# 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:

1. $R$  is an abelian group with respect to $+$  with neutral element $0$ .
2. $R$  is a monoid (that is, a group without inversion) with respect to $\cdot$  with neutral element $1$ .
3. 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:

1. $(I,+)$  is a subgroup of $(R,+)$ .
2. $\forall r\in R:rI\subseteq I$ , 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.

1. Reflexiveness: $a-a=0\in I$  since $I$  is an additive subgroup.
2. Symmetry: $a-b\in I\Leftrightarrow -(a-b)\in I$  since inverses are in the subgroup.
3. 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:

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