Irreducibility of Polynomials over Finite Fields: A Case Study of x^3 - 3x - 2 in Z/7

Understanding Irreducibility of Polynomials over Finite Fields: A Case Study of x3 - 3x - 2 in Z/7

Introduction

Polynomials over finite fields play a fundamental role in algebra and cryptography. One of the key questions in this context is determining the irreducibility of a polynomial. In this article, we explore the irreducibility of the polynomial x^3 - 3x - 2 € Z/7[x] over the field Z/7. We will show that this polynomial is irreducible by showing it has no roots in Z/7.

Degree and Linearity

Given the polynomial x^3 - 3x - 2 € Z/7[x], we observe that its degree is 3, which is less than 4. This simplifies our task because if a polynomial of degree less than 4 over a field is irreducible, it suffices to show that it has no linear factors. By the Factor Theorem, this is equivalent to showing that the polynomial has no roots in the field in question.

No Linear Factors

To prove that x^3 - 3x - 2 € Z/7[x] has no linear factors, we need to check whether it has any roots in Z/7. We do this by evaluating the polynomial for each element of Z/7.

We consider the elements 0, 1, 2, 3, 4, 5, 6 modulo 7:

x 0: 0^3 - 3*0 - 2 -2 equiv 5 (mod 7) x 1: 1^3 - 3*1 - 2 -4 equiv 3 (mod 7) x 2: 2^3 - 3*2 - 2 8 - 6 - 2 0 (mod 7) x 3: 3^3 - 3*3 - 2 27 - 9 - 2 16 equiv 2 (mod 7) x 4: 4^3 - 3*4 - 2 64 - 12 - 2 50 equiv 1 (mod 7) x 5: 5^3 - 3*5 - 2 125 - 15 - 2 108 equiv 4 (mod 7) x 6: 6^3 - 3*6 - 2 216 - 18 - 2 196 equiv 0 (mod 7)

As we see, x 2 and x 6 are roots of the polynomial, specifically x 6. Therefore, the polynomial is reducible over Z/7.

Quadratics and Cubics

Generally, quadratics and cubics are reducible if and only if they have a root in the field. In this case, we have found roots 2 and 6 for our cubic polynomial.

Discriminant and Roots

The discriminant of the polynomial x^3 - 3x - 2 is computed as follows:

Delta (-3)^2 - 4*1*(-2) 9 8 17 equiv 3 (mod 7)

Since the discriminant is 1 modulo 7, the roots of the polynomial can be found using the cubic formula. However, in characteristic 7, the discriminant simplifies the process as follows:

sqrt{Delta} sqrt{3} equiv 5 (mod 7)

The roots are then given by:

x frac{-(-3) pm sqrt{3}}{3*1} frac{3 pm 5}{3}

In Z/7, this evaluates to:

x frac{3 5}{3} frac{8}{3} equiv 2 (mod 7) x frac{3 - 5}{3} frac{-2}{3} equiv 6 (mod 7)

The polynomial indeed has roots at x 2 and x 6, confirming that it is reducible.

Conclusion

In summary, the polynomial x^3 - 3x - 2 € Z/7[x] is reducible over the field Z/7. This is demonstrated by finding the roots through direct evaluation and the discriminant. Understanding such properties is crucial for applications in algebra and cryptography.