Ideals and Quotient Rings

In chapter 5, we studied the congruence classes ring $F [x] / (p (x))$ . But we haven’t yet justified the notation, why symbolically looks like we are dividing a ring $F [x]$ by some subset $(p (x))$ . In this chapter, we will study the general situation. For a ring $R$ and a subset $I \subseteq R$ with special invariant properties (ideal), we can define the quotient ring $R / I$ and prove that it is a ring.

Ideals

Let’s recall the two key congruence relations:

$Z_{n} = Z / (n)$ , two integers $a, b$ are congruent modulo $n$ iff $n ∣ a - b$ iff $a - b \in (n)$ where:

(n) = {nk ∣ k \in Z}

$F [x] / (p (x))$ , two polynomials $f, g$ are congruent modulo $p (x)$ iff $p (x) ∣ f - g$ iff $f - g \in (p (x))$ where:

(p (x)) = {p (x) q (x) ∣ q (x) \in F [x]}

These two subsets have the following strong symmetric property:

if $a \in (n)$ , then any multiple of $a$ is also in $(n)$ , i.e. $ka \in (n)$ for any integer $k$ ,
if $f \in (p (x))$ , then any multiple of $f$ is also in $(p (x))$ , i.e. $f (x) q (x) \in (p (x))$ for any polynomial $q (x)$ .

It turns out this is all we need to generalize the congruence relations for any commutative rings $R$ and $I \subseteq R$ .

Definition

Let $R$ be a commutative ring. A subring $I \subseteq R$ is called an ideal if: For any $r \in R$ , $a \in I$ , we have $r a \in I$ (absorbing property).

Warning

An ideal must be a subring of $R$ , i.e. it must be closed under addition and multiplication. In particular, we also have $0 \in I$ .

Warning

In this chapter we will mostly work with unitary commutative rings. If we would like to define ideals in non-commutative rings, we need to make sure we have two-sided invariant property, i.e. $a r \in I$ and $r a \in I$ for any $r \in R$ and $a \in I$ . In some textbook, in the non-commutative setting, such an $I$ is called a two-sided ideal.

When verifying the ideal property, we need to first check the subring property, which itself contains 4 properties:

$0 \in I$ .
If $a \in I$ , then $- a \in I$ .
If $a, b \in I$ , then $a + b \in I$ .
If $a, b \in I$ , then $ab \in I$ .

Then we verify the absorbing property: 5. for any $r \in R$ , $a \in I$ , we have $r a \in I$ .

So in total we have 5 properties to check. It turns out the absorbing property is powerful enough to imply many of the other properties. We can reduce the verification to the following:

Theorem

Let $R$ be a commutative unitary ring and $I \subseteq R$ be a non-empty subset. Then $I$ is an ideal of $R$ if and only if:

For any $a, b \in I$ , $a - b \in I$ .

For any $r \in R$ , $a \in I$ , we have $r a \in I$ (absorbing property).

Proof of Theorem

We will shown with the given two properties in the theorem, we can derive the 5 properties of an ideal. The absorbing property is already given, so we only need to show the other 4 properties:

$0 \in I$ : Take $a \in I$ and $r = 0_{R}$ , we have $0_{R} a = 0_{R} \in I$ .
$- a \in I$ : Take $a \in I$ and $r = - 1_{R}$ , we have $- a = r a \in I$ .
$a + b \in I$ : Take $a, b \in I$ , we have $a, - b \in I$ , so $a - (- b) = a + b \in I$ .
$ab \in I$ : Take $a, b \in I$ , in particular $a \in I$ , $b \in R$ , we have $ab \in I$ by the absorbing property. $■$

Finitely Generated Ideals

As our standard examples, we have our ideals $(n)$ and $(p (x))$ . We notice we are using the single element to represent the ideal. In fact, this notation can be generalized to multiple generators.

Definition/Proposition

Let $R$ be a commutative unitary ring. An ideal $I \subseteq R$ is called finitely generated if there exists a finite set of elements $a_{1}, a_{2}, \dots, a_{n} \in R$ such that:
$I = (a_{1}, a_{2}, \dots, a_{n}) = {r_{1} a_{1} + r_{2} a_{2} + \dots + r_{n} a_{n} ∣ r_{i} \in R} .$
We denote such an ideal as $(a_{1}, a_{2}, \dots, a_{n}) \subset R$ .

