Problem 4 - A Book of Abstract Algebra

Question

In each of the following, a set \(A\) with operations of addition and multiplication is given. Prove that \(A\) satisfies all the axioms to be a commutative ring with unity. Indicate the zero element, the unity, and the negative of an arbitrary \(a\). \(A=\\{x+y \sqrt{2}: x, y \in \mathbb{Z}\\}\) with conventional addition and multiplication.

Step-by-Step Solution

Answer

The set \(A\) is a commutative ring with unity, with 0 as the additive identity and 1 as the multiplicative identity.

Step 1: Verify Closure Under Addition

To show that A is closed under addition, take two elements from A, say \( a = x_1 + y_1 \sqrt{2} \) and \( b = x_2 + y_2 \sqrt{2} \), where \( x_1, y_1, x_2, y_2 \in \mathbb{Z} \). Adding them gives: \( a + b = (x_1 + y_1 \sqrt{2}) + (x_2 + y_2 \sqrt{2}) = (x_1 + x_2) + (y_1 + y_2)\sqrt{2} \). Both \( x_1 + x_2 \) and \( y_1 + y_2 \) are integers, confirming that \( a + b \in A \). Thus, A is closed under addition.

Step 2: Verify Associativity of Addition

Associativity requires \((a + b) + c = a + (b + c)\) for all \(a, b, c \in A\). Using \( a = x_1 + y_1 \sqrt{2} \), \( b = x_2 + y_2 \sqrt{2} \), \( c = x_3 + y_3 \sqrt{2} \), calculate \((a + b) + c = [(x_1 + x_2) + (y_1 + y_2)\sqrt{2}] + c = [(x_1 + x_2) + x_3] + [(y_1 + y_2) + y_3]\sqrt{2} \), and \(a + (b + c) = a + [(x_2 + x_3) + (y_2 + y_3)\sqrt{2}] = [x_1 + (x_2 + x_3)] + [y_1 + (y_2 + y_3)]\sqrt{2} \). Both expressions simplify to the same result, confirming associativity.

Step 3: Verify Commutativity of Addition

We need to show \( a + b = b + a \) for any \( a, b \in A \). From earlier, \( a + b = (x_1 + x_2) + (y_1 + y_2)\sqrt{2} \) and \( b + a = (x_2 + x_1) + (y_2 + y_1)\sqrt{2} \). Since integer addition is commutative, \((x_1 + x_2) = (x_2 + x_1)\) and \((y_1 + y_2) = (y_2 + y_1)\), confirming addition is commutative in A.

Step 4: Identify Additive Identity

The additive identity \( 0 \) in A must satisfy \( a + 0 = a \) for all \( a \in A \). In terms of given form \( x + y \sqrt{2} \), this identity is \( 0 + 0 \sqrt{2} = 0 \). Adding \(0\) to any element \( x + y \sqrt{2} \), we get the same element, satisfying this property. Thus, \( 0 \) is the additive identity.

Step 5: Identify Additive Inverses

For any element \( a = x + y \sqrt{2} \), its additive inverse must satisfy \( a + (-a) = 0 \). The inverse is \( -x - y \sqrt{2} \), since \( (x + y \sqrt{2}) + (-x - y \sqrt{2}) = 0 \). As \(-x\) and \(-y\) are integers, the inverse is also an element of A.

Step 6: Verify Closure Under Multiplication

For closure under multiplication, consider \( a = x_1 + y_1 \sqrt{2} \) and \( b = x_2 + y_2 \sqrt{2} \). The product is \( a \times b = (x_1 + y_1 \sqrt{2})(x_2 + y_2 \sqrt{2}) = x_1x_2 + y_1x_2\sqrt{2} + x_1y_2\sqrt{2} + y_1y_2(\sqrt{2})^2 \). Simplifying gives \( x_1x_2 + 2y_1y_2 + (y_1x_2 + x_1y_2)\sqrt{2} \). As both coefficients are integers, \( a \times b \in A \).

Step 7: Verify Associativity of Multiplication

For associativity, show \( (a \times b) \times c = a \times (b \times c) \). Using associative properties of integers and the distributive nature of multiplication, both sides evaluate to \( x_1(x_2x_3 + 2y_2y_3) + 2y_1(y_2x_3 + x_2y_3) + 2y_1y_2(x_3 + y_3\sqrt{2})\sqrt{2} \). Hence, multiplication is associative in A.

Step 8: Verify Commutativity of Multiplication

To show commutativity, confirm \( a \times b = b \times a \) for any \( a, b \in A \). Since \( a \times b = x_1x_2 + 2y_1y_2 + (x_1y_2 + y_1x_2)\sqrt{2} \), and \( b \times a \) results in the same expression due to the commutative property of integers, multiplication is commutative.

Step 9: Identify Multiplicative Identity (Unity)

The unity in A is an element such that \( a \times 1 = a \) for all \( a \in A \). Observing \( 1 + 0 \sqrt{2} \) acts as this identity, since \( (x + y \sqrt{2}) \times 1 = x + y \sqrt{2} \). Thus, \( 1 \) is the multiplicative identity.

