06. Unique Factorisation

$a$ Divides $b$ Notation

Let $R$ be a integral domain

If $a,b\in \mathbb{R}$ then

$a$ divides $b$ or $a$ is a factor of $b$ if

$$\exists c\in R\text{ such that }b=a.c$$

Notationally written as

$$a\mid b$$

Note that $a\mid b⟺bR\subseteq aR$

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Higher Common Factor

Let $R$ be an integral domain
Let $a,b,c\in \mathbb{R}$

$c$ is the higher common factor of $a,b$ if

$$c\mid a\text{ and }c\mid b\text{ and }c\text{ is the highest common factor}$$

Notationally written as

$$c=\mathrm{hcf}(a,b)$$

Property of Common Factor $d$ be a common factor of $a$ and $b$ then

Let

$$d\mid c$$

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Least Common Multiple

Let $R$ be an integral domain
Let $a,b,c\in \mathbb{R}$

$c$ is the least common multiple of $a,b$ if

$$a\mid c\text{ and }b\mid c\text{ and }c\text{ is the lowest common factor}$$

Notationally written as

$$c=\mathrm{lcm}(a,b)$$

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Uniqueness of HCF and LCM lemma

Let $R$ be an integral domain
Let $a,b\in R$ then

If $\mathrm{hcf}\{a,b\}$ exists then it is unique up to units

If $\mathrm{lcm}\{a,b\}$ exists then is is unique up to units

If $R$ is a PID then both $\mathrm{hcf}\{a,b\}$ and $\mathrm{lcm}\{a,b\}$

Proof

If $g_1,g_2$ are two highest common factors then

$$g_{1}R\subseteq g_{2}R\text{ and }{g}_{2}R\subseteq g_{1}R$$

Hence

$$g_{1}R=g_{2}R$$

As $R$ is an integral domain then $g_1,g_2$ are associates (differ by unit)

Very similar proof for least common multiplies

If $R$ is a PID then $⟨a,b⟩$ is principal

As $⟨a,b⟩$ is the minimal principal ideal containing $a,b$ then
Any generator of $⟨a,b⟩$ is a higher common factor

Similarily

$$Ra\cap Rb$$

is principal so any generator of it is a least common multiple

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Unique Factorisation Domain

Let $R$ be a integral domain then

$R$ is a Unique Factorisation Domain (UFD) if

Every element of $R\setminus \{0\}$ is either a unit or can be written as a product of irreducible elements
where the factorisation into irreducible is unique up to reordering and units

Another property is if $r\in R$ is non-zero and not a unit then

Exists irreducible elements $p_1,\cdots ,p_k$ such that

$$r=p_1{p}_2\cdots p_k$$

Whenever $r=q_1{q}_2\dots q_l$ is another factorisation for $r$ then if $k=l$
There exists reordering of $q_j$s such that

$$q_j=u_j{p}_j\text{ where }{u}_j\in R\text{ is a unit}$$

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Equivalent Statements for an Unique Factorisation Domain lemma

Let $R$ be an integral domain

Equivalent Statements are

1. $R$ is a UFD

2. Both
   $(i)$ Every irreducible element is prime
   $(ii)$ Every nonzero non-unit $a\in R$ can be written as a product of irreducible

3. Every nonzero non-unit $a\in R$ can be written as a product of prime elements

Proof

Suppose $R$ is an UFD

Let $p$ be an irreducible

If $p$ divides $a.b$ where $a,b\in R$ then
If $a$ or $b$ is zero or a unit then done

Otherwise by assumption assume it may be written as a product of irreducible so

$$a=q_1\cdot q_k\text{ and }b=r_1\cdots r_l$$

where $k,l\ge 1$
As $a.b=p.d$ for some $d\in R$ then let

$$d=s_1\cdot s_m$$

as a product of irreducibles by uniqueness of factorisation of $a.b$ into irreducibles then
$p$ must be one of $q_i$s or $r_j$s up to units thus $p$ divides $a$ or $b$ as required

