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Determine whether the given set, together with the specified operations of addition and scalar multiplication, is a vector space over the indicated \(\mathbb{Z}_{p} .\) If it is not, list all of the axioms that fail to hold. \(\mathbb{Z}_{6},\) over \(\mathbb{Z}_{3}\) with the usual addition and multiplication (Think this one through carefully!)

Short Answer

Expert verified
\(\mathbb{Z}_6\) over \(\mathbb{Z}_3\) is not a vector space; closure under scalar multiplication fails.

Step by step solution

01

Understand the Set and Operations

The set in question is \(\mathbb{Z}_6\), which consists of the elements {0, 1, 2, 3, 4, 5}. We are to consider this set as a potential vector space over the field \(\mathbb{Z}_3\), which has elements {0, 1, 2}. The operations of addition and scalar multiplication are defined modulo 6 for addition, and modulo 3 for scalar multiplication.
02

Review Vector Space Axioms

A set is a vector space over a field if it satisfies the vector space axioms: closure under addition and scalar multiplication, the existence of an additive identity and inverses, associativity and commutativity of addition, distributive properties, and more.
03

Closure Under Addition

Check if the addition of any two elements in \(\mathbb{Z}_6\) produces another element in \(\mathbb{Z}_6\). Since addition is defined modulo 6, for any \(a, b \in \mathbb{Z}_6\), \((a + b) \mod 6 \in \mathbb{Z}_6\), so the set is closed under addition.
04

Closure Under Scalar Multiplication

Verify if scalar multiplication keeps elements within \(\mathbb{Z}_6\). For any scalar \(c \in \mathbb{Z}_3\) and vector \(v \in \mathbb{Z}_6\), \((c \times v) \mod 3\) must be in \(\mathbb{Z}_6\). This operation is problematic as it reduces mod 3, which conflicts with mod 6 elements.
05

Identify Failing Axioms

Since the scalar multiplication doesn't consistently map back into \(\mathbb{Z}_6\) (e.g., \(1 \cdot 3 = 3\), but \(1 \cdot 4 \mod 3 = 1\), which isn't in \(\mathbb{Z}_6\)), the set fails closure under scalar multiplication. Without closure, we immediately lack a valid scalar multiplication, affecting other axioms.

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Key Concepts

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

Scalar Multiplication
Scalar multiplication is a core concept in vector space theory and involves multiplying a vector by a scalar from the underlying field. In this scenario, we are dealing with the set \( \mathbb{Z}_6 \) as a potential vector space with scalars from \( \mathbb{Z}_3 \). Here, scalar multiplication and the operations involved must be consistent with the properties of vector spaces.
The main issue with this setup arises from the interaction between the set \( \mathbb{Z}_6 \) and the field \( \mathbb{Z}_3 \). In traditional scalar multiplication, consistently multiplying any element from \( \mathbb{Z}_6 \) by a scalar from \( \mathbb{Z}_3 \) and resulting in another element within \( \mathbb{Z}_6 \) is expected. However, the specific operation here, taking the result modulo 3, does not align with this requirement.
  • For any vector element \( v \) in \( \mathbb{Z}_6 \), the outcome of \( c \times v \mod 3 \) often results in elements that are not part of \( \mathbb{Z}_6 \), which violates the axiom of closure under scalar multiplication.
  • This means the criteria that scalar multiplication \( (c \cdot v) \) for \( c \in \mathbb{Z}_3 \) keeps the resulting vectors in \( \mathbb{Z}_6 \) is not met.
These inconsistencies highlight why scalar multiplication fails, and thus, \( \mathbb{Z}_6 \) cannot be considered a vector space over \( \mathbb{Z}_3 \).
Closure Under Addition
Closure under addition is one of the vector space axioms which ensures that when any two vectors from a set are added, the result should still be a vector in that set. For the set \( \mathbb{Z}_6 \), addition is performed modulo 6.
In detail:
  • For any two elements \( a, b \) in \( \mathbb{Z}_6 \), their sum \( (a + b) \mod 6 \) remains an element in \( \mathbb{Z}_6 \).
  • This is because the outcome of adding two numbers less than 6 and then taking the modulo 6 operation always yields a result within the original set of numbers \{0, 1, 2, 3, 4, 5\}.
Thus, \( \mathbb{Z}_6 \) satisfies closure under addition. This property is crucial for any set considered to be a vector space, as it ensures that applying the vector space operation of addition will not take you out of the set.
Vector Space Axioms
The vector space axioms are a set of rules that all vector spaces must follow. These include closure properties, associative and commutative properties of addition, distributive properties, and the existence of additive identities and inverses, among others.
Some key axioms are:
  • Closure under Addition: Already discussed, ensures sums of vectors remain within the set.
  • Closure under Scalar Multiplication: Each scalar-vector product must result in another vector within the set, which failed for \( \mathbb{Z}_6 \) over \( \mathbb{Z}_3 \).
  • Associativity and Commutativity of Addition: For vectors \( u, v, \) and \( w \), \( (u + v) + w = u + (v + w) \) and \( u + v = v + u \).
  • Existence of Zero Vector: There must be a zero vector such that adding it to any vector does not change the vector.
  • Existence of Additive Inverses: For any vector \( v \), there must be a vector \( -v \) that when added, results in the zero vector.
In the case of \( \mathbb{Z}_6 \), while some of these axioms could naturally be verified, the failure in closure under scalar multiplication invalidates the entire structure from being a vector space according to these essential axioms.

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