Step 10: Check Distributive Law

Verify \( a \times (b + c) = a \times b + a \times c \). Expanding both sides with the known form confirms they yield the same value by distributing multiplication over addition, hence satisfying the distributive law.

Key Concepts

Closure Under Addition

To understand closure under addition in the context of a commutative ring, consider two elements from the set in question. Let's denote these elements as \( a = x_1 + y_1 \sqrt{2} \) and \( b = x_2 + y_2 \sqrt{2} \), where \( x_1, y_1, x_2, \text{ and } y_2 \in \mathbb{Z} \). Adding these elements results in:

• \( a + b = (x_1 + y_1 \sqrt{2}) + (x_2 + y_2 \sqrt{2}) = (x_1 + x_2) + (y_1 + y_2)\sqrt{2} \).

For closure, the sum \((x_1 + x_2)\) and \((y_1 + y_2)\sqrt{2}\) must also be elements of the same set. In this case, because integer addition is closed in \( \mathbb{Z} \), both the sum of terms \( (x_1 + x_2) \text{ and } (y_1 + y_2) \) result in integers. Thus, the resulting expression remains in the set \( A \). This illustrates the principle that the set must remain intact under its operation—in this case, addition—ensuring every possible addition yields a result within the set, maintaining its integrity.

Associative Property

The associative property is essential in ensuring operations are consistent regardless of how they are grouped. For addition within a ring, the property states that for any elements \( a, b, \) and \( c \) in set \( A \), the equation \((a + b) + c = a + (b + c)\) holds. Let's break this down using similar elements like \( a = x_1 + y_1 \sqrt{2}, b = x_2 + y_2 \sqrt{2}, \) and \( c = x_3 + y_3 \sqrt{2} \). Substituting these into the expression gives:

• \((a + b) + c = [(x_1 + x_2) + x_3] + [(y_1 + y_2) + y_3]\sqrt{2} \)
• \(a + (b + c) = [x_1 + (x_2 + x_3)] + [y_1 + (y_2 + y_3)]\sqrt{2} \)

Since addition of integers is associative, both expressions reduce to the same result. This consistency regardless of grouping in the operation proves the associative property of addition is satisfied within the set. It assures the stability of calculations and correctness regardless of sequence.

Distributive Law

The distributive law connects addition and multiplication in a ring, ensuring that multiplication spreads over addition appropriately. The law requires that for any elements \( a, b, \) and \( c \) in \( A \), the equation \( a \times (b + c) = a \times b + a \times c \) holds true. Using expressions of form \( a = x_1 + y_1 \sqrt{2}, b = x_2 + y_2 \sqrt{2}, \) and \( c = x_3 + y_3 \sqrt{2} \), we compute:

• Left side: \( a \times (b + c) = (x_1 + y_1 \sqrt{2}) \times [(x_2 + x_3) + (y_2 + y_3)\sqrt{2}] \)
• Right side: \( a \times b + a \times c = (x_1 + y_1 \sqrt{2})(x_2 + y_2 \sqrt{2}) + (x_1 + y_1 \sqrt{2})(x_3 + y_3 \sqrt{2}) \)

Upon solving, both sides simplify to the same expression: \( x_1(x_2 + x_3) + y_1(y_2 + y_3) \cdot 2 + (x_1(y_2 + y_3) + y_1(x_2 + x_3))\sqrt{2} \). This demonstrates that multiplication distributes over addition consistently within the set, a key axiom characterizing rings.

Additive Identity

In every commutative ring, an element acts as the additive identity, essentially leaving any element unchanged when added to it. For a set to qualify as a commutative ring, it must include such an element — commonly \( 0 \). In our context, consider any element \( a = x + y \sqrt{2} \), where the additive identity is \( 0 \), represented as \( 0 + 0 \sqrt{2} \) in this algebraic structure. The principle can be shown as:

• \( a + 0 = (x + y \sqrt{2}) + (0 + 0 \sqrt{2}) = x + y \sqrt{2} = a \)

Here, the identity \( 0 \) successfully combines with any element \( a \) leaving it unchanged, satisfying the additive identity condition. This confirms that the set includes an additive identity, crucial for simplifying expressions and solving equations within the ring.

Multiplicative Identity

A multiplicative identity in a commutative ring is an element that leaves others unchanged when used as a multiplier. Commonly, this identity is \( 1 \). Within our set, it is crucial to identify this element to verify the ring structure. For an arbitrary element \( a = x + y \sqrt{2} \), the multiplicative identity is \( 1 \), which can be written in formulation as \( 1 + 0 \sqrt{2} \). The property to verify is:

• \( a \times 1 = (x + y \sqrt{2}) \times (1 + 0 \sqrt{2}) = x + y \sqrt{2} = a \)

Successfully, this confirms that \( 1 \) acts as the multiplicative identity, keeping any element intact when multiplied. The presence of a multiplicative identity reinforces the structure necessary for multiplicative operations, ensuring arithmetic coherency within any ring system.