/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Construct a finite field of 27 e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 23

Construct a finite field of 27 elements.

Short Answer

Expert verified
The finite field of 27 elements is constructed as \( F = \mathbb{F}_3[x] / \), where \( \mathbb{F}_3[x] \) is the polynomial ring over the finite field \( \mathbb{F}_3 \) and \( x^3 - x - 1 \) is an irreducible polynomial of degree 3 in \( \mathbb{F}_3[x] \).

Step by step solution

01

Identify the Prime

The size of the finite field is 27, which is \( 3^3 \). Here, the prime number (p) is 3 and the power (n) is 3.
02

Create Polynomial Ring

Create a polynomial ring \( \mathbb{F}_3[x] \) over the finite field with 3 elements.
03

Find Irreducible Polynomial

An irreducible polynomial in \( \mathbb{F}_3[x] \) of degree 3 needs to be found. One such polynomial is \(f(x) = x^3 - x - 1\). This polynomial is irreducible in \( \mathbb{F}_3[x] \) because it has no roots in \( \mathbb{F}_3 \). Therefore, \(f(x)\) is the generator of this field.
04

Construct Finite Field

Now, construct a finite field of size 27 as \( \mathbb{F}_3[x] / \). Here, the \( \) represents the ideal generated by \(f(x)\). This quotient field will have 27 distinct equivalence classes, each of which can be considered an 'element' of the constructed field.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Polynomial Rings
A polynomial ring is a mathematical structure formed from polynomials with coefficients in a particular field. In this exercise, the field is \( \mathbb{F}_3 \), which consists of the elements \{0, 1, 2\}. These elements represent the coefficients of the polynomials.
  • Polynomials in \( \mathbb{F}_3[x] \) are expressions like \( 2x^2 + x + 1 \).
  • The 'ring' implies that we can add, subtract, and multiply these polynomials just like integers, but division isn't always possible.
  • The coefficients of the polynomials are taken from the finite field \( \mathbb{F}_3 \), so operations are performed modulo 3.
For example, in \( \mathbb{F}_3[x] \), the polynomial \( x^3 + 2x^2 + 2 \) would be added to another polynomial by adding their respective coefficients and reducing each by modulo 3.
Irreducible Polynomials
An irreducible polynomial is a polynomial that cannot be factored into the product of two smaller degree polynomials over a given field. In the context of finite fields, finding an irreducible polynomial is crucial to constructing the field.
  • The irreducible polynomial acts as a 'building block' for creating a larger field from a smaller one.
  • In \( \mathbb{F}_3[x] \), the polynomial \( f(x) = x^3 - x - 1 \) is an example of an irreducible polynomial. It cannot be factored further over \( \mathbb{F}_3 \).
  • Checking if a polynomial is irreducible involves verifying that it has no roots in the field or cannot be expressed as a product of polynomials of lower degrees.
To check irreducibility, we can evaluate the polynomial at all elements of \( \mathbb{F}_3 \) (which are 0, 1, and 2) and ensure none of these values zero out the polynomial, confirming it's irreducible.
Quotient Fields
Quotient fields, also known as residue class fields, are constructed by dividing a polynomial ring by an ideal generated by an irreducible polynomial. In our example, this creates a finite field with a specific number of elements.
  • Such fields are denoted as \( \mathbb{F}_3[x] / \langle f(x) \rangle \), where \( f(x) \) is the irreducible polynomial. The classes or elements of this field are represented by the remainders of polynomial division by \( f(x) \).
  • Each element of the quotient field corresponds to a distinct equivalence class, and there are as many elements as the field size, 27 in this case.
  • These elements are combinations of lower-degree polynomials modulo the irreducible polynomial.
In practical terms, performing operations within the quotient fields means working with these equivalence classes. The irreducible polynomial acts as a constraint, ensuring all operations remain within the defined field, mirroring the finite nature of the coefficients.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

a) Find all roots of \(f(x)=x^{2}+4 x\) if \(f(x) \in \mathbf{Z}_{12}[x]\). b) Find four distinct linear polynomials \(g(x), h(x), s(x)\), \(t(x) \in \mathbf{Z}_{12}[x]\) so that \(f(x)=g(x) h(x)=s(x) t(x)\) c) Do the results in part (b) contradict the statements made in the paragraph following Example \(17.7 ?\)

a) If a projective plane has six lines through every point, how many points does this projective plane have in all? b) If there are 57 points in a projective plane, how many points lie on each line of the plane?

Give an example of a polynomial \(f(x) \in \mathbf{R}[x]\) where \(f(x)\) has degree 6 , is reducible, but has no real roots.

a) Rewrite the following \(4 \times 4\) Latin square in standard form. \(\begin{array}{llll}1 & 3 & 4 & 2 \\ 3 & 1 & 2 & 4 \\ 2 & 4 & 3 & 1 \\ 4 & 2 & 1 & 3\end{array}\) b) Find a \(4 \times 4\) Latin square in standard form that is orthogonal to the result in part (a). c) Apply the reverse of the process in part (a) to the result in part (b). Show that your answer is orthogonal to the given \(4 \times 4\) Latin square.

Find the orders \(n\) for all fields \(G F(n)\), where \(100 \leq\) \(n \leq 150\).
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.