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Chapter 19: Problem 2

Suppose \(6 x=0\) for every \(x \in A\). Prove that the function \(h(x)=3 x\) is a homomorphism from \(A\) to \(A\). If \(J=\\{x: 3 x=0\\}\) and \(B=\\{3 x: x \in A\\}\), explain why \(J\) is an ideal of \(A, B\) is a subring of \(A\), and \(A / J \cong B\).

Short Answer

Expert verified
The function \( h(x)=3x \) is a homomorphism. \( J \) is an ideal of \( A \), \( B \) is a subring of \( A \), and \( A/J \cong B \).

Step by step solution

01

Define Homomorphism

A function \( h: A \to A \) is a homomorphism if for all elements \( a, b \in A \), the following holds: \( h(a + b) = h(a) + h(b) \) and \( h(ab) = h(a)h(b) \). We need to verify these properties for the function \( h(x) = 3x \).
02

Verify Addition Property

Consider two elements \( a, b \in A \). For \( h(a + b) = 3(a + b) \) and \( h(a) + h(b) = 3a + 3b \), because addition is distributive, \( 3(a + b) = 3a + 3b \), which shows that \( h(a + b) = h(a) + h(b) \).
03

Verify Multiplication Property

Consider again the elements \( a, b \in A \). For \( h(ab) = 3(ab) \) and \( h(a)h(b) = (3a)(3b) = 9ab \), but since \( 6x = 0 \) for any \( x \in A \), thus \( 9ab = 3(ab) = 0 \). Hence, \( h(ab) = h(a)h(b) \).
04

Define the Ideal \( J \)

The set \( J = \{ x \in A : 3x = 0 \} \) is the kernel of the homomorphism \( h(x) \). It is an ideal of \( A \) because it is closed under addition and absorbing under multiplication by any element of \( A \).
05

Define the Subring \( B \)

\( B = \{ 3x : x \in A \} \) forms a subring because it contains the additive identity (if \( 3x = 0 \) for some \( x \) implies \( 0 \) is in \( B \)) and is closed under subtraction and multiplication by elements of \( A \).
06

Establish Isomorphism from \( A/J \) to \( B \)

To show \( A/J \cong B \), consider the map \( \varphi: A \to B \) given by \( \varphi(x + J) = 3x \). This map is well-defined, surjective, and injective, as \( \ker(\varphi) = J \), thus satisfying the criteria for the First Isomorphism Theorem.

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Key Concepts

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

Ring Homomorphism
A ring homomorphism is a concept in ring theory that captures the idea of a structure-preserving map between two rings. This map, denoted as \( h: A \to A \), ensures that operations within the ring are consistent. For a function to be considered a homomorphism, it must preserve two key properties of rings: addition and multiplication.

In simple terms, if you have two elements, \( a \) and \( b \), within the domain ring \( A \), then the following must hold:
  • Addition Property: \( h(a + b) = h(a) + h(b) \)
  • Multiplication Property: \( h(ab) = h(a)h(b) \)
For the function \( h(x) = 3x \) given in the exercise, both properties are verified. Thus, despite the multiplication resulting in zero (since \( 6x = 0 \) for all \( x \in A \)), the function remains a homomorphism.
Ideal in a Ring
An ideal is a special subset of a ring with certain properties making it crucial for understanding ring structure. The ideal \( J \) in our problem is defined as \( J = \{ x \in A : 3x = 0 \} \). This makes it the kernel of the homomorphism \( h(x) \).

To qualify as an ideal, the subset must satisfy:
  • Be closed under addition, meaning if \( a, b \in J \), then \( a + b \in J \).
  • Be closed under multiplication by any element from the ring \( A \), ensuring \( ra \in J \) for any \( r \in A \) and \( a \in J \).
The absorption property grants that multiplying an element of the ideal by any element of the ring still yields an element within the ideal, thereby affirming \( J \) as an ideal in the ring \( A \). This property is essential for forming quotient rings and applying the First Isomorphism Theorem.
Subring
Subrings are important as they represent a smaller ring within a larger ring that retains its ring properties. In the exercise, \( B = \{ 3x : x \in A \} \) is declared a subring of \( A \).

To be a subring, \( B \) must meet these conditions:
  • Contain the additive identity (zero). If \( 3x = 0 \) for some \( x \), then \( 0 \) is in \( B \).
  • Be closed under subtraction, meaning if \( a, b \in B \), then \( a - b \in B \). This follows from normal ring operations.
  • Be closed under ring multiplication. That means for any elements \( a, b \in B \), \( ab \in B \). In this situation, it includes elements of the form \( 3a \times 3b = 9ab \).
Thus, \( B \) inherits its operations from the larger ring \( A \) and functions as a standalone ring.
First Isomorphism Theorem
The First Isomorphism Theorem provides a powerful tool for connecting an ideal, its quotient, and another subring. It states that if \( h: A \to B \) is a homomorphism, then \( A/\ker(h) \cong \text{im}(h) \), where \( \ker(h) \) is the kernel and \( \text{im}(h) \) is the image.

In the context of our exercise:
  • The kernel \( J = \{ x \in A : 3x = 0 \} \) is the set of all elements mapping to zero, forming an ideal.
  • The image is the subring \( B = \{ 3x : x \in A \} \).
By applying the First Isomorphism Theorem, we conclude \( A/J \cong B \). This means there's a structural isomorphism between the quotient ring formed by the ideal and the subring \( B \), preserving the ring's properties. This theorem thus solidifies the connection between the abstract algebraic structures presented.

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Most popular questions from this chapter

Prove that \(x^{2}+10 y^{2}=n\) (where \(n\) is an integer) has no integer solutions if the last digit of \(n\) is \(2,3,7\), or 8 .

Prove that the sequence \(2,10,18,26, \ldots\) does not include the cube of any integer.

Let \(A\) be a commutative ring with unity, and \(J\) an ideal of \(A .\) Prove each of the following. 1 \(A / J\) is a commutative ring with unity.

An ideal \(J\) of a ring \(A\) is called semiprime iff it has the following property: For every \(a \in A\), if \(a^{n} \in J\) for some positive integer \(n\), then necessarily \(a \in J\). Prove that \(J\) is semiprime iff \(A / J\) has no nilpotent elements (except zero).

An ideal \(J\) of a ring \(A\) is called primary iff for all \(a, b \in A\), if \(a b \in J\), then either \(a \in J\) or \(b^{n} \in J\) for some positive integer \(n .\) Prove that every zero divisor in \(A / J\) is nilpotent iff \(J\) is primary.
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