04 - Ideals

Ideal

Let $R$ be a ring with subset $I\subseteq R$

$I$ is an ideal if it is a subgroup of $(R,+)$ and

$$\forall a\in I,r\in R⟹a.r\in I$$

written as $I◃R$

Kernel of a Homomorphism is an Ideal lemma

Let $R$ be a ring
Let $f:R\to S$ be a homomorphism then

$$\ker (f)\text{ is an Ideal}$$

Ideal Subset Criterion lemma

Let $R$ be a ring then
Let $I\subset R$ then

$I$ is an ideal $⟺$ $I$ is nonempty, closed under addition and multiplication by arbitrary elements of $R$

Additive Subgroup Generated by a Subset

Let $X,Y$ be any subsets of ring $R$
Let ${\mathcal{A}}_R(X)$ be the collection of subgroups of abelian group $(R,+)$ which contain $X$ defined by

$${\mathcal{A}}_R(X)=\{A:X\subseteq A\subseteq R\text{ and }(A,+)\le (R,+)\}$$

For arbitrary subset $X$ of $R$ define additive subgroup of $R$ generated by set $X$ by

$$⟨X{⟩}_{ab}=\bigcap _{A\in {\mathcal{A}}_R(X)}A=\left\{\sum _{k=1}^n{x}_k:n\in {\mathbb{Z}}_{\ge 0},x_k\in X\forall 1\le k\le n\right\}$$

with

$$\begin{array}{rl}X+Y& =⟨X\cup Y{⟩}_{ab}=⟨X{⟩}_{ab}+⟨Y{⟩}_{ab}\\ XY& =⟨X.Y{⟩}_{ab}\text{ where }X.Y=\{x.y:x\in X,y\in Y\}\end{array}$$

Properties of Ideals $R$ be a ring Let $I,J$ be ideals in $R$ Let $X$ be any subset in $R$

Let

Then the following are ideals

$$I+J,I\cap J,IX$$

with

$$IJ\subseteq I\cap J,I,J\subseteq I+J$$

Note that the infinite intersection of ideals is also an ideal

Proof $I+J$ is clearly an abelian subgroup of $R$ If $i\in I$, $j\in J$ and $r\in R$ then

$$r(i+j)=(r.i)+(r.j)\in I+J$$

Hence as $i+j\in I+J$ then $I+J$ is an ideal

If $y\in I\cap J$ and $r\in R$ then

$$\begin{array}{r}ry\in I(\text{as }y\in I)\\ ry\in J(\text{as }y\in J)\end{array}$$

Hence as $rj\in I\cap J$ then $I\cap J$ is an ideal

If ${\sum }_{k=1}^n{i}_k{x}_k\in IX$ and $r\in R$ then

$$r.\sum _{k=1}^n{i}_k{x}_k=\sum _{k=1}^n(r.i_k).x_k\in IX$$

with sum of two elements in $IX$ being in same sum form
Hence by Ideal subset criterion then $IX$ is an ideal

Generating Ideals

Let $T$ be any subset of ring $R$

Define Ideal generated by $T$ by

$$⟨T⟩=\bigcap _{T\subseteq I}I$$

where $I$ is an ideal

Characterisation of the Ideal Generated by a Subset lemma

Let $T\subseteq R$ be a nonempty subset of a ring then

$$⟨T⟩=R.T$$

Proof

By Properties of Ideals then $R.T$ is an ideal as $R$ is an ideal (of itself)
If $\{x_1,\cdots ,x_k\}\subseteq T\subseteq J$ and $r_1,\cdots ,r_k\in R$ where $J$ is an ideal then

$$r_k{x}_k\in J⟹\sum _{k=1}^n{r}_k{x}_k\in J$$

As $x_k,r_k$ and $n\in \mathbb{N}$ are arbitrary then

$$R.T\subseteq J$$

Hence

$$R.T\subseteq \bigcap _{I◃R,T\subseteq R}I$$

As $R.T$ is an ideal containing $T$ then intersection lies in $R.T$

Subring Generated by Subset

Let $T\subset R$ be a subset of ring $R$ then

Define Subring generated by subset by

$$⟨T{⟩}_s=\bigcap _{T\subseteq S}S$$

where subscript $s$ denotes subring

Principal Ideal

Principal Ideal is an Ideal generated by a single element

Generators of a Principal Ideas are Associates lemma

Let $R$ be an integral domain
Let $I$ be a Principal Ideal then

Generators of $I$ are associates ($a\in R$ is a generator if $I=⟨a⟩$)
Hence generators of principal ideal form single equivalence class of associate elements of $R$