For converse, use induction on minimal number $M(a)$ of irreducibles on factorisation of $a$ into irreducibles

If $M(a)=1$ then $a$ is irreducible and unique (by definition)
Suppose $M(a)>1$ and

$$a=p_1{p}_2\cdots p_M=q_1{q}_1\cdots q_k\text{ for irreducibles }{p}_i,q_j\text{ and }k\ge M$$

Hence

$$p_1\mid q_1\cdots q_k$$

As $p_1$ is prime then exists $q_j$ such that $p_1\mid q_j$
As $q_j$ is irreducible then exists unit $u_1\in R$ such that

$$q_j=u_1{p}_1$$

By reordering $q_l$s assume $j=1$ hence

$$(u_1^{-1}{p}_2)\cdots p_M=q_2{q}_2\cdots q_k$$

Hence $k-1=M-1$ hence $k=M$
Hence irreducibles occurring are equal up to reordering and units as required

Condition $(2)$ is equivalent to condition $(3)$ as
Prime elements are always irreducible need to check irreducibles are prime
Consider $a\in R$ is irreducible so

$$a=p_1{p}_2\cdots p_k$$

as a product of primes
So by definition of irreducibility then $k=1$ hence $p$ is prime

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Ascending Chain Condition on Ideals in a PID

Let $R$ be a PID
Suppose $\{I_n:n\in \mathbb{N}\}$ is a sequence of ideals such that

$$I_n\subseteq I_{n+1}$$

Then union

$$I=\bigcup _{n\ge 0}{I}_n$$

is an ideal and there exists $N\in \mathbb{N}$ such that

$$I_n=I_N=I\text{ for all }n\ge N$$

Note that a ring satisfying any nested ascending chain of Ideals that stabilises is called a Noetherian Ring

Proof

Let

$$I=\bigcup _{n\ge 1}{I}_n$$

For any two elements $p,q\in I$ then
Exists $k,l\in \mathbb{N}$ such that

$$p\in I_k\text{ and }q\in I_l$$

For any $r\in R$ then

$$r.p\in I_k\subset I$$

Let $n=\max \{k,l\}$ so

$$r,s\in I_n⟹r+s\in I_n\subset I$$

Hence $I$ is an ideal

As $R$ is a PID then

$$I=⟨c⟩\text{ for }c\in R$$

Hence exists $N$ such that $c\in I_n$ hence

$$I=⟨c⟩\subseteq I_n\subseteq I$$

Thus $I=I_N=I_n$ for all $n\ge N$

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04 - Principal Ideal Domain are Unique Factorisation Domains

Principal Ideal Domain are Unique Factorisation Domains

Let $R$ be a PID then

$R$ is a UFD

Proof

As irreducibles are prime in a PID then by

Suppose for contradiction $a=a_1\in R$ and $a$ is not a product of irreducible elements
As $a$ is not irreducible then

$$a=b.c\text{ where }b\text{ nor }c\text{ is a unit}$$

If both $b$ and $c$ can be written as a product of prime elements then so is $a$
Hence one of $b$ or $c$ cannot be written as a product of prime element

WLOG suppose it is $b$ and define $a_2=b$
Let

$$I_1=⟨a_2⟩\text{ and }{I}_2=⟨a_2⟩$$

Then

$$I_1⊈I_2$$

As $a_2$ cannot be irreducible then exists $a_3$ such that

$$I_2=⟨a_2⟩⊈⟨a_3⟩=I_3$$

Continuing to get nested sequence of ideals $I_k$ with each strictly bigger than previous
By Ascending chain condition on ideals in a PID, then it cannot happen if $R$ is a pid
Thus $a$ does not exist

By Equivalent statements for an unique factorisation domain then $R$ is a UFD

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Content of a Polynomial in $\mathbb{Z}[t]$

