/*! 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 6 Write down all monic cubic polyn... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 6

Write down all monic cubic polynomials in \(\mathbb{Z}_{2}[x]\), factorize them completely and construct a splitting field for each of them. Which of these fields are isomorphic?

Short Answer

Expert verified
All fields are isomorphic by extending to \(\mathbb{F}_4\).

Step by step solution

01

Understanding Monic Polynomials in \(\mathbb{Z}_2[x]\)

Monic polynomials in \(\mathbb{Z}_2[x]\) are polynomials where the leading coefficient is 1. A cubic polynomial has the form \(x^3 + ax^2 + bx + c\), where \(a, b, c \in \mathbb{Z}_2\). Since \(\mathbb{Z}_2 = \{0, 1\}\), the values \(a, b,\) and \(c\) can each be 0 or 1. We need to list all possible combinations.
02

Listing All Monic Cubic Polynomials

Based on the condition of being monic (leading coefficient is 1), the possible polynomials are: \(x^3, x^3 + x^2, x^3 + x, x^3 + 1, x^3 + x^2 + x, x^3 + x^2 + 1, x^3 + x + 1, x^3 + x^2 + x + 1\). So, there are 8 such polynomials to consider.
03

Factorization of Monic Cubic Polynomials

We'll attempt to factor each polynomial over \(\mathbb{Z}_2\):- \(x^3=(x)(x^2)\)- \(x^3 + x^2 = x^2(x + 1)\)- \(x^3 + x = x(x^2 + 1)\)- \(x^3 + 1 = (x + 1)(x^2 + x + 1)\)- \(x^3 + x^2 + x = x(x^2 + x + 1)\)- \(x^3 + x^2 + 1 \): Irreducible over \(\mathbb{Z}_2\)- \(x^3 + x + 1\): Irreducible over \(\mathbb{Z}_2\)- \(x^3 + x^2 + x + 1 = (x + 1)(x^2 + 1)\)
04

Constructing Splitting Fields

For each irreducible polynomial and those that have quadratic irreducible factors, we construct splitting fields:- \(x^3 + x^2 + 1\), \(x^3 + x + 1\) require extensions \(\mathbb{F}_4 = \mathbb{Z}_2(\alpha)\) where \(\alpha\) is a root such as \(\alpha^3 = \alpha^2 + 1\).- \(x^2 + x + 1\), appearing in \(x^3 + 1\) and \(x^3 + x^2 + x\), splits in an extension field, \(\mathbb{F}_4\), as well.
05

Identifying Isomorphic Fields

The splitting field of each irreducible or reduced factor to unique field operations always results in the field \(\mathbb{F}_4\) since \(\mathbb{F}_4\) can accommodate solutions to all such polynomials. Thus, they are all isomorphic.

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.

Polynomials over Finite Fields
Polynomials over finite fields, such as \( \mathbb{Z}_2[x] \), are fascinating due to their unique properties.These fields contain a limited set of numbers; for \( \mathbb{Z}_2[x] \), these numbers are just \(0\) and \(1\).In this context, a polynomial can be expressed like this:
  • For example, a cubic polynomial takes the form \(x^3 + ax^2 + bx + c\), with \(a, b, c \in \{0, 1\}\).
  • Due to the limited set of coefficients, operations such as addition and multiplication behave differently compared to real-number coefficients.
Understanding these behaviors is crucial for working with finite fields and utilizing their properties effectively.
Splitting Field
The concept of a splitting field is significant in polynomial algebra.A splitting field is essentially the smallest field extension within which a polynomial completely factors into linear factors.
  • For instance, if we have a polynomial that cannot be fully factored over its base field, we need a larger field.
  • This field, or splitting field, contains all the roots needed to express the polynomial as a product of linear terms.
For cubic polynomials from \( \mathbb{Z}_2[x] \), some needed a field like \( \mathbb{F}_4 \) for complete factorization because \( \mathbb{Z}_2[x] \) was not sufficient on its own.
Field Isomorphism
Field isomorphism is about understanding when two fields can be viewed as fundamentally the same structure from an algebraic perspective.
  • Two fields are isomorphic if there is a bijective (one-to-one and onto) map between them that preserves addition and multiplication.
  • This means that even though two fields may appear different externally or have different elements, they can essentially "do the same things" with those elements.
In the context of splitting fields for polynomials in \( \mathbb{Z}_2[x] \), fields such as \( \mathbb{F}_4 \) are isomorphic because they accommodate solutions to the polynomials in question equally.
Monic Polynomials
A monic polynomial is a type of polynomial where the leading coefficient is always \(1\).
  • In \( \mathbb{Z}_2[x] \), naming a polynomial monic simplifies its expression since the highest power term is always make with coefficient \(1\).
  • For example, given \(x^3 + ax^2 + bx + c\), if no leading coefficient is specified, its understood to be \(1\) by default.
Monic polynomials are especially important because they often represent the simplest form of the polynomials that need analysis or simplification, especially when finding factorization in finite fields.

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

Find a splitting field extension for \(x^{3}-5\) over \(\mathbb{Z}_{7}, \mathbb{Z}_{11}\) and \(\mathbb{Z}_{13} .\)

Find a splitting field extension \(K: \mathbb{Q}\) for each of the following polynomials over \(\mathbb{Q}: x^{4}+1, x^{4}+4,\left(x^{4}+1\right)\left(x^{4}+4\right),\left(x^{4}-1\right)\left(x^{4}+4\right)\). In each case determine the degree \([K: \mathbb{Q}]\) and find \(\alpha\) such that \(K=Q(\alpha) .\)

Suppose that \(L: K\) is a splitting field extension for a polynomial of degree \(n\). Show that \([L: K]\) divides \(n !\)

Suppose that \(M: L\) and \(L: K\) are extensions, and that \(\alpha \in M\) is algebraic over \(K\). Does \([L(\alpha): L]\) always divide \([K(\alpha): K]\) ?

Suppose that \(K\) is a field over which \(x^{n}-1\) splits, and suppose that \(K(t): K\) is a simple transcendental extension. Show that \(x^{n}-t\) is irreducible in \(K(t)[x]\) (Exercise 5.4). Construct a splitting field extension for \(x^{n}-t\) by considering another simple transcendental extension \(K(s): K\) and a monomorphism \(i: K(t) \rightarrow K(s)\) which fixes \(K\) and sends \(t\) to \(s^{n}\).
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Statistics

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.