Proofs in Ring Theory

Rings

Evaluating these roles, we observe that properties (1) through (4) establish the integers as an Abelian group under addition. Property (5) extends this structure into the realm of multiplication. Crucially, property (6), the Distributive Law, serves as the “logical glue” of the system. Without it, addition and multiplication would remain two disparate operations, providing the essential framework required to define a Ring.

In this abstract context, the symbols $0$ and $-a$ must be viewed as algebraic roles rather than literal integers. The “additive identity” ($0$) is the unique neutral element under addition, and its existence in axiom R3 guarantees that every ring is a non-empty set.

Similarly, the “additive inverse” ($-a$) is simply the element that satisfies the identity requirement, regardless of its form.

Rings may manifest as familiar number sets, arrays of data, or mappings between spaces, provided their operations are closed and meet the prescribed criteria.

The Ring of Even Integers ($2\mathbb{Z}$)

The set of even integers

$$2\mathbb{Z} = \{2k : k \in \mathbb{Z}\}$$

is closed under addition and multiplication, as the sum and product of any two even integers are always even.

This illustrates that a subset of a ring can function as a ring itself, even if it lacks certain features of the parent ring (such as a multiplicative identity).

Matrix and Function Rings ($M_n(\mathbb{R})$ and $F_{\mathbb{R}}$)

$M_n(\mathbb{R})$ is the set of $n \times n$ matrices (for $n \ge 2$) under matrix addition and multiplication. The additive identity is the zero matrix $Z$, and the additive inverse of $A$ is $-A$.

This confirms that ring elements need not be “numbers” and that multiplication in a ring is not required to be commutative.

$F_{\mathbb{R}}$ is the set of all real-valued functions $f: \mathbb{R} \to \mathbb{R}$. Addition and multiplication are defined pointwise. The additive identity is the zero function $f_0(x) = 0$.

This establishes that rings can consist of mappings, expanding the theory into functional analysis.

Elementary Properties of Rings

While the core axioms (R1–R6) define a ring, we classify higher-order structures by testing for additional properties: R7 (Commutativity of Multiplication) and R8 (Existence of Unity).

Certain properties are not merely observed but are derived from the ring axioms.

Proof

The additive identity is unique

Suppose $0$ and $0’$ are both additive identities for $R$. Since $0$ is an identity, $0’ + 0 = 0’$. Since $0’$ is an identity, $0 + 0’ = 0$.

By commutativity (R1), $0’ + 0 = 0 + 0’$, implying $0’ = 0$.

The inverse of any ring element is unique

Suppose $x$ has additive inverses $-x$ and $x’$. Then

$$x + (-x) = 0$$

and

$$x + x' = 0$$

Thus:

$$-x = -x + 0 = -x + (x + x') = (-x + x) + x' = 0 + x' = x'$$

Proof

Suppose $1$ and $1’$ are both unities in $R$.

Since $1$ is unity,

$$1 \cdot 1' = 1'$$

Since $1’$ is unity,

$$1 \cdot 1' = 1$$

Thus, 1 = 1'.

Proof

If $a + b = a + c$, add the unique additive inverse $-a$ to both sides:

$$(-a) + (a + b) = (-a) + (a + c)$$

By R2,

$$((-a) + a) + b = ((-a) + a) + c$$

which simplifies to

$$0 + b = 0 + c$$

and thus $b = c$.

Proof

For $a \in R$, observe

$$a \cdot 0 + 0 = a \cdot 0 = a \cdot (0 + 0) = a \cdot 0 + a \cdot 0$$

By The Cancellation Law, we cancel $a \cdot 0$ from both sides, yielding $0 = a \cdot 0$.

Proof

We must show that $(-a)b$ is the additive inverse of $ab$.

We verify this by showing their sum is the additive identity:

$$ab + (-a)b = (a + (-a))b$$

by R6. This simplifies to $0 \cdot b$, which is 0 by The Multiplicative Property of Zero. Thus,

$$(-a)b = -(ab)$$

These elementary properties establish a consistent “arithmetic” within any ring, mirroring the structural behavior of integers.

Subrings

Identifying subrings—subsets of a ring that are rings in their own right—is a powerful tool because subrings inherit the “inherited” properties (associativity, commutativity, distributivity) of the parent ring.

Proof

Suppose $S$ is a nonempty subset of $R$ closed under subtraction and multiplication.

Identity:

Since $S$ is nonempty, there exists $s \in S$. Closure under subtraction implies

$$s - s = 0 \in S$$

Inverses

For any $a \in S$, the closure under subtraction implies

$$0 - a = -a \in S$$

Addition

For $a, b \in S$, we have $-b \in S$, so

$$a - (-b) = a + b \in S$$

$S$ is closed under addition.

Inheritance

Since $S \subseteq R$, the operations in $S$ are the same as in $R$.

Therefore, property R1 (commutativity of addition), R2 (associativity of addition), R5 (associativity of multiplication), and R6 (distributive laws) are automatically satisfied in $S$.

