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Chapter 16: Problem 16

If \(R\) is a field, show that the only two ideals of \(R\) are \(\\{0\\}\) and \(R\) itself.

Short Answer

Expert verified
Question: Prove that the only two ideals of a field \(R\) are \(\{0\}\) and \(R\) itself. Solution: We first showed that \(\{0\}\) and \(R\) are indeed ideals of \(R\). Then, we considered any ideal \(I\) distinct from \(\{0\}\) and \(R\) itself. We demonstrated that \(I\) must contain the element 1, and since \(I\) is an ideal, this implies that every element in \(R\) is contained in \(I\). Thus, \(I = R\). Therefore, the only two ideals of a field \(R\) are \(\{0\}\) and \(R\) itself.

Step by step solution

01

Recall Definitions: Field and Ideal

A field is a commutative ring with unity in which every nonzero element has an inverse. An ideal of a ring \(R\) is a nonempty subset \(I\) of \(R\) such that for any \(a, b \in I\), \(a-b \in I\) and for any \(a \in I\) and \(r \in R\), \(ra \in I.\)
02

Showing {0} is an Ideal

To show that \(\{0\}\) is an ideal, we first observe that it is a nonempty subset of \(R\). Now, consider any \(a, b \in \{0\}\). Since the only element in \(\{0\}\) is \(0\), then \(a = b = 0\). Therefore, \(a - b = 0 - 0 = 0 \in \{0\}\). Additionally, for any \(r \in R\), \(r \cdot a = r \cdot 0 = 0 \in \{0\}\). So, \(\{0\}\) is indeed an ideal of \(R\).
03

Showing R is an Ideal

To show that \(R\) is an ideal, we observe that it is a nonempty subset of itself. Now, consider any \(a, b \in R\). Since \(R\) is a commutative ring, \(a - b \in R\). Additionally, for any \(r \in R\), \(r \cdot a \in R\). So, \(R\) is indeed an ideal of itself.
04

Proving that {0} and R are the only two ideals

Let \(I\) be an ideal in \(R\) distinct from the zero ideal \(\{0\}\) and \(R\) itself. Since \(I \neq \{0\}\), there must exist an element \(a \in I\) such that \(a \neq 0\). Now, recall that since \(R\) is a field, every nonzero element has a multiplicative inverse. Meaning, there exists an element \(a^{-1} \in R\) such that \(a \cdot a^{-1} = 1\). Now, as \(I\) is an ideal, for any element \(r \in R\) and \(a \in I\), \(ra \in I\). Therefore, \(a^{-1} \cdot a = 1 \in I\). Now, take any element \(b \in R\), then \(1 \cdot b = b \in I\). Since \(b\) is an arbitrary element from \(R\), we have shown that every element of \(R\) is contained in \(I\). Thus, \(I = R\). Therefore, the only two ideals of a field \(R\) are \(\{0\}\) and \(R\) itself.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Field Theory
Field theory explores the fascinating world of fields within abstract algebra, a crucial area that serves as the foundation of modern mathematics. A field is a set equipped with two operations鈥攁ddition and multiplication鈥攆ollowing specific rules. These rules ensure that a field is essentially an environment where addition, subtraction, multiplication, and division (except by zero) are all possible.

For instance, the set of real numbers forms a field, as does the set of complex numbers. To better understand the behavior of ideals in fields, let's recall that a field is a special kind of ring, one that requires every nonzero element to have a multiplicative inverse. This key characteristic of fields heavily influences the structure of their ideals.

In the context of the given exercise, we see that in a field, there are only two possible ideals. Recognizing the concepts from field theory gives us the prerequisite knowledge to fully comprehend why this is true, emphasizing the uniqueness and restricted nature of ideals in this setting.
Ring Theory
Diving into ring theory, we encounter rings which are algebraic structures combining elements under addition and multiplication. Rings are more general than fields because they don't require the existence of multiplicative inverses for every nonzero element.

An ideal within a ring is a subgroup that absorbs multiplication by any element in the ring, thus maintaining its structure under both ring operations. Ideals play a critical role in ring theory because they are the building blocks for constructing quotient rings, much like how normal subgroups are used to build quotient groups.

The original exercise illustrates a fundamental fact about ideals in the context of a field: due to the field's defining property that every nonzero element must have an inverse, the behavior of ideals is greatly restricted, leading to the intriguing result that only trivial ideals exist.
Commutative Algebra
Commutative algebra is a branch of algebra that studies commutative rings, their ideals, and modules over them. This field goes a step beyond ring theory by focusing on rings where the order in which you multiply matters not鈥攖he multiplication operation is commutative.

Fields are a subset of commutative rings, where not only is multiplication commutative, but every nonzero element has a reciprocal. This restriction is so significant that when we discuss ideals in a field, the structure of commutative algebra constrains them to the two simplest forms: the zero ideal and the field itself, as seen in our exercise. The commutative nature of multiplication in fields reinforces why any non-zero ideal must necessarily be the whole field.
Abstract Algebra
Abstract algebra is a vast and generalized study of algebraic structures such as groups, rings, and fields. It's here that we develop the language and intuition for understanding how different algebraic systems behave. The exercise provided is a classic result in the study of abstract algebra, showcasing the distinctive properties of ideal structure in fields.

While discussing ideals in a field, we don't just depend on the properties of the field itself, but we're also invoking foundational notions from abstract algebra. This problem reveals the underlying unifying themes of abstract algebra: the interactions between the elements of a structure and the operational laws that govern them dictate the overall behavior of the system. This highlights the limited nature of ideal diversity in fields and contrasts it with the rich structure of ideals in more general rings.

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Most popular questions from this chapter

Prove that \(\mathbb{Z}[\sqrt{3} i]=\\{a+b \sqrt{3} i: a, b \in \mathbb{Z}\\}\) is an integral domain.

Let \(R\) be a commutative ring. An element \(a\) in \(R\) is nilpotent if \(a^{n}=0\) for some positive integer \(n\). Show that the set of all nilpotent elements forms an ideal in \(R\).

Let \(R\) be the ring of \(2 \times 2\) matrices of the form $$ \left(\begin{array}{ll} a & b \\ 0 & 0 \end{array}\right) $$ where \(a, b \in \mathbb{R}\). Show that although \(R\) is a ring that has no identity, we can find a subring \(S\) of \(R\) with an identity.

Prove the Correspondence Theorem: Let \(I\) be an ideal of a ring \(R\). Then \(S \rightarrow S / I\) is a one-to-one correspondence between the set of subrings \(S\) containing \(I\) and the set of subrings of \(R / I\). Furthermore, the ideals of \(R\) correspond to ideals of \(R / I\).

Let \(a\) be any element in a ring \(R\) with identity. Show that \((-1) a=-a .\)
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