# Abstract Algebra/Rings, ideals, ring homomorphisms

## Basic definitions

Definition 10.1:

A ring is a set ${\displaystyle R}$ together with two binary operations ${\displaystyle +:R\times R\to R}$ and ${\displaystyle \cdot :R\times R\to R}$ and two special elements, the unit ${\displaystyle 1}$ and the zero ${\displaystyle 0}$, such that:

1. ${\displaystyle R}$ is an abelian group with respect to ${\displaystyle +}$ with neutral element ${\displaystyle 0}$.
2. ${\displaystyle R}$ is a monoid (that is, a group without inversion) with respect to ${\displaystyle \cdot }$ with neutral element ${\displaystyle 1}$.
3. The distributive laws hold: ${\displaystyle a\cdot (b+c)=a\cdot b+a\cdot c}$, ${\displaystyle (b+c)\cdot a=b\cdot a+c\cdot a}$.

Examples 10.2:

• The whole numbers ${\displaystyle \mathbb {Z} }$ with respect to usual addition and multiplication are a ring.
• Every field is a ring.
• If ${\displaystyle R}$ is a ring, then all polynomials over ${\displaystyle R}$ form a ring. This example will be explained later in the section on polynomial rings.

Definition 10.3:

Let ${\displaystyle R}$ be a ring. A left ideal of ${\displaystyle R}$ is a subset ${\displaystyle I\subseteq R}$ such that the following two things hold:

1. ${\displaystyle (I,+)}$ is a subgroup of ${\displaystyle (R,+)}$.
2. ${\displaystyle \forall r\in R:rI\subseteq I}$, where ${\displaystyle rI=\{ri|i\in I\}}$ (closedness by left multiplication).

Replacing closedness by left multiplication by closedness by right multiplication, we can define right ideals, and then both-sided ideals. If ${\displaystyle I\subseteq R}$ is a both-sided ideal of ${\displaystyle R}$, we write ${\displaystyle I\leq R}$.

We'll now show an important property of the set of all ideals of a given ring, namely that it's inductive. This means:

Definition 10.4:

Let ${\displaystyle (S,\leq )}$ be a partially ordered set (that is, the usual conditions transitivity, reflexivity and anti-symmetry are satisfied). ${\displaystyle S}$ is called inductive if and only if every ascending chain of elements of ${\displaystyle S}$ (that is, a sequence ${\displaystyle (s_{n})_{n\in \mathbb {N} }}$ in ${\displaystyle S}$ such that ${\displaystyle s_{1}\leq s_{2}\leq s_{3}\leq \cdots \leq s_{k}\leq \cdots }$) has an upper bound (that is, an element ${\displaystyle b\in S}$ such that ${\displaystyle \forall n\in \mathbb {N} :b\geq s_{n}}$).

With this definition, we observe:

Theorem 10.5:

If a commutative ring ${\displaystyle R}$ is given, the set of all ideals ${\displaystyle I}$ of ${\displaystyle R}$, partially ordered by inclusion (i.e. ${\displaystyle S=\{I\subset R|I\leq R\}\subset 2^{R}}$, where we use the convention of Donald Knuth and denote the power set of a set ${\displaystyle T}$ by ${\displaystyle 2^{T}}$) is inductive.

Proof:

If

${\displaystyle I_{1}\subseteq I_{2}\subseteq I_{3}\subseteq \cdots \subseteq I_{k}\subseteq \cdots }$

is an ascending chain of ideals, we set

${\displaystyle J:=\bigcup _{n\in \mathbb {N} }I_{n}}$

and claim that ${\displaystyle J\leq R}$. Indeed, if ${\displaystyle a,b\in J}$, find ${\displaystyle m,n\in \mathbb {N} }$ such that ${\displaystyle a\in I_{n}}$ and ${\displaystyle b\in I_{m}}$. Then set ${\displaystyle N:=\max\{m,n\}}$, so that ${\displaystyle a,b,a+b\in I_{N}\subseteq J}$ since ${\displaystyle I_{N}\leq R}$. Similarly, if ${\displaystyle a\in J}$ and ${\displaystyle r\in R}$, pick ${\displaystyle n\in \mathbb {N} }$ such that ${\displaystyle a\in I_{n}}$, whence ${\displaystyle ra\in I_{n}\subseteq J}$ since ${\displaystyle I_{n}\leq R}$.${\displaystyle \Box }$

## Residue class rings

Definition and theorem 10.4:

Let ${\displaystyle R}$ be a ring, and ${\displaystyle I\leq R}$. Then we define a relation ${\displaystyle \sim _{I}}$ on ${\displaystyle R}$ as follows:

${\displaystyle a\sim _{I}b:\Leftrightarrow a-b\in I}$.

This relation is an equivalence relation, and an equivalence class ${\displaystyle [a]}$ shall be denoted by ${\displaystyle a+I}$ for ${\displaystyle a\in R}$. If we define an addition

${\displaystyle (a+I)+(b+I):=(a+b)+I}$

and a multiplication

${\displaystyle (a+I)\cdot (b+I):=a\cdot b+I}$,

then these two are well-defined (i. e. independent of the choice of the representatives ${\displaystyle a}$ and ${\displaystyle b}$) and turn ${\displaystyle R/I}$ into a ring, called the residue class ring with respect to the ideal ${\displaystyle I}$.

Proof:

First, we check that ${\displaystyle \sim _{I}}$ is an equivalence relation.

1. Reflexiveness: ${\displaystyle a-a=0\in I}$ since ${\displaystyle I}$ is an additive subgroup.
2. Symmetry: ${\displaystyle a-b\in I\Leftrightarrow -(a-b)\in I}$ since inverses are in the subgroup.
3. Transitivity: Let ${\displaystyle a-b\in I}$ and ${\displaystyle b-c\in I}$. Then ${\displaystyle a-c=a-b+(b-c)\in I}$, since a subgroup is closed under the group operation.

Then we check that addition and multiplication are well-defined. Let ${\displaystyle a+I=a'+I}$ and ${\displaystyle b+I=b'+I}$. Then

${\displaystyle a+b-(a'+b')=a+b-(a+i+b+j)=-i-j\in I}$ for certain ${\displaystyle i,j\in I}$.

Furthermore,

${\displaystyle a\cdot b-a'\cdot b'=a\cdot b-a\cdot b-a\cdot j-i\cdot b-i\cdot b}$

for these same ${\displaystyle i,j\in I}$; this is in ${\displaystyle I}$ by closedness by left and right multiplication.

The ring axioms directly carry over from the old ring ${\displaystyle R}$.${\displaystyle \Box }$

## Ring homomorphisms

Definition 10.5:

Let ${\displaystyle R,S}$ be rings. A ring homomorphism between the two is a map

${\displaystyle \varphi :R\to S}$

such that:

1. For all ${\displaystyle a,b\in R}$ ${\displaystyle \varphi (a+b)=\varphi (a)+\varphi (b)}$ and ${\displaystyle \varphi (a\cdot b)=\varphi (a)\cdot \varphi (b)}$.
2. ${\displaystyle \varphi (1_{R})=1_{S}}$ (${\displaystyle 1_{R}}$ is the unit of ${\displaystyle R}$ and ${\displaystyle 1_{S}}$ of ${\displaystyle S}$).