Problem 38 - A Computational Introduction to Number Theory and Algebra

Question

Let \(R\) be a ring. An ideal \(I\) of \(R\) is called prime if \(I \subsetneq R\) and if for all \(a, b \in R, a b \in I\) implies \(a \in I\) or \(b \in I .\) An ideal \(I\) of \(R\) is called maximal if \(I \subsetneq R\) and there are no ideals \(J\) of \(R\) such that \(I \subsetneq J \subsetneq R\). Show that: (a) an ideal \(I\) of \(R\) is prime if and only if \(R / I\) is an integral domain; (b) an ideal \(I\) of \(R\) is maximal if and only if \(R / I\) is a field; (c) all maximal ideals of \(R\) are also prime ideals.

Step-by-Step Solution

Answer

In summary, we have shown that: 1. An ideal I of R is prime if and only if R/I is an integral domain. (Proof for (a) - Part 1 and Part 2) 2. An ideal I of R is maximal if and only if R/I is a field. (Proof for (b) - Part 1 and Part 2) 3. All maximal ideals of R are also prime ideals. (Proof for (c)) These results show the close relationship between prime ideals, maximal ideals, and the quotient ring structures they induce. By understanding these relationships, we can better analyze ring structures and their properties.

Step 1: Proof for (a) - Part 1

Let \(I\) be a prime ideal of the ring \(R\). We want to show that \(R/I\) is an integral domain. In other words, we must prove that \(R/I\) has no zero divisors. Let \(a + I, b + I \in R/I\) such that \((a+I)(b+I) = 0 + I\). This means that \(ab + I = 0 + I\). Notice that \((a+I)(b+I) = ab+I\), so \(ab \in I\). Since \(I\) is a prime ideal, this implies that either \(a \in I\) or \(b \in I\). Consequently, either \(a+I = 0+I\) or \(b+I=0+I\). Therefore, \(R/I\) is an integral domain.

Step 2: Proof for (a) - Part 2

Conversely, let \(R/I\) be an integral domain where \(I \subsetneq R\). We want to show that \(I\) is a prime ideal of \(R\). Let \(a, b \in R\) be such that \(ab \in I\). Then, we have \((a+I)(b+I) = ab + I = 0+ I\) in the quotient ring \(R/I\). Since \(R/I\) is an integral domain, this implies that either \(a+I = 0+I\) or \(b+I = 0+I\). Thus, either \(a \in I\) or \(b \in I\), which is the definition of a prime ideal. Therefore, an ideal \(I\) of \(R\) is prime if and only if \(R / I\) is an integral domain.

Step 3: Proof for (b) - Part 1

Let \(I\) be a maximal ideal of the ring \(R\). We want to show that \(R/I\) is a field. To do this, we need to show that every nonzero element of \(R/I\) has an inverse. Consider an arbitrary nonzero element \(a+I \in R/I\). Since \(I\) is a maximal ideal, there is no ideal \(J\) such that \(I \subsetneq J \subsetneq R\). This means that the ideal \(I+Ra\) (generated by \(I\) and \(a\)) must be equal to \(R\). So, there exists \(r, s \in R\) such that \(1 = r + sa \in I+Ra\). This implies that \(rs + as \in I+Ra\) or \(1 - rs \in I+Ra\). Then, in \(R/I\), we have \((1 - r + I)(s + I) = (1 - rs)+I=1 + I\). Hence, we found the inverse of \(s+I\), which is \(1-r+I\). Since \(a+I\) was arbitrary, this shows that \(R/I\) is a field.

Step 4: Proof for (b) - Part 2

Conversely, let \(R/I\) be a field with \(I \subsetneq R\). Suppose there exists an ideal \(J\) of \(R\) such that \(I \subsetneq J \subsetneq R\). Pick an arbitrary \(a \in J\), such that \(a \notin I\). In the quotient ring \(R/I\), consider the element \(a+I\) that must be nonzero as \(a \notin I\). Since \(R/I\) is a field, there must exist an inverse for \(a+I\), say \((a+I)^{-1} = b+I\). Then, \((a+I)(b+I) = 1+I\), which means \(ab - 1 \in I\). Now, notice that \(ab \in J\) since \(J\) is an ideal. Thus, \(ab - 1 + I = (ab-1) + I = 1+I\) implies that \((ab-1) \in J\). As a result, adding \(1\) to both sides, we get that \(1 \in J\), making \(J=R\). This means that there are no ideals between \(I\) and \(R\), which implies that \(I\) is a maximal ideal. Therefore, an ideal \(I\) of \(R\) is maximal if and only if \(R / I\) is a field.