Proof

if $I=\{0\}=⟨0⟩$ then result follows

Assume $I\ne 0$
Let $a,b$ be generators of $I$ so

$$I=⟨a⟩=⟨b⟩$$

As $a\in ⟨b⟩$ then

$$\exists r\in R\text{ with }a=r.b$$

Similarly as $b\in ⟨a⟩$ then

$$\exists s\in R\text{ with }b=s.a$$

Then

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

Hence as $R$ is an integral domain

$$a(1-r.s)=0⟹r.s=1(\text{as }a\ne 0)$$

Hence $r$ and $s$ are units

If $I=⟨a⟩$ and $b=u.a$ where $u\in R^{\times }$ then $b\in ⟨a⟩=I$ thus $⟨B⟩\subseteq U$
But if $x\in I$ then $x=r.a$ where $r\in R$ hence

$$x=r.(u^{-1}.b)=(r.u^{-1}).b$$

Thus $x\in ⟨B⟩$ thus $I=⟨B⟩$

Ideals in Domain corollary

Let $R$ be a domain then

Ideals in $R$ are the kernels of the set of homomorphisms with domain $R$

Proof

By Ring Structure of Quotient Group then Kernel of Ring homomorphism is an ideal
If $I$ is an ideal and $q:R\to R/I$ is quotient map then q is a ring homomorphism with

$$\ker (q)=I$$

Image of an Ideal under Ring Homomorphism lemma

Let $\varphi :R\to S$ be a surjective homomorphism of rings
Let $I$ be an ideal of $R$ so $I◃R$ then

$$\varphi (I)=\{s\in S:\exists i\in I,s=\varphi (i)\}$$

is an ideal in $S$

Similarly let $J$ be an ideal of $S$ so $J◃S$ then

$${\varphi }^{-1}(J)=\{r\in R:\varphi (r)inJ\}$$

is an ideal in $R$

Thus $\varphi$ induces a pair of maps

$$\{\text{Ideals in }R\}\stackrel{\varphi }{\underset{{\varphi }^{-1}}{\rightleftarrows }}\{\text{Ideals in }S\}$$

Proof

$\varphi (I)$ is an additive subgroup of $S$ since $\varphi$ is a homomorphism of additive groups
If $s\in S$ and $j=\varphi (i)\in \varphi (I)$ then by surjectivity of $\varphi$

$$\exists r\in R\text{ such that }\varphi (r)=s$$

hence

$$s.j=\varphi (r).\varphi (i)=\varphi (r.i)\in \varphi (I)\text{ as }r.i\in I$$

Consider subset

$${\varphi }^{-1}(J)=\{x\in R:\varphi (x)\in J\}$$

If $x,y\in {\varphi }^{-1}(J)$ then

$$\varphi (x+y)=\varphi (x)+\varphi (y)\in J$$

since $J$ is an additive subgroup then ${\varphi }^{-1}(J)$ is an additive subgroup

If $r\in R$ and $x\in {\varphi }^{-1}(J)$ then

$$\varphi (r.x)=\varphi (r).\varphi (x)\in J\text{ as }\varphi (x)\in H$$

Correspondence of Ideals under a Surjective Homomorphism

Let $\varphi :R\to S$ be a surjective ring homomorphism
Let $K=\ker (\varphi )$

1. If $J◃S$ then

$$\varphi ({\varphi }^{-1}(J))=J$$

2. If $I◃R$ then

$${\varphi }^{-1}(\varphi (I))=I+K$$

Proof

If $f:X\to Y$ is a map of sets and $Z\subseteq Y$ then

$$f(f^{-1}(Z))=Z\cap \mathrm{im}(f)$$

As $\varphi$ is surjective then for any subset $J\subseteq S$ then

$$\varphi ({\varphi }^{-1}(J))=J$$

If $I◃R$ then $0\in \varphi (I)$ hence

$$K=\ker (\varphi )={\varphi }^{-1}(0)\subseteq {\varphi }^{-1}(\varphi (I))$$

Hence $I+K$ being the ideal generated by $I$ and $K$ must lie in ideal ${\varphi }^{-1}(\varphi (I))$
If $x\in {\varphi }^{-1}(\varphi (I))$ then there exists $i\in I$ such that

$$\varphi (x)=\varphi (i)⟹\varphi (x-i)=0$$

Hence

$$x=i+(x-i)\in I+K$$

Hence ${\varphi }^{-1}(\varphi )(I)\subseteq I+K$

Correspondence Theorem for Rings corollary

Let $R$ be a ring
Let $I$ be an ideal in $R$ so $I◃R$
Let $q:R\to R/I$ be quotient map

Suppose $J$ is an ideal then $q(J)$ is an ideal in $R/I$
Suppose $K$ is an ideal in $R/I$ then

$$q^{-1}(K)=\{r\in R:q(r)\in K\}$$

is an ideal in $R$ containing $I$

Moreover, correspondences give bijection between ideals in $R/I$ and ideals in $R$ containing $I$

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 $R$

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 $b\in I$

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

$$I=⟨a⟩$$

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

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 $f=f^{\prime }$