Question

This exercise explores some examples of prime and maximal ideals. Show that: (a) in the ring \(\mathbb{Z},\) the ideal \\{0\\} is prime but not maximal, and that the maximal ideals are precisely those of the form \(p \mathbb{Z},\) where \(p\) is prime; (b) in an integral domain \(D,\) the ideal \\{0\\} is prime, and this ideal is maximal if and only if \(D\) is a field; (c) if \(p\) is a prime, then in the ring \(\mathbb{Z}[X],\) the ideal \((X, p)\) is maximal, while the ideals \((X)\) and \((p)\) are prime, but not maximal; (d) if \(F\) is a field, then in the ring \(F[X, Y],\) the ideal \((X, Y)\) is maximal, while the ideals \((X)\) and \((Y)\) are prime, but not maximal.

Short answer

ns the element 1, and therefore \(I\) must be equal to \(\mathbb{Z}\). Hence, \((p)\) is maximal. Now suppose that \(I\) is a maximal ideal in \(\mathbb{Z}\). Let \(a \in I\) be a non-zero element with the smallest absolute value. Since \(I\) is a proper ideal, \(a \neq 1\). Let \(p\) be a prime divisor of \(a\). Then, there exist \(b, c \in \mathbb{Z}\) such that \(a = pb\). Since \(a \in I\) and \(I\) is an ideal, \(pb \in I\). By the definition of prime ideals, this implies that either \(p \in I\) or \(b \in I\). If \(b \in I\), then \(\gcd(a, p) = 1\), because \(p\) is prime. Therefore, there exist \(x, y \in \mathbb{Z}\) such that \(ax + py = 1\). Since both \(a\) and \(p\) are in \(I\), and \(I\) is an ideal, \(1 \in I\), which contradicts the fact that the ideal is maximal. Thus, we must have \(p \in I\). Now, let \(J = (p) = \{pn | n \in \mathbb{Z}\}\). Since \(p \in I\), \(J \subseteq I\). Furthermore, \(I\neq J\), because we assumed \(I\) is maximal. If we had \(I\) properly contained in \(J\), that would contradict the maximality of \(I\). Thus, \(I = J\), and the maximal ideal \(I\) is of the form \(p \mathbb{Z}\), where \(p\) is prime. #Phase 2: Investigating {0} as a Prime and Maximal Ideal in an Arbitrary Integral Domain# #tag_title#Part (b): In an integral domain \(D\), show that the ideal {0} is prime#tag_content#Let \(D\) be an integral domain, and let \(a, b \in D\) such that \(ab \in \{0\}\), which means that \(ab = 0\). Since \(D\) is an integral domain, it has no zero divisors, so if \(ab = 0\), then either \(a = 0\) or \(b = 0\). Therefore, at least one of \(a\) and \(b\) is an element of \(\{0\}\), and \(\{0\}\) is a prime ideal in \(D\). #tag_title#Part (b): Show that the ideal {0} is maximal in \(D\) if and only if \(D\) is a field#tag_content#Suppose \(D\) is a field. Let \(I\) be an ideal in \(D\) such that \(\{0\} \subseteq I\). If \(I\) contains a non-zero element \(a\), then since \(D\) is a field, the element \(a\) must have an inverse \(a ^{-1}\in D\), and thus \((a)(a^{-1}) = 1 \in I\). Since \(I\) contains the identity element, it must be equal to the entire field \(D\). Thus, in a field, the only ideal containing \(\{0\}\) is \(D\) itself, which makes \(\{0\}\) a maximal ideal. Now, suppose that \(\{0\}\) is maximal in \(D\). To show that \(D\) is a field, we must demonstrate that every non-zero element in \(D\) has a multiplicative inverse. Let \(a \in D\) be a non-zero element. Then, the principal ideal generated by \(a\), denoted \((a)\), is a proper ideal since \(a \neq 0\) and \((a) \subseteq D\). Since the ideal \(\{0\}\) is maximal, any ideal that contains \(\{0\}\) and is not equal to \(\{0\}\) must be equal to the entire domain \(D\). Thus, \((a) = D\). Since \((a) = D\), there exists an element \(b \in D\) such that \(ab = 1\), which means that \(a\) has a multiplicative inverse \(a^{-1} = b\). Therefore, every non-zero element in \(D\) has a multiplicative inverse, and \(D\) is a field. #Phase 3 and 4: Investigating Prime and Maximal Ideals in Polynomial Rings# #tag_title#Parts (c) and (d): Prime and Maximal Ideals in \(\mathbb{Z}[X]\) and\(F[X, Y]\)#tag_content#For parts (c) and (d), we can use a general approach based on the properties of prime and maximal ideals in polynomial rings. The key idea is that prime ideals correspond to irreducible polynomials, and maximal ideals correspond to prime elements and irreducible polynomials in the coefficient rings. For example, in part (c), we can prove that ideals such as \((X)\) or \((p, X)\), where \(p\) is a prime number, are prime or maximal in \(\mathbb{Z}[X]\) by showing that the respective polynomials are irreducible or that their coefficients are prime elements in \(\mathbb{Z}\). Similarly, in part (d), we can study the prime and maximal properties of ideals like \((X, Y)\) in \(F[X, Y]\) by investigating the irreducibility of the polynomials \(X\) and \(Y\) or by studying the properties of the coefficients in the field \(F\). In both cases, we may also apply the correspondences between prime (maximal) ideals and irreducible (prime) polynomials in factorization domains, helping us deduce the properties we're looking for.

