# Abstract Algebra/Rings, ideals, ring homomorphisms
## Basic definitions

Definition 10.1:
A ring is a set $R$ together with two binary operations $+:R\times R\to R$ and $\cdot :R\times R\to R$ and two special elements, the unit $1$ and the zero $0$, such that:
* $R$ is an abelian group with respect to $+$ with neutral element $0$.
* $R$ is a monoid (that is, a group without inversion) with respect to $\cdot$ with neutral element $1$.
* The distributive laws hold: $a\cdot (b+c)=a\cdot b+a\cdot c$, $(b+c)\cdot a=b\cdot a+c\cdot a$.
Examples 10.2:
* The whole numbers $\mathbb {Z}$ with respect to usual addition and multiplication are a ring.
* Every field is a ring.
* If $R$ is a ring, then all polynomials over $R$ form a ring. This example will be explained later in the section on polynomial rings.
Definition 10.3:
Let $R$ be a ring. A left ideal of $R$ is a subset $I\subseteq R$ such that the following two things hold:
* $(I,+)$ is a subgroup of $(R,+)$.
* $\forall r\in R:rI\subseteq I$, where $rI=\{ri|i\in I\}$ (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 $I\subseteq R$ is a both-sided ideal of $R$, we write $I\leq R$.
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 $(S,\leq )$ be a partially ordered set (that is, the usual conditions transitivity, reflexivity and anti-symmetry are satisfied). $S$ is called inductive if and only if every ascending chain of elements of $S$ (that is, a sequence $(s_{n})_{n\in \mathbb {N} }$ in $S$ such that $s_{1}\leq s_{2}\leq s_{3}\leq \cdots \leq s_{k}\leq \cdots$) has an upper bound (that is, an element $b\in S$ such that $\forall n\in \mathbb {N} :b\geq s_{n}$).
With this definition, we observe:
Theorem 10.5:
If a commutative ring $R$ is given, the set of all ideals $I$ of $R$, partially ordered by inclusion (i.e. $S=\{I\subset R|I\leq R\}\subset 2^{R}$, where we use the convention of Donald Knuth and denote the power set of a set $T$ by $2^{T}$) is inductive.
Proof:
If
    $I_{1}\subseteq I_{2}\subseteq I_{3}\subseteq \cdots \subseteq I_{k}\subseteq \cdots$
is an ascending chain of ideals, we set
    $J:=\bigcup _{n\in \mathbb {N} }I_{n}$
and claim that $J\leq R$. Indeed, if $a,b\in J$, find $m,n\in \mathbb {N}$ such that $a\in I_{n}$ and $b\in I_{m}$. Then set $N:=\max\{m,n\}$, so that $a,b,a+b\in I_{N}\subseteq J$ since $I_{N}\leq R$. Similarly, if $a\in J$ and $r\in R$, pick $n\in \mathbb {N}$ such that $a\in I_{n}$, whence $ra\in I_{n}\subseteq J$ since $I_{n}\leq R$.$\Box$

## Residue class rings

Definition and theorem 10.4:
Let $R$ be a ring, and $I\leq R$. Then we define a relation $\sim _{I}$ on $R$ as follows:
    $a\sim _{I}b:\Leftrightarrow a-b\in I$.
This relation is an equivalence relation, and an equivalence class $[a]$ shall be denoted by $a+I$ for $a\in R$. If we define an addition
    $(a+I)+(b+I):=(a+b)+I$
and a multiplication
    $(a+I)\cdot (b+I):=a\cdot b+I$,
then these two are well-defined (i. e. independent of the choice of the representatives $a$ and $b$) and turn $R/I$ into a ring, called the residue class ring with respect to the ideal $I$.
Proof:
First, we check that $\sim _{I}$ is an equivalence relation.
* Reflexiveness: $a-a=0\in I$ since $I$ is an additive subgroup.
* Symmetry: $a-b\in I\Leftrightarrow -(a-b)\in I$ since inverses are in the subgroup.
* Transitivity: Let $a-b\in I$ and $b-c\in I$. Then $a-c=a-b+(b-c)\in I$, since a subgroup is closed under the group operation.
Then we check that addition and multiplication are well-defined. Let $a+I=a'+I$ and $b+I=b'+I$. Then
    $a+b-(a'+b')=a+b-(a+i+b+j)=-i-j\in I$ for certain $i,j\in I$.
Furthermore,
    $a\cdot b-a'\cdot b'=a\cdot b-a\cdot b-a\cdot j-i\cdot b-i\cdot b$
for these same $i,j\in I$; this is in $I$ by closedness by left and right multiplication.
The ring axioms directly carry over from the old ring $R$.$\Box$

## Ring homomorphisms

Definition 10.5:
Let $R,S$ be rings. A ring homomorphism between the two is a map
    $\varphi :R\to S$
such that:
* For all $a,b\in R$ $\varphi (a+b)=\varphi (a)+\varphi (b)$ and $\varphi (a\cdot b)=\varphi (a)\cdot \varphi (b)$.
* $\varphi (1_{R})=1_{S}$ ($1_{R}$ is the unit of $R$ and $1_{S}$ of $S$).