Problem 3 - 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 \(a\). \(A\) is the set \(\mathbb{Q} \times \mathbb{Q}\) of ordered pairs of rational numbers, and the operations are the following addition \(\oplus\) and multiplication \(\odot:\) $$ \begin{aligned} &(a, b) \oplus(c, d)=(a+c, b+d) \\ &(a, b) \odot(c, d)=(a c-b d, a d+b c) \end{aligned} $$

Step-by-Step Solution

Answer

Set \(A\) is a commutative ring with unity \((1, 0)\), zero \((0, 0)\), and negative \((-a, -b)\) for \((a, b)\).

Step 1: Closure Property

For a set to be a ring, it must be closed under the operations of addition and multiplication. For addition, given any two elements \((a, b), (c, d)\) in \(A = \mathbb{Q} \times \mathbb{Q}\), the sum is \((a + c, b + d)\), which is also an element of \(\mathbb{Q} \times \mathbb{Q}\). For multiplication, the result of the operation \((ac - bd, ad + bc)\) is again in \(\mathbb{Q} \times \mathbb{Q}\) since it consists of rational numbers.

Step 2: Associativity of Addition and Multiplication

We need to check if the operations are associative. For addition:\[((a, b) \oplus (c, d)) \oplus (e, f) = (a+c, b+d) \oplus (e, f) = (a+c+e, b+d+f)\]\[(a, b) \oplus ((c, d) \oplus (e, f)) = (a, b) \oplus (c+e, d+f) = (a+c+e, b+d+f)\]Both are equal. Similarly, check associativity for multiplication:\[((a, b) \odot (c, d)) \odot (e, f) = (ac-bd, ad+bc) \odot (e, f) = ((ac-bd)e - (ad+bc)f, (ac-bd)f + (ad+bc)e)\]\[(a, b) \odot ((c, d) \odot (e, f)) = (a, b) \odot (ce-df, cf+de) = (a(ce-df) - b(cf+de), a(cf+de) + b(ce-df))\]With simplification, both are equal, verifying associativity.

Step 3: Commutativity of Addition and Multiplication

For a commutative ring, both operations must be commutative. For addition:\[(a, b) \oplus (c, d) = (a+c, b+d) = (c+a, d+b) = (c, d) \oplus (a, b)\]For multiplication:\[(a, b) \odot (c, d) = (ac-bd, ad+bc) = (ca-db, cb+da) = (c, d) \odot (a, b)\]Indeed, both addition and multiplication are commutative.

Step 4: Existence of Additive Identity (Zero Element)

The additive identity or zero element must satisfy \((a, b) \oplus (0_A) = (a, b)\) for all \((a, b)\). The pair \((0, 0)\) satisfies:\[(a, b) \oplus (0, 0) = (a+0, b+0) = (a, b)\]Thus, \((0, 0)\) is the additive identity.

Step 5: Existence of Additive Inverses (Negative)

For every element \((a, b)\), there exists an element \((-a, -b)\) such that:\[(a, b) \oplus (-a, -b) = (a-a, b-b) = (0, 0)\]Hence, \((-a, -b)\) is the additive inverse.

Step 6: Existence of Multiplicative Identity (Unity)

The multiplicative identity or unity must satisfy \((a, b) \odot (1_A) = (a, b)\) for all \((a, b)\). The pair \((1, 0)\) satisfies:\[(a, b) \odot (1, 0) = (a\cdot1 - b\cdot0, a\cdot0 + b\cdot1) = (a, b)\]Thus, \((1, 0)\) is the multiplicative identity.

Key Concepts

Closure Property

The closure property is a fundamental concept in algebra, ensuring that applying an operation on members of a set will result in a member of the same set. For instance, in the context of a commutative ring, we must verify that both addition and multiplication operations keep results within the set.

Consider the set \(A = \mathbb{Q} \times \mathbb{Q}\), which represents ordered pairs of rational numbers. When two elements \((a, b)\) and \((c, d)\) undergo the given operation of addition \((a, b) \oplus (c, d) = (a+c, b+d)\), the resulting pair \((a+c, b+d)\) also lies within \(\mathbb{Q} \times \mathbb{Q}\). This confirms the closure of \(A\) under addition.

For multiplication, \((a, b) \odot (c, d) = (ac-bd, ad+bc)\) results in pairs like \((ac-bd, ad+bc)\), which remain ordered pairs of rational numbers. Therefore, \(A\) is closed under multiplication too. Closure assures us that our operations will not leave the set, maintaining its integrity.

Associative Property

The associative property is all about the way in which operands are grouped during an operation. In a ring, both addition and multiplication need to satisfy this property to ensure consistent results regardless of grouping.

For addition, consider how grouping affects the sum when adding more than two elements. Let's take three pairs \((a, b)\), \((c, d)\), and \((e, f)\):

• \(((a, b) \oplus (c, d)) \oplus (e, f) = (a+c+e, b+d+f)\)
• \((a, b) \oplus ((c, d) \oplus (e, f)) = (a+c+e, b+d+f)\)

Both expressions simplify to the same result, proving that addition is associative in \(A\).

Similarly, for multiplication, consider:

• \(((a, b) \odot (c, d)) \odot (e, f)\) and
• \((a, b) \odot ((c, d) \odot (e, f))\)

Both these operations simplify and result in the same pair due to the properties of multiplication, hence multiplication is associative as well. Associativity ensures the outcome is consistent no matter how the elements are grouped.

Commutative Property

The commutative property describes how operands can be switched without affecting the outcome of an operation. In a commutative ring, both addition and multiplication must be commutative.

For addition, if you have two pairs \((a, b)\) and \((c, d)\), the order of addition doesn't matter:

• \((a, b) \oplus (c, d) = (a+c, b+d)\)
• \((c, d) \oplus (a, b) = (c+a, d+b)\)

Both these expressions give the same result, demonstrating commutativity in addition.

For multiplication, it works in a similar way. Using the same pairs, the results of the operations are:

• \((a, b) \odot (c, d) = (ac-bd, ad+bc)\)
• \((c, d) \odot (a, b) = (ca-db, cb+da)\)

These results are identical through simplification, confirming that multiplication is commutative. The commutative property allows flexibility and interchangeability in operations, a key feature of ease in computations.

Additive Identity

The additive identity is an element which, when added to any other element in the set, returns the same element. It's like the effect of adding zero in regular arithmetic.

In the commutative ring \(A = \mathbb{Q} \times \mathbb{Q}\), the additive identity is the pair \((0, 0)\). Why? Because when you add \((0, 0)\) to any element \((a, b)\), you get:

• \((a, b) \oplus (0, 0) = (a, b)\)

This simple operation leaves \((a, b)\) unchanged, confirming \((0, 0)\) as the additive identity. It's an essential component of the ring, ensuring that there is always a 'zero-like' element that doesn't affect other elements when added.

Multiplicative Identity

The multiplicative identity is an element that, when used in multiplication with any other element of the set, yields that same element. It's akin to multiplying by one in normal arithmetic.

In our commutative ring \(A = \mathbb{Q} \times \mathbb{Q}\), the multiplicative identity is represented by the pair \((1, 0)\). This is shown by multiplying any element \((a, b)\) by \((1, 0)\):

• \((a, b) \odot (1, 0) = (a\cdot1 - b\cdot0, a\cdot0 + b\cdot1) = (a, b)\)

This confirms the pair \((1, 0)\) acts as a one-like element, thus serving as the multiplicative identity. It's crucial for the ring to function as it maintains the integrity of other elements when involved in multiplication.