This definition needs a justification, we must show the set ${r_{1} a_{1} + r_{2} a_{2} + \dots + r_{n} a_{n} ∣ r_{i} \in R}$ is indeed an ideal. Take any $r \in R$ , any element $x \in (a_{1}, a_{2}, \dots, a_{n})$ , we can write:

x = r_{1} a_{1} + r_{2} a_{2} + \dots + r_{n} a_{n}

and

r x = r_{1} r a_{1} + r_{2} r a_{2} + \dots + r_{n} r a_{n} .

Since $r_{i} r \in R$ , we have $r x \in (a_{1}, a_{2}, \dots, a_{n})$ . $■$

In our standard examples, we have ideals generated by a single element, e.g. $(n)$ and $(p (x))$ . These will be called principal ideals:

Definition

An ideal $I \subseteq R$ is called a principal ideal if there exists an element $a \in R$ such that:
$I = (a) = {r a ∣ r \in R} .$

We spend a lot of time only studying the principal ideals $(n) \subset Z, (p (x)) \subset F [x]$ in chapter 2 and chapter 5. One may wonder why we don’t study the general case. The reason is, for $Z$ and $F [x]$ , the ideals are all principal ideals. In fact, we have the following theorem:

Theorem

Let $a_{1}, a_{2}, \dots, a_{k} \in Z$ , not all zero. Let $d = g cd (a_{1}, a_{2}, \dots, a_{k})$ . Then:
$(a_{1}, a_{2}, \dots, a_{k}) = (d) .$

proof of the theorem

To show two subset of $Z$ are equal, we need to show:

$(a_{1}, a_{2}, \dots, a_{k}) \subseteq (d)$ .
$(d) \subseteq (a_{1}, a_{2}, \dots, a_{k})$ .

$(a_{1}, a_{2}, \dots, a_{k}) \subseteq (d)$ : Take any $x \in (a_{1}, a_{2}, \dots, a_{k})$ , we can write:

x = r_{1} a_{1} + r_{2} a_{2} + \dots + r_{k} a_{k} .

By the definition of $d$ , we have $d ∣ r_{1} a_{1} + r_{2} a_{2} + \dots + r_{k} a_{k}$ , so $x \in (d)$ .

$(d) \subseteq (a_{1}, a_{2}, \dots, a_{k})$ : By Bezout’s Theorem, we can write:

d = r_{1} a_{1} + r_{2} a_{2} + \dots + r_{k} a_{k}

for some integers $r_{1}, r_{2}, \dots, r_{k}$ . Take any $x \in (d)$ , we can write:

x = s d = s (r_{1} a_{1} + r_{2} a_{2} + \dots + r_{k} a_{k}) = s d r_{1} a_{1} + s d r_{2} a_{2} + \dots + s d r_{k} a_{k}

for some integer $s$ . Thus $x \in (a_{1}, a_{2}, \dots, a_{k})$ . $■$

A similar arument can be made for $F [x]$ . In fact, we have the following theorem:

Theorem

Let $p_{1} (x), p_{2} (x), \dots, p_{k} (x) \in F [x]$ be polynomials in $F [x]$ , not all zero. Let $d (x) = g cd (p_{1} (x), p_{2} (x), \dots, p_{k} (x))$ . Then:
$(p_{1} (x), p_{2} (x), \dots, p_{k} (x)) = (d (x)) .$

The finitely generated ideal $(a_{1}, a_{2}, \dots, a_{k})$ is the minimal ideal containing $a_{1}, a_{2}, \dots, a_{k}$ . We have:

Proposition

Let $I$ be an ideal of $R$ . Suppose $a_{1}, a_{2}, \dots, a_{k} \in I$ . Then:
$(a_{1}, a_{2}, \dots, a_{k}) \subseteq I .$

Congruence in Abstract Rings

Now we can define the congruence relation in abstract rings.

