How to Find the Minimal Polynomial of √2 - 1 over Q

Introduction

Determining the minimal polynomial of an algebraic number over the rational field Q is a fundamental task in algebra. In this article, we will explore the steps to find the minimal polynomial of √2 - 1 over the rational field Q. This involves understanding the algebraic properties of √2 - 1 and the techniques used to derive the minimal polynomial.

Understanding the Algebraic Nature of √2 - 1

First, let's understand why √2 - 1 is indeed an algebraic number. An algebraic number is a root of a non-zero polynomial equation with rational coefficients. In this case, we will show that √2 - 1 satisfies a polynomial equation with rational coefficients.

Step 1: Expressing √2 - 1 as a Root of a Polynomial

Let ( r sqrt{2} - 1 ). If we square both sides of the equation, we get:

[ r^2 (sqrt{2} - 1)^2 2 - 2sqrt{2} 1 3 - 2sqrt{2} ]

Next, we express ( 3 - 2sqrt{2} ) in terms of ( r ):

[ 3 - 2sqrt{2} 3 - 2(r 1) 3 - 2r - 2 1 - 2r ]

Thus, we have:

[ r^2 1 - 2r ]

This is a polynomial equation, and it shows that ( r sqrt{2} - 1 ) is a root of ( x^2 2x - 1 ).

Step 2: Verifying Irreducibility

To confirm that ( x^2 2x - 1 ) is the minimal polynomial, we need to show that it is irreducible over the rational field Q. A polynomial is irreducible if it cannot be factored into the product of two non-constant polynomials with rational coefficients.

We will use the Rational Root Theorem to check if ( x^2 2x - 1 ) has any rational roots. The Rational Root Theorem states that any rational root of the polynomial ( a_nx^n a_{n-1}x^{n-1} ldots a_0 ) must be a factor of the constant term divided by a factor of the leading coefficient. For ( x^2 2x - 1 ), the constant term is -1 and the leading coefficient is 1. The possible rational roots are ( pm 1 ).

Testing these possible roots:

[ 1^2 2 cdot 1 - 1 1 2 - 1 2 eq 0 ]

[ (-1)^2 2(-1) - 1 1 - 2 - 1 -2 eq 0 ]

Since neither 1 nor -1 is a root, the polynomial ( x^2 2x - 1 ) has no rational roots. Therefore, it is irreducible over the rational field Q.

Hence, the polynomial ( x^2 2x - 1 ) is the minimal polynomial of ( sqrt{2} - 1 ) over the rational field Q.

Conclusion

In summary, the minimal polynomial of ( sqrt{2} - 1 ) over the rational field Q is ( x^2 2x - 1 ). This polynomial is irreducible, and thus it is the unique monic polynomial with rational coefficients that has ( sqrt{2} - 1 ) as a root.

References

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