Step-by-step solution

Part (a): In \(\mathbb{Z}\), show that the ideal \\{0\\} is prime but not maximal

In \(\mathbb{Z}\), if \(a\) and \(b\) are integers such that \(ab \in \{0\}\), then \(ab = 0\). By the definition of integers, this implies that either \(a = 0\) or \(b = 0\). So at least one of \(a\) and \(b\) is an element of \\{0\\}. Thus, \\{0\\} is a prime ideal. To show that \\{0\\} is not maximal, consider the ideal \((2) = \\{2n | n \in \mathbb{Z}\\}\). We can see that \\{0\\} \(\subset (2)\), but \((2)\) is not the whole ring, because it doesn't contain odd integers. Thus, the \\{0\\} ideal is not maximal.

Part (a): Show that maximal ideals are precisely those of the form \(p \mathbb{Z}\), where \(p\) is prime

Let \(p\) be a prime number. The ideal \(p \mathbb{Z}\) contains all multiples of \(p\). Suppose there is an ideal \(I\) such that \((p) \subset I \subset \mathbb{Z}\). Let \(a \in I\) be an integer not divisible by \(p\). Since \(a\) and \(p\) are relatively prime, there exist integers \(x, y\) such that \(ax + py = 1\). This shows that \(I\) contai...

Key concepts

Ring Theory

Ring theory is a fascinating area in abstract algebra that focuses on rings, which are sets equipped with two binary operations: addition and multiplication. For a set to be considered a ring, it must satisfy several conditions:

• The set must be an abelian group under addition.
• Multiplication must be associative.
• The set must have a multiplicative identity, often denoted as 1, even though not all rings require this.
• Distributive laws must hold: for any elements \(a, b,\) and \(c\) in the ring, \(a(b + c) = ab + ac\) and \((b + c)a = ba + ca\).

Rings can be commutative or non-commutative. In a commutative ring, the order of multiplication does not affect the result (i.e., \(ab = ba\)).

Prime and maximal ideals are special subsets of rings. These ideals play a central role in understanding the structure of the ring. For instance, the ring of integers \(\mathbb{Z}\) exhibits interesting properties regarding these ideals, showcasing foundational aspects of ring theory.

Integral Domain

An integral domain is a special kind of ring with additional properties. It can be seen as a stepping stone between general rings and fields. Here are the key characteristics of an integral domain:

• It's a commutative ring with unity (an element that acts as a multiplicative identity).
• It has no zero divisors. This means if the product of two elements is zero, then at least one of the elements must be zero.

These properties ensure that integral domains have a structure closely resembling the integers. For example, within an integral domain, the zero ideal \(\{0\}\) is always a prime ideal. This ties back to the concept of prime numbers in integers. They are the building blocks, similar to how prime ideals function within rings. An integral domain becomes a field if the only ideals it contains are the zero ideal and the whole domain itself.

This provides an elegant connection between the structure of an integral domain and fields, emphasizing how closely related these algebraic structures are.

Field Theory

Fields are one of the most important structures in algebra. They are not just rings but have additional properties that make them ideal for various mathematical applications. Here's what fields are all about:

• Fields are commutative rings with unity.
• Every non-zero element has a multiplicative inverse. This means for any non-zero element \(a\), there exists an element \(b\) such that \(ab = 1\).

Fields are integral domains with the property that you can "divide" by any non-zero element. Examples of fields include the set of rational numbers \(\mathbb{Q}\), real numbers \(\mathbb{R}\), and complex numbers \(\mathbb{C}\).

In fields, all non-zero elements form a group under multiplication. This specific property allows fields to describe a broad range of phenomena in mathematics, from solving polynomial equations to geometry and beyond.

Ideal Theory

Ideal theory explores the structure and classification of ideals within a ring, providing insights into how a ring can be decomposed and analyzed. Here are the key points about ideal theory:

• An ideal is a special subset of a ring, closed under addition and absorbing under multiplication by elements of the ring.
• Prime ideals have the property that if a product of two elements is in the ideal, at least one of those elements must be in the ideal.
• Maximal ideals are those that are as large as possible without being the entire ring, meaning there are no other ideals between them and the ring.

Understanding these concepts helps in determining the factorization within rings. In a ring like \(\mathbb{Z}[X]\) or \(F[X, Y]\), prime and maximal ideals give insights into the algebraic structure and potential divisions of the set.

Learning about ideals holds importance not just for theoretical considerations, but for practical applications in number theory, algebraic geometry, and beyond.