Problem 1 - A Book of Abstract Algebra

Question

Prove that each of the following is a subring of the indicated ring. \(\\{x+\sqrt{3} y: x, y \in \mathbb{Z}\\}\) is a subring of \(\mathbb{R}\)

Step-by-Step Solution

Answer

The given set is a subring of \(\mathbb{R}\) because it satisfies all subring criteria.

Step 1: Understand a Subring

To prove that a set is a subring, you need to demonstrate three things: it includes the additive identity, it is closed under addition and multiplication, and it contains the additive inverses.

Step 2: Identify Elements and Operation

The set given is \(\{x + \sqrt{3}y : x, y \in \mathbb{Z}\}\). This is a subset of the real numbers \(\mathbb{R}\), where each element has the form \(x + \sqrt{3}y\). Here, both \(x\) and \(y\) are integers.

Step 3: Check Additive Identity

The additive identity in \(\mathbb{R}\) is 0. We find that if we choose \(x = 0\) and \(y = 0\), we get \(0 + \sqrt{3} \times 0 = 0\), which is in the set. Hence, the set includes the additive identity.

Step 4: Verify Closure Under Addition

Choose two elements \(a = x_1 + \sqrt{3}y_1\) and \(b = x_2 + \sqrt{3}y_2\) from the set. Their sum is \(a + b = (x_1 + x_2) + \sqrt{3}(y_1 + y_2)\). Since both \(x_1 + x_2\) and \(y_1 + y_2\) are integers, the sum is also in the set. Thus, the set is closed under addition.

Step 5: Verify Closure Under Multiplication

Take two elements, \(a = x_1 + \sqrt{3}y_1\) and \(b = x_2 + \sqrt{3}y_2\). Their product is \((x_1 + \sqrt{3}y_1)(x_2 + \sqrt{3}y_2) = x_1x_2 + 3y_1y_2 + \sqrt{3}(x_1y_2 + x_2y_1)\). The terms \(x_1x_2 + 3y_1y_2\) and \(x_1y_2 + x_2y_1\) are both integers, so the product is in the set, showing closure under multiplication.

Step 6: Check Additive Inverses

For an element \(a = x + \sqrt{3}y\), its additive inverse is \(-a = -x - \sqrt{3}y\). Since \(-x\) and \(-y\) are integers, \(-a\) is also in the set. Therefore, the set contains the additive inverse of each of its elements.

Step 7: Final Step: Conclusion

Since the set \(\{x + \sqrt{3}y : x, y \in \mathbb{Z}\}\) contains the additive identity, is closed under addition and multiplication, and includes the additive inverses, it fulfills all the criteria to be a subring of \(\mathbb{R}\).

Key Concepts

Real Numbers

Real numbers are an essential concept in mathematics. This vast set includes all rational and irrational numbers. Real numbers can be represented on a number line and comprise numbers like \( -2, 0, 3.5, \sqrt{2}, \ ext{and} \pi \).
Real numbers have some critical properties:

• Ordered: You can compare two real numbers to determine which is larger.
• Complete: There are no gaps on the real number line.
• Dense: Between any two real numbers, there is another real number.

Understanding real numbers is foundational for more advanced mathematical concepts, such as rings and fields.

Closure Under Addition

Closure under addition is an essential property for a subring. It means that if you take any two elements from the set and add them, their sum will also be in that set.
Let's consider the set \(\{x + \sqrt{3}y : x, y \in \mathbb{Z}\}\). If we choose two elements like \(a = x_1 + \sqrt{3}y_1\) and \(b = x_2 + \sqrt{3}y_2\), their sum is

• \(a + b = (x_1 + x_2) + \sqrt{3}(y_1 + y_2)\).
• The integers \(x_1 + x_2\) and \(y_1 + y_2\) ensure that the result is still of the form \(x + \sqrt{3}y\).

Thus, the set remains closed under addition, which is a requirement to qualify as a subring.

Closure Under Multiplication

Closure under multiplication is another significant property, and it's necessary for a subring. It implies that the product of any two elements from the set is also within the same set.
Consider two elements \(a = x_1 + \sqrt{3}y_1\) and \(b = x_2 + \sqrt{3}y_2\) from the set. Their product is calculated as:

• \((x_1 + \sqrt{3}y_1)(x_2 + \sqrt{3}y_2) = x_1x_2 + 3y_1y_2 + \sqrt{3}(x_1y_2 + x_2y_1)\).
• The expressions \(x_1x_2 + 3y_1y_2\) and \(x_1y_2 + x_2y_1\) are integers, which means the product belongs to the set.

Hence, closure under multiplication is satisfied, reinforcing that this set is indeed a subring.

Additive Identity

The additive identity in mathematics is the number 0. It is a fundamental requirement for subrings since the set must always contain this element.
In our example, the set is \(\{x + \sqrt{3}y : x, y \in \mathbb{Z}\}\). To find if the set contains the additive identity:

• We choose \(x = 0\) and \(y = 0\).
• The result is \(0 + \sqrt{3} \times 0 = 0\).

This verifies that the additive identity 0 is included in the set, satisfying one more condition for the set to be a subring. The additive identity plays a key role in maintaining the structure of the set in operation with addition.