Definition

Let $R$ be a commutative unitary ring and $I \subseteq R$ be an ideal. Two elements $a, b \in R$ are said to be congruent modulo $I$ if:
$a - b \in I .$
We denote this relation as:
$a \equiv b mod I .$

The key observation is, this abstract congruence relation is still an equivalence relation. We can check the three properties:

Reflexive: $a - a = 0 \in I$ , so $a \equiv a mod I$ .
Symmetric: If $a \equiv b mod I$ , then $a - b \in I$ , so $b - a = - (a - b) \in I$ , thus $b \equiv a mod I$ .
Transitive: If $a \equiv b mod I$ and $b \equiv c mod I$ , then $a - b \in I$ and $b - c \in I$ , so $a - c = (a - b) + (b - c) \in I$ , thus $a \equiv c mod I$ .

This makes us be able to define the congruence classes:

Definition

Let $R$ be a commutative unitary ring and $I \subseteq R$ be an ideal. The congruence class of $a \in R$ modulo $I$ is defined as:
$[a] = {b \in R ∣ b \equiv a mod I} = {b \in R ∣ a - b \in I} .$
We denote the set of all congruence classes as:
$R / I = {[a] ∣ a \in R} .$

The set $R / I$ actually has a ring structure. We can define the addition and multiplication as follows:

Definition/Theorem

Let $R$ be a commutative unitary ring and $I \subseteq R$ be an ideal. The quotient ring $R / I$ is defined as:
$[a] + [b] = [a + b], [a] [b] = [ab]$
where $[a], [b] \in R / I$ are congruence classes.

proof of the definition

We need to show the addition and multiplication are well-defined. This means, if $[a_{1}] = [a_{2}]$ and $[b_{1}] = [b_{2}]$ , then $[a_{1} + b_{1}] = [a_{2} + b_{2}]$ and $[a_{1} b_{1}] = [a_{2} b_{2}]$ .

Addition: Assume $[a_{1}] = [a_{2}]$ and $[b_{1}] = [b_{2}]$ , then $a_{1} - a_{2} \in I$ and $b_{1} - b_{2} \in I$ . We have:

a_{1} + b_{1} - (a_{2} + b_{2}) = (a_{1} - a_{2}) + (b_{1} - b_{2}) \in I .

Thus $[a_{1} + b_{1}] = [a_{2} + b_{2}]$ . 2. Multiplication: Assume $[a_{1}] = [a_{2}]$ and $[b_{1}] = [b_{2}]$ , then $a_{1} - a_{2} \in I$ and $b_{1} - b_{2} \in I$ . We can write:

a_{1} = a_{2} + i_{1}, b_{1} = b_{2} + i_{2}

for some $i_{1}, i_{2} \in I$ . We have:

a_{1} b_{1} - a_{2} b_{2} = (a_{2} + i_{1}) (b_{2} + i_{2}) - a_{2} b_{2} = a_{2} i_{2} + b_{2} i_{1} + i_{1} i_{2} .

Notice that $a_{2} i_{2}, b_{2} i_{1}, i_{1} i_{2} \in I$ by the absorbing property. Thus:

a_{1} b_{1} - a_{2} b_{2} \in I .

Thus $[a_{1} b_{1}] = [a_{2} b_{2}]$ . $■$

Notation for Congruence Classes

So far, we used the notation $[a]$ to denote the congruence class of $a$ , this is inherited from the notation of congruence classes in $Z$ and $F [x]$ . In the abstract setting, we will also use following notation:

Notation

Let $R$ be a commutative unitary ring and $I \subseteq R$ be an ideal. We will use the notation:
$a + I = [a] = {b \in R ∣ b \equiv a mod I}$
to denote the congruence class of $a$ modulo $I$ . We also consider $a + I$ as the notation for an element of $R / I$ .

This notation is very useful, we could formally introduce the rules:

$r \cdot I = I$ for any $r \in R$ .
$I + I = I$ .
$I \cdot I = I$ . (This rule is only correct for computations in $R / I$ .) We can formally multiply two elements using distributive property:

(a + I) (b + I) = ab + a I + b I + I^{2} = ab + I .

As long as we realize both sides as elements of $R / I$ , formal calculations and simplification using above rules will be correct.

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Ideals and Quotient Rings

Ideals

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Congruence in Abstract Rings

proof of the definition

Notation for Congruence Classes

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