Question

Let \(K\) be a vector space over \(\mathbb{Z}_{3}\) with basis \(\\{1, t\\},\) so \(K=\left\\{a+b t \mid a, b,\right.\) in \(\left.\mathbb{Z}_{3}\right\\} .\) It is known that \(K\) becomes a field of nine elements if we define \(t^{2}=-1\) in \(\mathbb{Z}_{3} .\) In each case find the inverse of the element \(x\) of \(K\) : a. \(x=1+2 t\) b. \(x=1+t\)

Short answer

The inverses are: a. \( 2 + t \); b. \( 2 + 2t \).

Step-by-step solution

Understanding the structure of the field

The vector space \(K\) over \(\mathbb{Z}_3\) with basis \(\{1, t\}\) becomes a field with 9 elements when we define \(t^2 = -1\). This means that each element in the field \(K\) can be expressed as \( a + bt \) where \( a, b \in \mathbb{Z}_3 \). The elements in \(\mathbb{Z}_3\) are \{0, 1, 2\}.

Finding the inverse of \(x = 1 + 2t\)

To find the inverse of \( x = 1 + 2t \), we express its inverse as \( y = a + bt \) such that \((1 + 2t)(a + bt) = 1\). Expanding this product gives \( a + (2a + b)t + 2bt^2 \). Substituting \( t^2 = -1 \), we get \( a + (2a + b)t - 2b \). Equating coefficients with 1 gives: 1. \( a - 2b = 1 \) 2. \( 2a + b = 0 \).Solving these simultaneous equations in \( \mathbb{Z}_3 \): Substitute \( b = -2a \) into \( a - 2(-2a) = 1 \), resulting in \( a + 4a = 1 \), or \( 5a = 1 \). Since \(5 \equiv 2\), we solve \( 2a = 1 \). The solution in \(\mathbb{Z}_3\) is \(a = 2\). Substituting \(a = 2\) back gives \(b = -2(2) = 1\). Thus, the inverse is \( y = 2 + t \).

Finding the inverse of \(x = 1 + t\)

We find the inverse of \( x = 1 + t \) by expressing its inverse as \( y = c + dt \) such that \((1 + t)(c + dt) = 1\). Expanding gives \( c + (c + d)t + dt^2 \). Substituting \( t^2 = -1 \) gives \( c + (c + d)t - d \). Equating coefficients with 1, we get:1. \( c - d = 1 \)2. \( c + d = 0 \).Solving these equations in \( \mathbb{Z}_3 \): From the second equation, \( d = -c \), substitute into the first equation to get \( c - (-c) = 1 \) or \( 2c = 1 \). Solving \( 2c = 1 \) in \(\mathbb{Z}_3\), we find \( c = 2 \). Then \( d = -2 = 1 \). Thus, the inverse is \( y = 2 + 2t \).

Key concepts

Vector Space Over Finite Field

A vector space over a finite field is a mathematical structure built upon a finite field, such as \( \mathbb{Z}_3 \), which consists of all integers modulo 3. This vector space can carry out operations similar to those in traditional arithmetic. In our context, the vector space \( K \) is constructed over \( \mathbb{Z}_3 \), with a specific basis \{ 1, t \}.
Each element within this vector space can be represented in the form \( a + bt \) where both \( a \) and \( b \) are elements of \( \mathbb{Z}_3 = \{0, 1, 2\} \). The beauty of this setup lies in its finite nature, which simplifies operations and enables a compact representation of mathematical entities.
By introducing the condition \( t^2 = -1 \) within this vector space, \( K \) becomes not just any vector space, but a field—a set wherein every non-zero element possesses a multiplicative inverse. This transformation makes \( K \) a field with exactly 9 elements. Hence, \( K \) not only represents all possible linear combinations of \( 1 \) and \( t \), but it also adheres to additional quintessential field properties.

Inverse Elements in Fields

The inverse of an element in a field is what, when multiplied by the original element, yields the multiplicative identity of the field, usually denoted as 1. In the field \( K \), to find the inverse of an element \( x \, x = a + bt \, \) involves solving \((a + bt)(c + dt) = 1 \). Here, \( c \) and \( d \) are coefficients to be determined.
Consider the element \( x = 1 + 2t \. \) We wish to find its multiplicative inverse, \( y = c + dt \, \) by ensuring \( (1 + 2t)(c + dt) = 1 \). Expanding this product and simplifying using the condition \( t^2 = -1 \, \) provides a system of equations in terms of \( c \) and \( d \) over \( \mathbb{Z}_3 \. \)
Solving these equations yields values for \( c \) and \( d \, \) thus giving the inverse. These calculations highlight how the principles of finding inverses in finite fields mirror those in larger number systems, just within a more limited set.

Finite Field Arithmetic

Finite field arithmetic involves operations—addition, subtraction, multiplication, and division—restricted to a finite set of elements. In \( K \, \) arithmetic is governed by the rules of modulo arithmetic, specifically mod 3 in this case.
Each arithmetic operation involves finding a result within \( \mathbb{Z}_3 \, \) either natively or through operations like expanding expressions. This arithmetic naturally deals with conditions such as \( t^2 = -1 \, \) which shape the identities in the field.
For example, when multiplying elements like \( (1 + t)(c + dt) \, \) expansion and substitution of \( t^2 = -1 \) are crucial steps. Every operation results in values adjusted to stay within the finite field by applying modulo 3. Hence, each arithmetic step embodies precision, influenced heavily by the constraints of finiteness, especially apparent when reducing expressions like polynomial identities or solving equations of inverses.