03 - Integral Domains

Integral Domain

Let $R$ be a ring then

$R$ is an Integral Domain if it is not the zero ring and has no zero divisors

Product is Zero Property $a,b\in R$ then

For

$$a.b=0⟹a=0\text{ or }b=0$$

Cancellation Property $x,y,z\in R$ then If

For

$$x.y=x.y$$

Then

$$x=0\text{ or }y=z$$

Proof

$$x.y=x.z⟺x.(y-z)=0$$

then result follows by definition

Subrings of Integral Domains lemma

Let $R$ be an Integral Domain
Let $S$ be any subring of $R$ then

$$S\text{ is also an Integral Domain}$$

Characteristic of Integral Domains

Let $R$ be a integral domain then

Characteristic of $R$

$$\mathrm{char}(R)=0\text{ or }p\text{ where prime }p\in Z$$

Proof

If $\mathrm{char}(R)=n>0$ then

$$\mathbb{Z}/n\mathbb{Z}\text{ is a subring of }R$$

So if $n=a.b$ where $a,b\in \mathbb{Z}$ where $a,b>1$ then

$$n_R=0⟺a_R.b_R=0$$

Hence $a_R$ and $b_R$ are both zero divisors as $a_R\ne 0$ and $b_R\ne 0$

So as $R$ is an integral domain then $\mathrm{char}(R)$ is either zero or prime

Euclidean Domain

Let $R$ be an integral domain
Let $N:R\setminus \{0\}\to \mathbb{N}$ be a function

$R$ is a Euclidean Domain if for any $a,b\in R$ with $b\ne 0$ then

$$\exists q,r\in R\text{ such that }a=b.q+r$$

then either

$$r=0\text{ or }N(r)<N(b)$$

Euclidean Domain for $\mathbb{Z}[i]$ lemma

Let $R=\mathbb{Z}[i]$
Let $N:R\to \mathbb{N}$ by $N(z)=a^2+b^2$ where $z=a+ib\in \mathbb{Z}[i]$ then

$$(R,N)\text{ is an Euclidean Domain}$$

Proof

As $N$ is the restriction of square of modulus function on $\mathbb{C}$ so

$$N(z.w)=N(z).N(w)$$

Write $|z{|}^2$ instead of $N(z)$ for $z\in \mathbb{C}\setminus \mathbb{Z}[i]$
Suppose $s,t\in \mathbb{Z}[i]$ and $t\ne 0$ then

$$\frac{s}{t}\in \mathbb{C}$$

So let

$$\frac{s}{t}=u+iv\text{ where }u,v\in Q$$

Let $a,b\in \mathbb{Z}$ such that

$$|u-a|\le \frac{1}{2}\text{ and }|v-b|\le \frac{1}{2}$$

Hence for $q=a+ib$ then

$${\left|\frac{s}{t}-q\right|}^2\le \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$

Thus

$$N(s-qt)\le \frac{1}{2}N(t)$$

Hence

$$r=s-qt\in \mathbb{Z}[i]$$

Thus either

$$r=0\text{ or }N(r)\le \frac{1}{2}N(t)<N(t)$$

Ideals of Euclidean Domains are principal lemma

Let $(R,N)$ be an Euclidean Domain then

Any ideal of $R$ is principal

Proof

If $I$ is a non-zero ideal then

Let $d\in I$ such that $N(d)$ is minimal

If $m\in I$ then

$$m=q.d+r$$

where

$$r=0\text{ or }N(r)<N(d)$$

But

$$r=m-q.d\in I$$

Hence by minimality of $N(d)$ then $r=0$ so $m=q.d$ so

$$I\subseteq R.d$$

Since $d\in I$ then

$$Rd\subseteq I$$

Hence $I=Rd$

Principal Ideal Domain

Let $R$ be an integral domain

If every ideal $I$ in $R$ is principal so exists $a\in R$ such that $I=⟨a⟩$ then

$R$ is known as a Principal Ideal Domain (PID)

Irreducible

Let $R$ be an integral domain

Non-zero element $r\in R$ is irreducible if whenever $r=a.b$ then

$$\text{exactly one of }a\text{ or }b\text{ is a unit}(\text{not both})$$

Note that this ensures $r$ is not a unit

Reducible Element

Let $R$ be an integral domain

None-zero element $r\in R$ is reducible if it is not irreducible

Equivalent Statements for PIDs

Let $R$ be a PID
Let $d\in R\setminus \{0\}$ then

Equivalent Statements are

1. $R.d=⟨d⟩$ is a prime ideal

2. $d$ is irreducible in $R$

3. $R.d$ is a maximal ideal in $R$

Proof

Proof $(1)⟹(2)$

If $d=a.b$ then as $d\in R.d$ is prime then

$$a\in R.d\text{ or }b\in R.d$$

WLOG assume $a\in R.d$ then
Then there is some $r\in R$ with $a=r.d$ so

$$d=a.b=(r.b).d$$

Hence

$$(1-r.b).d=0$$

As $R$ is an integral domain and $d\ne 0$ then

$$r.b=1⟹b\in R^{\times }$$

Note that as $R.d$ is a proper ideal and $R.a\subseteq R.d$ then $a$ is not a unit

Proof $(2)⟹(3)$

Suppose $d$ is irreducible then

$$R.d\subseteq I◃R$$

As $R$ is a PID then

$$I=R.a\text{ for some }a\in R$$

Hence

$$R.d\subseteq R.a⟹d=a.b\text{ for some }b\in R$$

As $d$ is irreducible then either one of $a$ or $b$ is a unit
If $a$ is unit then

$$R.a=R$$

But if $b$ is a unit then $d$ and $a$ are associates and hence generate same ideal so

$$R.d=I$$

Hence

$$R.d\text{is a maximal ideal}$$

Proof $(3)⟹(1)$
Already shown a maximal idea is prime

$a$ 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 $a\mid b⟺bR\subseteq aR$

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

$$d\mid c$$

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

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

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

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

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

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

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 $n\ge N$

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 $\{{\pm }_1\}$)

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

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

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

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

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 $\mathbb{Z}[t]$

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