04. Prime and Maximal Fields, Euclidean Domains and PIDs

Maximal Ideal

Let $R$ be a ring
Let $I$ be an ideal of $R$

$I$ is a maximal ideal if it is not strictly contained in any proper ideal of $R$

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Prime Ideal

Let $R$ be a ring
Let $I$ be an ideal of $R$

$I$ is prime ideal if

1. $I\ne R$
2. For all $a,b\in R$ then if $a.b\in I$ then either $a\in I$ or $b\in I$

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Prime Element

Let $I$ be a prime ideal of ring $R$

$a\in I$ is a prime element if $a$ is a generator of $I$ so

$$I=⟨a⟩$$

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Characterisation of Prime and Maximal Ideals via Quotients

Let $I$ be an ideal in ring $R$

$I$ is a prime ideal if and only if $R/I$ is an integral domain
$I$ is maximal if and only if $R/I$ is a field

In particular, maximal ideal is always prime

Proof

Suppose $a,b\in R$

As $(a+I)(b+I)=0+I$ if and only if $a.b\in I$

Suppose $R/I$ is an integral domain then

$$a+I=0\text{ or }b+I=0$$

Hence either $a$ or $b$ is in $I$ hence $I$ is prime

Converse argument is very similar

As a field is a ring with no non-trivial ideals and are integral domains then
Result follows from Correspondence Theorem for Rings

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Degree of Polynomials of Ring

Let $R$ be a ring

Suppose $f\in R[t]$ is non-zero then

$$f=\sum _{i=0}^n{a}_i{t}^i$$

where $a_n\ne 0$ then

$$\mathrm{deg}(f)=n$$

where $a_n$ is the leading coefficient of $f$

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Degree of Product of Polynomials in Integral Domain

Let $R$ be an integral domain then

For $f,g\in R[t]$ then

$$\mathrm{deg}(f.g)=\mathrm{deg}(f)+\mathrm{deg}(g)$$

Note that this implies $R[t]$ is also an integral domain

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Division Algorithm lemma

Let $R$ be a ring with

$$f=\sum _{i=0}^n{a}_i{t}^i\in R[t]\text{ where }{a}_n\in R^{\times }$$

Then if $g\in R[t]$ is any polynomial then

Exists $q,r\in R[t]$ such that

$$r=0\text{ or }\mathrm{deg}(r)<\mathrm{deg}(f)\text{ and }g=q.f+r$$

Proof

Induction on $\mathrm{deg}g$
As $a_n\in R^{\times }$

If $h\in R[t]\setminus \{0\}$ then

$$\mathrm{deg}(f.h)=\mathrm{deg}(f)+\mathrm{deg}(h)$$

Hence if $\mathrm{deg}(g)<\mathrm{deg}(f)$ then $q=0$ and hence $r=g$

Otherwise suppose

$$g=\sum _{j=0}^m{b}_j{t}^j\text{ where }{b}_m\ne 0\text{ and }m=\mathrm{deg}(g)\ge n\ge \mathrm{deg}(f)$$

Then $a_n^{-1}{b}_m{t}^{m-n}.f$ has leading erm $b_m{t}^m$

has leading term $b_m{t}^m$ has leading term $b_n{t}^m$
As

$$h=g-a_n^{-1}{b}_m{t}^{m-n}f$$

as $\mathrm{deg}(h)<\mathrm{deg}(q)$

Then by induction there exists unique $q^{\prime },r^{\prime }$ such that $h=q^{\prime }.f+r^{\prime }$
Setting

$$q=a_n^{-1}{b}_n{t}^{m-n}+q^{\prime }\text{ and }r=r^{\prime }$$

then

$$g=q.f+r$$

As $q$ and $r$ are uniquely determined by $q^{\prime }$ and $r^{\prime }$ then they are also unique

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Ideals of a Field are Principal lemma

Let $k$ be a field
Let $I$ be a nonzero ideal in $k[t]$

Then there exists unique monic polynomial $f$ such that

$$I=⟨f⟩$$

with all ideals in $k[t]$ being principal

Proof

Let $I$ be a non-zero ideal in $k[t]$

Let $f\in I$ be of minimal degree and rescale to make it monic

Suppose $I=⟨f⟩$

If $g\in I$ then by Division algorithm then

$$g=q.f+r\text{ where }r=0\text{ or }\mathrm{deg}(r)<\mathrm{deg}(f)$$

Then

$$r=g-q.f\in I$$

By minimality of degree of $f\in I$ then $r=0$ hence $g=q.f$
Uniqueness follows as if

$$I=⟨f⟩=⟨f^{\prime }⟩⟹f=a.f^{\prime }\text{ and }{f}^{\prime }=b.f$$

for some polynomials $a,b\in k[t]$

But $f=a.f^{\prime }=(ab).f$ hence $a,b$ must be degree zero so $z,b\in k$
As $f$ and $f^{\prime }$ is monic then

$$a=b=1$$

Hence $f=f^{\prime }$

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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

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

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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

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

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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 $I=Rd$

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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)

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Reducible Element

Let $R$ be an integral domain

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

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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

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