Step 5: Proof for (c)

We already showed that an ideal \(I\) of \(R\) is maximal if and only if \(R / I\) is a field, and that an ideal of \(R\) is prime if and only if \(R / I\) is an integral domain. Since every field is also an integral domain, every maximal ideal of \(R\) being a field implies that it is also an integral domain. Therefore, all maximal ideals of \(R\) are also prime ideals.

Key Concepts

Ring Theory

Ring Theory is a branch of abstract algebra that deals with the study of rings. Rings are mathematical structures that generalize the concept of arithmetic used in the integers. They have two basic operations: addition and multiplication, and they need to satisfy certain properties for these operations to behave well.

A ring is defined as a set equipped with two binary operations. These are usually denoted as addition (+) and multiplication (·). For a structure to be considered a ring, it must satisfy several key properties:

• Associative Property: Both the addition and multiplication operations must be associative. This means \(a + (b + c) = (a + b) + c\) and \(a(bc) = (ab)c\) for all elements \(a, b, c\) in the ring.
• Distributive Property: Multiplication must distribute over addition. This is expressed as \(a(b + c) = ab + ac\) and \((a + b)c = ac + bc\).
• Identity Elements: Every ring has an additive identity (usually denoted as 0) such that for any element a in the ring, \(a + 0 = a\). Some rings also have a multiplicative identity (usually denoted as 1) such that \(a \cdot 1 = a\).
• Existence of Additive Inverse: For every element in the ring, there must exist another element that, when added, will yield the additive identity (0).

Rings can vary greatly in nature depending on the additional properties they might possess, such as commutativity of multiplication or the existence of a multiplicative inverse for every non-zero element.

Fields

Fields are specific types of rings that exhibit more refined structures, allowing for division besides addition and multiplication. A field is essentially a commutative ring with unity (multiplicative identity is present), where every non-zero element has a multiplicative inverse.

Let's break down the properties that make a field distinct:

• Commutativity: Both addition and multiplication in a field must be commutative. This means that \(a + b = b + a\) and \(a \cdot b = b \cdot a\) for all elements \(a, b\) in the field.
• Existence of Multiplicative Inverses: Every non-zero element \(a\) in a field has an inverse \(b\) such that \(a \cdot b = 1\). This allows for dividing by any non-zero element.
• No Zero Divisors: If the product of two elements is zero, then at least one of the elements must be zero. This property is crucial for ensuring that division is well-defined.

Fields form the foundation of many areas in mathematics. For example, the set of rational numbers \( ext{ℚ}\), real numbers \( ext{ℝ}\), and complex numbers \( ext{ℂ}\) are all familiar examples of fields.

Integral Domains

Integral Domains are types of commutative rings characterized by the absence of zero divisors. This means in an integral domain, if \(a \cdot b = 0\), then either \(a = 0\) or \(b = 0\). This property is crucial as it closely resembles our understanding of integers under multiplication.

Integral Domains have the following core characteristics:

• Commutativity: The multiplication is commutative, i.e., \(a \cdot b = b \cdot a\) for all elements.
• Identity Element: There is a multiplicative identity, usually denoted as 1, such that for any element \(a\), \(a \cdot 1 = a\).
• No Zero Divisors: This is the defining feature. If the product of two elements is zero, then one of those elements has to be zero.

In practice, integral domains are essential for constructing fields. Any integral domain can be used to create a field by forming fractions much like the transition from integers to rational numbers.

For instance, the set of integers \( ext{ℤ}\) is a classic example of an integral domain. By taking fractions \( ext{ℚ}\), we obtain the field of rational numbers.