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 $S$).