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Chapter 26: Problem 26

Let \(R\) be a commutative ring and let \(a \in R\). Show that \(I_{a}=\\{x \in R \mid a x=0\\}\) is an ideal of \(R\).

Short Answer

Expert verified
The set \(I_a\) is an ideal of \(R\) because it satisfies the properties of a subring, absorption, and closure.

Step by step solution

01

Verify Subset Criterion

To show that a set is an ideal, we first need to establish that it is a subset of the ring. Since all elements of \(I_a\) are defined as \(x \in R\) such that \(a \cdot x = 0\), it is clear that \(I_a \subseteq R\). Thus, the subset criterion is satisfied.
02

Check for Additive Closure

For two elements \(x, y \in I_a\), we must have \(x + y \in I_a\). Since \(a \cdot x = 0\) and \(a \cdot y = 0\), it follows that:\[a \cdot (x + y) = a \cdot x + a \cdot y = 0 + 0 = 0.\]Hence, \(x + y \in I_a\), meaning the set is closed under addition.
03

Confirm Existence of Additive Inverse

For any \(x \in I_a\), we need \(-x \in I_a\). Since \(a \cdot x = 0\), it follows that:\[a \cdot (-x) = -(a \cdot x) = -0 = 0.\]Thus, \(-x \in I_a\), confirming the existence of additive inverses.
04

Verify Absorption by Ring Multiplication

For \(r \in R\) and \(x \in I_a\), we need to show \(r \cdot x \in I_a\). Since \(a \cdot x = 0\), then for any \(r \in R\),\[a \cdot (r \cdot x) = r \cdot (a \cdot x) = r \cdot 0 = 0.\]Therefore, \(r \cdot x \in I_a\), satisfying the absorption property.

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Key Concepts

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

Commutative Ring
A commutative ring is a fascinating structure from algebra that expands the concept of both addition and multiplication outside of basic number systems. In a commutative ring, operations of addition and multiplication have versatile rules. Here's what you need to know:
  • For **addition**, any two elements added together remain within the ring, and the process is both associative (grouping doesn't matter) and commutative (order doesn't matter).
  • For **multiplication**, the operations are similar. However, commutativity means that changing the order of multiplication does not alter the result. This is a key feature and is crucial in ring theory, setting commutative rings apart from more complex structures where this might not hold.
Within a commutative ring, you can add any elements and multiply them together, always respecting the closure properties that keep everything within the set. These properties make commutative rings useful in various mathematical and applied fields.
Additive Closure
Additive closure is a basic yet important concept in algebra when dealing with subsets of rings. It refers to the idea that when you add two elements from a subset, the sum is still a part of the subset. Let's break it down:
  • Consider two elements, say **x** and **y**, within a subset. Add them together, and you'll find their sum also belongs to the subset.
  • This property is crucial for structures to qualify as subgroups of a ring, particularly when proving if a subset is an ideal.
In our specific case with the set \( I_a \), we have already shown that if **x** and **y** are such that \( a \cdot x = 0 \) and \( a \cdot y = 0 \), then \( a \cdot (x + y) = 0 \), confirming that the subset containing these elements is indeed closed under addition. This closure makes it robust against combining its elements, maintaining their integrity within the structure.
Subset Criterion
The subset criterion is the groundwork for determining if a set is an ideal in a mathematical ring. It is one of the simplest yet essential rules:
  • A subset of a ring must naturally belong to the larger ring set from which it derives.
  • Verification requires checking that every element of the subset comes from the original set鈥攊n our scenario, all elements \( x \) in \( I_a \) arise from \( R \) itself, since \( I_a = \{x \in R \mid a \cdot x = 0 \} \).
By establishing the subset criterion, you ensure the subset's propriety as a legitimate component of a ring. It's about confirming that what's in your subset truly belongs and follows the original set's larger rules and properties.
Additive Inverse
The concept of an additive inverse is critical in algebra, particularly when discussing group and ring operations. An additive inverse relates to each element coupled with a unique counterpart, such that:
  • When an element is added to its inverse, it results in the neutral element of addition, which is zero.
  • For any element **x**, its inverse is **-x**, and their sum is \( x + (-x) = 0 \).
In proving that \( I_a \) is an ideal, we needed to show that for every element included, its additive inverse also remains within that set. The idea is that if \( a \cdot x = 0 \), then \( a \cdot (-x) = -(a \cdot x) = 0 \), confirming the presence of inverses within our ideal. Thus, it fortifies the spatial consistency of the subset under the group operations.

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Most popular questions from this chapter

Find all ideals \(N\) of \(\mathbb{Z}_{12}\). In each case compute \(Z_{12} / N\); that is, find a known ring to which the quotient ring is isomorphic.

Let \(\phi\) be a homomorphism of a ring \(R\) with unity onto a nonzero ring \(R^{\prime}\). Let \(u\) be a unit in \(R\). Show that \(\delta(u)\) is a unit in \(R^{\prime}\).

Find a subring of the ring \(2 \times 2\) that is not an ideal of \(Z \times 2\).

Let \(R=|a+b \sqrt{2}| a, b \in \mathbb{Z})\) and let \(R\) ' consist of all \(2 \times 2\) matrices of the form \(\left[\begin{array}{cc}a & 2 b \\ b & a\end{array}\right]\) for \(a, b \in \mathbb{Z}\). Show that \(R\) is a subring of \(\mathbb{R}\) and that \(R^{\prime}\) is a subring of \(M_{2}(Z)\). Then show that \(\phi: R \rightarrow R^{\prime}\), where \(\phi(a+b \sqrt{2})=\) \(\left[\begin{array}{cc}a & 2 b \\ b & a\end{array}\right]\) is an isomorphism.

Show that \(\phi: \mathrm{C} \rightarrow M_{2}(\mathbb{R})\) given by $$ \phi(a+b i)=\left(\begin{array}{rr} a & b \\ -b & a \end{array}\right) $$ for \(a, b \in \mathbb{R}\) gives an isomorphism of \(\mathbb{C}\) with the subring \(\phi[\mathrm{C}]\) of \(M_{2}(\mathrm{R})\).
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