Let $f\in \mathbb{Z}[t]$ so $f={\sum }_{i=0}^n{a}_i{t}^i$ then

Content $c(f)$ of $f$ is defined as

$$c(f)=\mathrm{hcf}\{a_0,a_1,\cdots ,a_n\}$$

so it is the highest common factor of the coefficients of $f$

With property holding for non-zero $f$ then

$$f=c(f).f_1$$

where $f_1$ has content $1$

Note that we define $c(f)>0$ so it is unique (as units in $\mathbb{Z}$ are $\{{\pm }_1\}$)

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Gauss Property lemma

Let $f,g\in \mathbb{Z}[t]$ then

$$c(f.g)=c(f).c(G)$$

Proof

Suppose $c(f)=c(g)=1$
Let $p\in \mathbb{N}$ is prime

For each prime homomorphism $\mathbb{Z}[t]\to {\mathbb{F}}_p[t]$ defined by

$${\varphi }_p\left(\sum _{i=0}^n{a}_i{t}^i\right)=\sum _{i=0}^n\overline{a_i}{t}^i\text{ where }{\overline{a}}_i\text{ denotes }{a}_i+p\mathbb{Z}\in {\mathbb{F}}_p$$

Hence

$$\ker ({\varphi }_p)=p\mathbb{Z}[t]$$

Thus

$$p\mid c(f)⟺{\varphi }_p(f)=0$$

As ${\mathbf{F}}_p$ is a field and ${\mathbf{F}}_p[t]$ is an integral domain then as ${\varphi }_p$ is a homomorphism then

$$\begin{array}{rl}p\mid c(f.g)& ⟺{\varphi }_p(f.g)=0⟺{\varphi }_p(f).{\varphi }_p(g)=0\\ & ⟺{\varphi }_p(f)=0\text{ or }{\varphi }_p(g)=0\\ & ⟺p\mid c(f)\text{ or }p\mid c(g)\end{array}$$

Hence $c(f.g)=1$ if $c(f)=c(g)=1$

Let $f,g\in \mathbb{Z}(t)$
Let $f=a.f^{\prime }$ and $g=b.g^{\prime }$ where $f^{\prime },g^{\prime }\in \mathbb{Z}[t]$ such that $c(f^{\prime })=c(g^{\prime })=1$ then

$$f.g=(a.b).(f^{\prime }{g}^{\prime })$$

As $c(f^{\prime }{g}^{\prime })=1$ then

$$c(f.g)=c(f).c(g)$$

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Uniqueness of Content of Polynomial in $\mathbb{Q}[t]$ lemma

Let $f\in \mathbb{Q}[t]$ be nonzero

Then there exists unique $\alpha \in {\mathbb{Q}}_{>0}$ such that

$$f=\alpha f^{\prime }\text{ where }{f}^{\prime }\in \mathbb{Z}[t]\text{ and }c(f^{\prime })=1$$

Hence $c(f)=\alpha$

With property for $f,g\in \mathbb{Q}[t]$

$$c(f.g)=c(f).c(g)$$

Proof

Let

$$f=\sum _{i=0}^n{a}_i{t}^i\text{ where }{a}_i=\frac{b_i}{c_i}\text{ for }{b}_i,c_{i}in\mathbb{Z}$$

with $\mathrm{hcf}(b_i,c_i)=1$ for all $i$ where $0\le i\le n$

Let $d\in {\mathbb{Z}}_{>0}$ such that $d{a}_i\in \mathbb{Z}$ so $df\in \mathbb{Z}[t]$
For example

$$d=\mathrm{lcm}\{c_i:0\le i\le n\}\text{ or }d=\prod _{i=0}^n{c}_i$$

Let $c(f)=c(d.f)/d$ then

$$f=c(f).f^{\prime }\text{ where }{f}^{\prime }=(d.f)/c(d.f)\text{ with }c(f^{\prime })=1$$

