Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 3: Problem 3
Determine the factorization of \(x^{9}+x+1\) over \(\mathbb{F}_{2}\).
Short Answer
Expert verified
The factorization of \(x^{9}+x+1\) over \(\mathbb{F}_{2}\) is \((x^3 + x + 1)^3\).
Step by step solution
01
Understand the Polynomial
The polynomial to be factored is \(x^{9}+x+1\). In this polynomial, the coefficients are 1 and the exponents in decreasing powers of x are 9, 1 and 0. We have to factorize this polynomial over the set \{0, 1\}, as we are working over \(\mathbb{F}_{2}\).
02
List all possible factors
For a polynomial of degree \(n\) over \(\mathbb{F}_{2}\), the possible factors are all the irreducible polynomials of degree that divide \(n\). The possible degrees that divide 9 are 1, 3 and 9 itself. We need to list the irreducible polynomials of these degrees. For degree 1, there are two: \(x\) and \(x+1\). For degree 3, there are two: \(x^3+x+1\) and \(x^3+x^2+1\). It's also possible that the entire polynomial is irreducible, so \(x^{9}+x+1\) is included as a possible factor.
03
Test the Factors
We test potential factors using polynomial division. Given that we are in \(\mathbb{F}_{2}\), the operations are mod 2; meaning that \(1+1=0\). We discover that given \(x^9 + x + 1\) divided by \(x^3 + x + 1\) doesn't have a remainder. Thus, \(x^3 + x + 1\) is a factor of the original polynomial.
04
Continue the Factorization
After extracting the factor \(x^3 + x + 1\), we get another polynomial of degree 3, by dividing the original polynomial \(x^9 + x + 1\) by \(x^3 + x + 1\). We repeat the division process with the new polynomial and find that it can also be factored by \(x^3 + x + 1\). This shows that \(x^9 + x + 1\) is equivalent to \((x^3 + x + 1)^3\).
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.
Finite Fields
Finite fields, often denoted as \(mathbb{F}\), are mathematical structures with a finite number of elements. A common finite field used in polynomial factorization is \(mathbb{F}_2\), which consists only of the elements 0 and 1. Finite fields are interesting as they exhibit all the properties of larger number systems like the real numbers but only within a limited set of elements.
The operations of addition and multiplication in finite fields differ slightly due to their cyclic nature:
The operations of addition and multiplication in finite fields differ slightly due to their cyclic nature:
- In \(mathbb{F}_2\), addition is simply the bitwise XOR operation. Thus, \(1 + 1 = 0\).
- Multiplication follows standard binary rules.
Irreducible Polynomials
An irreducible polynomial over a field is a polynomial that cannot be factored into the product of smaller degree polynomials over the same field. When working in fields like \(mathbb{F}_2\), identifying irreducible polynomials becomes a key step in tasks such as factorization.
For instance, in attempting to factor \(x^9 + x + 1\) over \(mathbb{F}_2\), we assess its decomposition by considering polynomials of degrees that divide the highest degree, such as 1 and 3. In this exercise, some potential factors include:
For instance, in attempting to factor \(x^9 + x + 1\) over \(mathbb{F}_2\), we assess its decomposition by considering polynomials of degrees that divide the highest degree, such as 1 and 3. In this exercise, some potential factors include:
- For degree 1: \(x\) and \(x + 1\)
- For degree 3: \(x^3 + x + 1\) and \(x^3 + x^2 + 1\)
Polynomial Division
Polynomial division is a process much like long division but with polynomials. It allows us to divide one polynomial by another, often to determine if one is a factor of the other. In the context of finite fields, such as \(mathbb{F}_2\), division takes place under specific modulus, meaning calculations must be done with special care.
In our specific problem with \(x^9 + x + 1\), the division is tested with possible factors like \(x^3 + x + 1\). The dividend is continuously divided by polynomial divisors until either:
In our specific problem with \(x^9 + x + 1\), the division is tested with possible factors like \(x^3 + x + 1\). The dividend is continuously divided by polynomial divisors until either:
- The remainder is zero, indicating the divisor is a factor.
- A non-zero remainder is found, suggesting the divisor is not a factor.
Modulo Operations
Modulo operations are key in simplifying polynomial arithmetic, especially in finite fields. When dealing with polynomials under modulo 2 operations, the calculations are easier than they may seem because of 2's characteristic:
- The numbers wrap around after reaching 2, so \(1 + 1 = 0\).
- This is akin to how integers behave in modular arithmetic, where numbers reset after reaching a certain modulus.
