03. Ideals and Quotients

Kernel of Ring

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

Kernel of $f$ is

$$\ker (f)=\{r\in R:f(r)=0\}$$

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Image of Ring

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

Image of $f$ is

$$\mathrm{im}(f)=\{s\in S:\exists r\in R\text{ such that }f(r)=s\}$$

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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 $I◃R$

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Kernel of a Homomorphism is an Ideal lemma

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

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

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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 $R$

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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

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

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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

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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

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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 $R.T$

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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

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Principal Ideal

Principal Ideal is an Ideal generated by a single element

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Associates

Let $a,b\in R$ where $R$ is a ring then

$a,b$ are associates if

$$\exists u\in {\mathbb{R}}^{\times }\text{ such that }a=u.b$$

Note that it is an equivalence class

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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 $I=⟨B⟩$

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3.1 The Quotient Construction

Quotient

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

Quotient Group $(R/I,+)$ is defined through

Equivalence Relation $r\sim s$ if $r-s\in I$ with equivalence class cosets

$$\{r+I:r\in R\}$$

Operation

$$\begin{array}{rl}(r_1+I)+(r_2+I)& =(r_1+r_2)+I\\ (r_1+I)\times (r_2+I)& =r_1.r_2+I\end{array}$$

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01 - Ring Structure of Quotient Group

Quotient Ring Construction

Let ideal $I$ in ring $R$

Datum $(R/I,+,\times ,0+I,1+I)$ defines ring structure on $R/I$

Map $q:R\to R/I$ by

$$q(r)=r+I$$

is a surjective ring homomorphism

with $\ker Q=I$

Note that $q:R\to R/I$ is called the quotient homomorphism

Proof - $\ker Q=I$

$$\ker Q=\{r\in R:q(r)=0\}$$

Where $q(r)=0$ means

$$q(r)=r+I+0+I$$

Thus $r\in I$ so $\ker (q)=I$

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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

$$\ker (q)=I$$

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02 - Universal Property of Quotients

Universal Property of Quotients

Let $R$ be a ring with ideal $I$ of $R$
Consider Quotient Homomorphism $q:R\to R/I$

If $\varphi :R\to S$ is a ring homomorphism such that

$$I\subseteq \ker (\varphi )$$

then there is unique ring homomorphism

$$\overline{\varphi }:R/I\to S$$

such that

$$\overline{\varphi }\circ q=\varphi$$

and

$$\ker (\overline{\varphi })=\ker (\varphi )/I=\{m+I:m\in \ker (\varphi )\}$$

Proof

Since $q$ is surjective then

$$\overline{\varphi }(q(r))=\varphi (r)\text{ for }r\in R$$

uniquely determines values of $\overline{\varphi }$ such that $\overline{\varphi }$ is unique if it exists

If $r-r^{\prime }\in I$ then

$$I\subseteq \ker (\varphi )⟹0=\varphi (r-r^{\prime })=\varphi (r)-\varphi (r^{\prime })$$

Hence $\varphi$ is constant on $I$-cosets

Thus induces map

$$\overline{\varphi }(m+I)=\varphi (m)$$

As $\overline{\varphi }$ is a homomorphism following from definition of ring structure on quotient $R/I$

For kernel of $\overline{\varphi }$ then

$$\overline{\varphi }(r+I)=\varphi (r)=0⟺r\in \ker (\varphi )$$

Hence $r+I\in \ker (\varphi )/Iz$

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Isomorphism Theorem corollary

First Isomorphism Theorem

Let $\varphi :R\to S$ is a homomorphism then

$\varphi$ induces isomorphism

$$\overline{\varphi }:R/\ker (\varphi )\to \mathrm{im}(\varphi )$$

Second Isomorphism Theorem

Let $R$ be a ring
Let $A$ be a subring of $R$
Let $I$ be an ideal of $R$ then

$$(A+I)/I\cong A/(A\cap I)$$

Third Isomorphism Theorem

Let $I_1\subseteq I_2$ be ideals in ring $R$ then

$$(R/I_1)/(I_2/I_1)\cong R/I_2$$

Proof

First Isomorphism Theorem
Apply Universal Property of Quotients to $I=\ker (\varphi )$ then

$$\ker (\overline{\varphi })=\ker (\varphi )/\ker (\varphi )=0$$

Hence $\overline{\varphi }$ is injective and hence induces isomorphism onto image from

$$\overline{\varphi }\circ q=\varphi =\mathrm{im}(\varphi )$$

Second Isomorphism Theorem
As $A$ is a subring and $I$ is an ideal then $A+I$ is a subring of $R$ containing $I$
Let $q:R\to R/I$ be quotient map

Then $q$ restricts to homomorphism $p$ from $A$ to $R/I$ with image $(A+I)/I$
By First Isomorphism Theorem then kernel of $p$ is $A\cap I$
(As if $a\in A$ with $p(a)=0$ then $a+I=0+I$ so $a\in I$ so $a\in A\cap I$)

Third Isomorphism Theorem
Let $q_i:R\to R/I_j$ for $j=1,2$
By Universal Property of Quotients for $q_2$ then exists homomorphism

$$\overline{q_2}:R/I_1\to R/I_2$$

induced by map $q_2:R\to R/I_2$ with kernel

$$\ker (q_2)/I_1=I_2/I_2$$

and $\overline{q_2}\circ q_1=q_2$

Hence $\overline{q_2}$ is surjective as $q_2$ is hence result follows by First Isomorphism Theorem

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03 - Direct Sum Decomposition of Quotients

Direct Sum Decomposition of Quotients

Let $R$ be a ring
Let $I,J$ be ideals of $R$ such that $I+J=R$ then

$$R/(I\cap J)\cong R/I\oplus R/J$$

Proof

Consider quotient maps

$$\begin{array}{rl}{q}_1& :R\to R/I\\ q_2& :R\to R/J\end{array}$$

Define

$$q:R\to R/I\oplus R/J$$

by

$$q(r)=(q_1(r),q_2(r))$$

By First Isomorphism Theorem then result follows

Showing Subjectivity
Suppose $(r+I,s+J)\in R/I\oplus R/J$ then
Since $R=I+J$ then

$$r=i_1+j_2\text{ and }s=i_2+j_2\text{ where }{i}_1,i_2\in I\text{ and }{j}_1,j_2\in J$$

But

$$r+I=j_1+I\text{ and }s+J=i_2+J$$

So

$$q(j_1+j_2)=(r+I,s+J)$$

Showing $\ker (q)=I\cap J$
if $q(r)=0$ then $q_1(r)=0$ and $q_2(r)=0$ so

$$r\in I\text{ and }r\in J⟹r\in I\cap J$$

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3.2 Images and Preimages of Ideals

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

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

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

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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 $I$

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