Since $S$ contains the additive identity and inverses and is closed under both operations, $S$ is a ring.

Gaussian Integers ($G$)

The set

$$\{a + bi : a, b \in \mathbb{Z}\}$$

is a subring of $\mathbb{C}$. The difference and product of any two Gaussian integers yield another Gaussian integer, satisfying the test.

Intersection of Subrings

If $S_1$ and $S_2$ are subrings, $S_1 \cap S_2$ is a subring because the zero element exists in both, and closure under subtraction and multiplication is maintained within the intersection.

Integral Domains

The transition from general rings to integral domains marks a shift toward reclaiming the “Cancellation Law”, that elegant property of integers where common factors can be eliminated from equations. In many abstract rings, this property is absent due to the existence of “zero divisors,” which we must catalog and understand to appreciate the stability of more restricted domains.

While our intuition from \mathbb{Z} and \mathbb{R} suggests that a product can only be zero if at least one factor is zero, abstract ring theory presents us with various “pathological” counter-examples.

Consider $\mathbb{Z}_6$, in this system, the interaction of composite factors yields

$$[2] \cdot [3] = [6] = [0] and [4] \cdot [3] = [12] = [0]$$

Thus, $[2], [3]$, and $[4]$ are identified as zero divisors.

On the ring $F_{\mathbb{R}}$ consider $f(x) = 1$ for $x \in \mathbb{Q}$ and $0$ for $x \notin \mathbb{Q}$, and $g(x) = 0$ for $x \in \mathbb{Q}$ and $1$ for $x \notin \mathbb{Q}$.

Though neither function is the zero function, their product $(f \cdot g)(x) = 0$ for all $x$, rendering them zero divisors.

Proof by Contradiction

Suppose $1 = 0$. For any nonzero element a \in R,

$$a = a \cdot 1 = a \cdot 0 = 0$$

This forces $a = 0$ for all elements, contradicting the definition of a nontrivial ring.

Familiar systems such as $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ meet this standard. However, $2\mathbb{Z}$ fails this classification; despite being commutative and devoid of zero divisors, it lacks the requisite “unity” (the element $1$).

The utility of a ring for solving equations is dictated by the Cancellation Law of Multiplication.

Proof

Assume $R$ has no zero divisors and let $ab = ac$ with $a \neq 0$. Then $ab - ac = 0$, implying $a(b - c) = 0$.

Since $a \neq 0$ and zero divisors are absent, $b - c$ must be $0$, which yields $b = c$.

Assume the Cancellation Laws hold. Let $ab = 0$. If $a \neq 0$, we observe $ab = a \cdot 0$. By canceling $a$, we find $b = 0$. Thus, $R$ possesses no zero divisors.

Fields

Proof

Let $b$ and $c$ be inverses of $a$. Then

$$b = b \cdot 1 = b(ac) = (ba)c = 1 \cdot c = c$$

This uniqueness is a direct consequence of the associative law.

Case Study: The Field of Complex Numbers ($\mathbb{C}$)

To prove $\mathbb{C}$ is a field, we must derive an inverse for any $x = a + bi \neq 0$.

We seek $y = c + di$ such that $(a+bi)(c+di) = 1 + 0i$. Expanding this product yields the system:

1. $ac - bd = 1$
2. $ad + bc = 0$

To isolate $c$, multiply (1) by a and (2) by $b$:

$$ a^2c - abd = a abd + b^2c = 0 $$

Summing these gives $(a^2 + b^2)c = a$, hence $c = \frac{a}{a^2 + b^2}$.

To isolate $d$, multiply (1) by $-b$ and (2) by $a$:

$$ -abc + b^2d = -b a^2d + abc = 0 $$

Summing these gives $(a^2 + b^2)d = -b$, hence $d = \frac{-b}{a^2 + b^2}$.

Since $a+bi \neq 0$ implies $a^2+b^2 \neq 0$, the inverse

$$ x^{-1} = \frac{a}{a^2 + b^2} + \frac{-b}{a^2 + b^2}i $$

always exists.

The Hierarchical Relationship

Proof

Let $F$ be a field. If $ab = 0$ and $a \neq 0$, we multiply by

$$a^{-1}: a^{-1}(ab) = a^{-1} \cdot 0 \implies (a^{-1}a)b = 0 \implies 1b = 0 \implies b = 0$$

Since $b$ must be zero, $F$ contains no zero divisors.

Summary Table

Ring	Commutative	Unity	No Zero Divisors	Integral Domain	Field
$\mathbb{Z}$	Yes	Yes	Yes	Yes	No
$\mathbb{Q}$	Yes	Yes	Yes	Yes	Yes
$\mathbb{Z}_n$ (composite)	Yes	Yes	No	No	No
$\mathbb{Z}_p$ (prime)	Yes	Yes	Yes	Yes	Yes
$2\mathbb{Z}$	Yes	No	Yes	No	No