finitely generated module over Euclidean Domain $R$ then

Exists integer $s$ and nonzero non-units $c_1,c_2,\cdots ,c_r\in R$ with $c_1\mid c_2\mid \cdots \mid c_r$ such

$$M\cong \left(\underset{i=1}{\overset{r}{⨁}}R/c_{i}R\right)\oplus R^s$$

Proof

As $R$ is a PID then there is presentation for $M$ with

$$\begin{array}{rl}\psi & :R^m\to R^n\\ \varphi & :R^n\to m\end{array}$$

where $m\le n$ and $\psi$ is injective and $\varphi$ is surjective so

$$M\cong R^n/\mathrm{im}(\psi )$$

If $A$ is the matrix of $\psi$ with respect to standard bases of $R^m$ and $R^n$ then

By Smith Normal Form then
There is normal form for matrices over Euclidean Domain so $A$ can be transformed into
Diagonal Matrix $D$ with diagonal entries $d_1\mid d_2\mid \cdots \mid d_m$ using EROs and ECOs
As EROs and ECOs correspond to changing bases in $R^n$ and $R^m$ then

Exists basis for $R^n$ and $R^m$ with respect to $\psi$ which has matrix $D$
Let $\{f_1,\cdots ,f_n\}$ denote the basis of $R^n$ so image of $\psi$ has basis

$$\{d_1{f}_1,\cdots ,d_m{f}_m\}$$

Define map

$$\theta :R^n\to \left(\underset{i=1}{\overset{m}{⨁}}R/d_{i}R\right)\oplus R^n$$

For $m={\sum }_{i=1}^n{a}_i{f}_i\in M$ then

$$\theta \left(\sum _{i=1}^n{a}_i{f}_i\right)i$$

where $\theta$ is surjective and $\ker (\varphi )$ is submodule generated by

$$\{d_i{f}_i:1\le i\le m\}=(a_1+d_{1}R,\cdots ,a_m+d_{m}R,a_{m+1},\cdots ,a_n)$$

which is $\mathrm{im}(\psi )$

By First Isomorphism Theorem then

$$M\cong R^k/\mathrm{im}(\psi )\cong \left(\underset{i=1}{\overset{m}{⨁}}R/d_{i}R\right)\oplus R^n$$

As $\psi$ is injective then $d_i$ are all nonzero
If $d_i$ is a unit then $R/d_{i}R=0$

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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

$$\mathrm{rk}(M/M^{\mathrm{tor}})$$

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

$$R^s\oplus (\underset{i=1}{\overset{r}{⨁}}R/c_{i}R)$$

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 $c_i\mid c_k$ then $c_k(a)=0$
If $m\in N$ then say $m=(a_1,\cdots ,a_k)$ then

$$a_i\in R/c_{i}R⟹c_k(a_1,\cdots ,a_m)=(c_k{a}_1,\cdots ,c_k{a}_k)=(0,\cdots ,0)$$

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 $M^{\mathrm{tor}}=N$

Thus $M$ is torsion free if an only if $M=F$ is free
By Second Isomorphism Theorem then

$$F\cong M/M^{\mathrm{tor}}$$

Hence $s=\mathrm{rk}(F)=\mathrm{rk}(M/M^{\mathrm{tor}})$

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Structure Theorem for Finitely Generated Abelian Groups corollary

Let $A$ be a finitely generated abelian group then

Exists integer $r\in {\mathbb{Z}}_{\ge 0}$ and integers $c_1,c_2,\cdots ,c_k\in \mathbb{Z}$ greater than $1$ such that

$$c_1\mid c_2\mid \cdots \mid c_k\text{ and }A\cong {\mathbb{Z}}^r\oplus (\mathbb{Z}/c_1\mathbb{Z})\oplus \cdots \oplus (\mathbb{Z}/c_k\mathbb{Z})$$

where $s,c_1,\cdots ,c_k$ are uniquely determined

Proof

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

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Chinese Remainder Theorem for PIDs

Let $d_1,\cdots ,d_k$ be a set of pairwise coprime elements of PID such that

$$\mathrm{hcf}\{d_i,d_j\}=1\text{ for }i\ne j$$

Then

$$R/(d_1{d}_2\cdots d_k)R=\underset{i=1}{\overset{k}{⨁}}R/d_{i}R$$

Proof

As $d_i$s are pairwise coprime then for $i<k$ define

$$c_i=d_{i+1}\cdots d_k$$

Hence for each $i$ then $\mathrm{hcf}\{d_i,c_i\}=1$
Thus for $1\le i\le k-1$ then

$$d_{i}R+c_{i}R=R\text{ and }{d}_{i}R\cap c_{i}R=d_i{c}_{i}R$$

Hence by Chinese Remainder Theorem and induction on $k$ then

$$R/(d_1\cdots d_k)R=R/(d_1{c}_{1}R)\cong R/d_{1}R\oplus R/c_{1}R\cong \underset{i=1}{\overset{k}{⨁}}R/d_{i}R$$

where use induction on factor $R/c_{1}R$ as $c_1=d_2\cdots d_k$

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10 - Structure Theorem in Primary Decomposition Theorem Form

Structure Theorem in Primary Decomposition Theorem Form

Let $R$ be a Euclidean Domain
Let $M$ be a finitely generated $R$-module

Then there are irreducibles $p_1,\cdots p_k\in R$ and integers $s,r_i$ where $1\le i\le k$ such that

$$M\cong \underset{i=1}{\overset{k}{⨁}}(R/p_i^{r_i}R)\oplus R^s$$

where pairs $(p_i,r_i)$ are uniquely determined up to units

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