Let ${\alpha }_i=m_i/n_i$ for positive integers $m_i,n_i$ then

$$n_2.(m_1.f_1)=n_1.(m_2{f}_2)\in \mathbb{Z}[t]$$

Hence

$$c(n_2.(m_1{f}_1))=c(n_2).(m_1{f}_1)=n_2.m_{1}c(n_1.(m_2{f}_2))=c(n_1).c(m_2{f}_2)=n_1{m}_2$$

So ${\alpha }_1=\frac{m_1}{n_1}=\frac{m_2}{n_2}={\alpha }_2$
Hence $f={\alpha }_1.f_1={\alpha }_2.f_2$ implies ${\alpha }_1={\alpha }_2$ where $f_1$ and $f_2$ have content $1$

Checking multiplicity so if $f,g\in \mathbb{Q}[t]\setminus \{0\}$ then
If $d_1,d_2\in \mathbb{Z}$ such that $d_1.f,d_2.g\in \mathbb{Z}[t]$ then

$$c(f.g)=\frac{c(d_1{d}_{2}f.g)}{d_1{d}_2}=\frac{c((d_1.f).(d_{2}g))}{d_1{d}_2}=\frac{c(d_{1}f)}{d_1}.\frac{c(d_2.g)}{d_2}=c(f).c(g)$$

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Property of Prime Elements

1. Suppose $f\in \mathbb{Z}[t]\subset \mathbb{Q}[t]$ is nonzero and $f=g.h$ where $g,h\in \mathbb{Q}[t]$ then
   Exists $a\in \mathbb{Q}$ such that $(\alpha .g),({\alpha }^{-1}h)\in \mathbb{Z}[t]$ hence

$$f=(\alpha .g)({\alpha }^{-1}h)$$

is a factorisation of $f$ in $\mathbb{Z}[t]$

2. Suppose $f\in \mathbb{Q}[t]$ is irreducible and $c(f)=1$ then
   $f$ is a prime element of $\mathbb{Z}[t]$

3. Let $p\in \mathbb{Z}$ be a prime number then
   $p$ is a prime element in $\mathbb{Z}[t]$

Proof

By Uniqueness of content of polynomial in $\mathbb{Q}[t]$ then

$$g=c(g).g_1\text{ and }h=c(h).h_1$$

where $g_1,h_1\in \mathbb{Z}[t]$ with content $1$

Then $c(f)=c(g)c(h)$ so as $f\in \mathbb{Z}[t]$ then

$$c(g).c(h)\in \mathbb{Z}[t]$$

Let $\alpha =c(h)$ so

$$f=(\alpha .g).({\alpha }^{-1}.h)\text{ where }\alpha .g=(c(g).c(h)).g_1\text{ and }{\alpha }^{-1}h=h_1\in \mathbb{Z}[t]$$

If $f\in \mathbb{Q}[t]$ has $c(f)=1$ then $f\in \mathbb{Z}[t]$
So $(1)$ is proved

For $g,h\in \mathbb{Z}[t]$ then
If $f\mid g.h$ in $\mathbb{Z}[t]$ then $f\mid g.h$ in $\mathbb{Q}[t]$

As $\mathbb{Q}[t]$ is a PID then irreducibles are prime so

$$f\mid g\text{ or }f\mid h$$

WLOG suppose $f\mid g$ then

$$g=f.k\text{ for }k\in \mathbb{Q}[t]$$

By Uniqueness of content of polynomial in $\mathbb{Q}[t]$ then

$$k=c(k).k^{\prime }\text{ where }{k}^{\prime }\in \mathbb{Z}[t]$$

Hence

$$c(g)=c(f).c(k)=c(k)\text{ as }c(f)=1$$

However $g\in \mathbb{Z}[t]$ so

$$c(g)=c(k)\in \mathbb{Z}$$

Hence

$$k\in \mathbb{Z}[t]$$

Thus $f$ divides $g$ in $\mathbb{Z}[t]$
Hence $(2)$ is proved

As homomorphism ${\varphi }_p:\mathbb{Z}[t]\to {\mathbb{F}}_p[t]$ has kernel $p\mathbb{Z}[t]$
Then as ${\mathbb{F}}_p[t]$ is an integral domain then ideal $p\mathbb{Z}[t]$ is prime
Hence $p$ is a prime element of $\mathbb{Z}[t]$

