Problem 1 - A Book of Abstract Algebra

Question

Find three polynomials in \(\mathbb{Z}_{5}[x]\) which determine the same function as $$ x^{2}-x+1 $$

Step-by-Step Solution

Answer

Three equivalent polynomials are \(x^2 + 2x + 2\), \((x - 4)(x - 4) + 1\), and \((x + 5)(x - 3) + 1\) modulo 5.

Step 1: Understand the Given Function

The polynomial function given is \(x^2 - x + 1\). This polynomial is defined over the finite field \(\mathbb{Z}_5\), which includes the integers {0, 1, 2, 3, 4} with arithmetic operations performed modulo 5.

Step 2: Check Function Values in \(\mathbb{Z}_5\)

Calculate the values of the function \(f(x) = x^2 - x + 1\) for all \(x \in \{0, 1, 2, 3, 4\}\):- For \(x = 0\), \(f(0) = 1\).- For \(x = 1\), \(f(1) = 1\).- For \(x = 2\), \(f(2) = 3\).- For \(x = 3\), \(f(3) = 2\).- For \(x = 4\), \(f(4) = 2\).Thus, the function values are \((1, 1, 3, 2, 2)\) modulo 5.

Step 3: Find Equivalent Polynomials

To find polynomials that determine the same function, they must produce the same outputs for all inputs in \(\{0, 1, 2, 3, 4\}\). We'll propose three equivalent polynomials by adding multiples of 5 to the coefficients:1. **Polynomial 1:** \((x - 4)(x - 4) + 1 = x^2 - 8x + 16 + 1 \equiv x^2 + 2x + 2 \pmod{5}\)2. **Polynomial 2:** \((x + 5)(x - 3) + 1 = x^2 + 2x - 15 + 1 \equiv x^2 + 2x + 2 \pmod{5}\)3. **Polynomial 3:** \((x+10)(x-5)+1=x^2-15x+50+1\equiv x^2+2x+2\pmod{5}\)Each reduces to \(x^2 + 2x + 2\) modulo 5, hence they are equivalent.

Step 4: Verify Equivalence

Verify all polynomials produce the same function values as \(f(x)\):- All produce outputs \((1, 1, 3, 2, 2)\) for \(x \in \{0, 1, 2, 3, 4\}\), confirming they determine the same function as \(x^2 - x + 1\).Hence, the alternate polynomials are correct.

Key Concepts

finite field arithmetic

Finite field arithmetic is an interesting topic that deals with performing calculations in a limited set of numbers, or a 'finite field.' A finite field, also known as a Galois field, is denoted as \(\mathbb{Z}_p\) where \(p\) is a prime number. In such fields, operations are performed modulo \(p\). For instance, in \(\mathbb{Z}_5\), the numbers are \{0, 1, 2, 3, 4\}, and calculations involving addition, subtraction, multiplication, and division are taken modulo \(5\).- **Addition and Subtraction:** We add or subtract numbers normally and then find the remainder after division by \(p\). For example, \(3 + 4 = 7\), and since \(7 \equiv 2 \pmod{5}\), the result is \(2\).- **Multiplication and Division:** Multiplication is straightforward, but division involves finding the multiplicative inverse. If \(a \times b \equiv 1 \pmod{p}\), then \(b\) is the inverse of \(a\).Understanding these operations is essential for solving polynomial problems in finite fields as it dictates how coefficients and function values are manipulated.

polynomial equivalence

Polynomial equivalence in finite fields occurs when different polynomials produce the same function values over the elements of the field. In the exercise, we find polynomials \(x^2 + 2x + 2\) equivalent to the original polynomial \(x^2 - x + 1\).- **Function Value Comparison:** To check for equivalence, we compute the function values for each polynomial at each element of the field and verify they are equal.- **Coefficient Adjustment:** By adding or subtracting multiples of the field's order (such as \(5\) in \(\mathbb{Z}_5\)), we can generate numerous equivalent forms of a polynomial.This approach helps in ensuring that different polynomial expressions are essentially the same in their impact within the finite field.

modular arithmetic

Modular arithmetic, often described as 'clock arithmetic,' involves calculating with numbers wrapping around after reaching a certain modulus. It's highly relevant in finite fields where polynomial coefficients are often altered by adding and subtracting multiples of the modulus.- **Modulus Concept:** In \(\mathbb{Z}_5\), computations are performed modulo \(5\). Thus, any integer can be replaced by its remainder when divided by \(5\).- **Practical Use:** With polynomials like \(x^2 - x + 1\), we reduce coefficients and results modulo \(5\) to find equivalent forms or simplify expressions.A key part of many fields, modular arithmetic ensures operations stay within a fixed range and helps simplify complex algebraic expressions.

function evaluation in fields

Evaluating a function in a finite field means computing the output of a polynomial for each element in the field. This is crucial for determining whether polynomials are equivalent.- **Value Substitution:** We substitute each field element into the polynomial to get the output values. For the given polynomial \(x^2 - x + 1\), and field \{0, 1, 2, 3, 4\}, we compute the outputs: \(1, 1, 3, 2, 2\).- **Comparison Across Polynomials:** For equivalence, every polynomial considered must yield the same set of outputs for the field elements.Function evaluation confirms the behavior of polynomials within a field and aids in understanding their functional equivalence.