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 $ℤ$ 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 $(R, +)$ .
$\forall r \in R : r I \subseteq I$ , where $r I = {r i | i \in I}$ (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 $(s_{n})_{n \in ℕ}$ in $S$ such that $s_{1} \leq s_{2} \leq s_{3} \leq \dots \leq s_{k} \leq \dots$ ) has an upper bound (that is, an element $b \in S$ such that $\forall n \in ℕ : b \geq s_{n}$ ).

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 \dots \subseteq I_{k} \subseteq \dots

is an ascending chain of ideals, we set

J := ⋃_{n \in ℕ} I_{n}

and claim that $J \leq R$ . Indeed, if $a, b \in J$ , find $m, n \in ℕ$ 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 $n \in ℕ$ such that $a \in I_{n}$ , whence $r a \in I_{n} \subseteq J$ since $I_{n} \leq R$ . $◻$

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 $R$ . $◻$

Ring homomorphisms

Definition 10.5:

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

φ : R \to S

such that:

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