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05 - Ring of Polynomials with Integer Coefficients is a UFD

Ring of Polynomials with Integer Coefficients is a UFD

Let $R=\mathbb{Z}[t]$ then

$R$ is a UFD

Proof

Since $\mathbb{Z}[t]$ is a integral domain as $\mathbb{Z}$ is also an integral domain
Let $f\in \mathbb{Z}[t]$ then

$$f=a.f^{\prime }\text{ where }c(f^{\prime })=1$$

Since $\mathbb{Z}$ is a UFD then
$a$ is factorisable into a product of prime elements of $\mathbb{Z}$ (prime in $\mathbb{Z}[t]$)
Hence assume $c(f)=1$

As $f$ is an element of $\mathbb{Q}[t]$ then it is a product of prime elements in $\mathbb{Q}[t]$ then

$$f=p_1{p}_2\cdots p_k$$

By Uniqueness of content of polynomial in $\mathbb{Q}[t]$ then

$$p_i=a_i{q}_i\text{ where }{a}_i\in \mathbb{Q}\text{ and }{q}_i\in \mathbb{Z}[t]\text{ and }c(q_i)=1$$

By Property of prime elements (3) as $q_i$ is prime in $\mathbb{Z}[t]$ and

$$f=(a_1\cdots a_k)(q_1\cdots q_k)$$

By comparing contents then

$$(a_1\cdots a_k)=1$$

So then by Equivalent statements for an unique factorisation domain so $R$ is UFD

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06 - UFD Property of Polynomial Ring

UFD Property of Polynomial Ring

Let $R$ be a UFD then

Polynomial Ring $R[T]$ is also a UFD

Generally if $R$ is a UFD then

$$R[t_1,\cdots ,t_n]$$

which is the ring of polynomials in $n$ variables with coefficients in $R$ is also a UFD

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6.1 Irreducible Polynomials

Eisenstein's Criterion lemma

Let $f\in \mathbb{Z}[t]$ with $c(f)=1$ with

$$f=a_n{t}^n+a_{n-1}{t}^{n-1}+\cdots +a_{1}t+a_0$$

If there exists prime $p\in \mathbb{Z}$ such that

$$p\mid a_i\text{ for all }0\le i\le n-1$$

and $p$ does not divide $a_n$ and $p^2$ does not divide $a_0$ then

$f$ is irreducible in $\mathbb{Z}[t]$ and $\mathbb{Q}[t]$

Proof

Since $c(f)=1$ then irreducibility in $\mathbb{Z}[t]$ and $\mathbb{Q}[t]$ are equivalent

Let ${\varphi }_p:\mathbb{Z}[t]\to {\mathbb{F}}_p[t]$ be a quotient map

Suppose $f=g.h$ is a factorisation for $f$ in $\mathbb{Z}[t]$ where $0<\mathrm{deg}(g)=k<n$ then

$${\varphi }_p(f)={\varphi }_p(g).{\varphi }_p(h)$$

By assumption then ${\varphi }_p(f)={\overline{a}}_n{t}^n$ (for $m\in \mathbb{Z}$ then $\overline{m}$ for $m+p\mathbb{Z}$ so the image of $m$ in ${\mathbb{F}}_p$)

As ${\mathbb{F}}_p[t]$ is UFD then
$t$ is irreducible and ${\overline{a}}_n$ is a unit
Hence up to units then

$${\varphi }_p(g)=t^k,{\varphi }_p(h)=t^{n-k}$$

But then constant terms of $g$ and $h$ must be divisible by $p$ hence $a_0$ is divisible by $p^2$
Hence contradicting assumption

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