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Chapter 6: Problem 17

Show that the minimal polynomial for \(\sqrt[3]{2}\) is \(x^{3}-2\)

Short Answer

Expert verified
The minimal polynomial for \sqrt[3]{2} is \ x^{3} - 2 \ because \sqrt[3]{2} \ satisfies \theta^{3} = 2 \ and it is irreducible over the rational numbers.

Step by step solution

01

Identify the expression

Consider the expression for which the polynomial is to be found: \ \( \theta = \sqrt[3]{2} \ \)
02

Rewrite the expression

Rewrite \sqrt[3]{2} in a form that can be used in polynomial equation: \( \theta = \sqrt[3]{2} \implies \theta^{3} = 2 \ \)
03

Formulate the polynomial equation

The minimal polynomial can be expressed as \( f(\theta) = 0 \). In this case, \( f(x) = x^{3} - 2 \), because \sqrt[3]{2} \ satisfies this polynomial equation. So, \( f(\theta) = 0 \implies \theta^{3} - 2 = 0 \)
04

Verify polynomial is minimal

To ensure that \( x^{3} - 2 \) is minimal, one must confirm that it is irreducible over the rational numbers (which is true if it cannot be factored further into polynomials with rational coefficients). \( x^{3}-2 \) is indeed irreducible over \mathbb{Q}\, ensuring it is the minimal polynomial for \sqrt[3]{2} \.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cubic Equations
A cubic equation is a polynomial equation of degree three. This means it is of the form \( ax^3 + bx^2 + cx + d = 0 \), where \( a \), \( b \), \( c \), and \( d \) are constants, and \( a \) is not zero because that would reduce the degree.
Here, we are working with the cubic polynomial \( x^3 - 2 \).

In our case, \( a = 1 \), \( b = 0 \), \( c = 0 \), and \( d = -2 \). 路
The solution to the cubic equation \( x^3 - 2 = 0 \) is \( x = \sqrt[3]{2} \).

Cubic equations can have one real root and two complex roots. Even if they are irreducible, real solutions are still possible.
Irreducible Polynomial
An irreducible polynomial is a polynomial that cannot be factored into polynomials of lower degree with rational coefficients.
It's essential when finding the minimal polynomial, to verify its irreducibility.

Our polynomial in question is \( x^3 - 2 \). To check if it is irreducible:
  • This polynomial does not factor over the rationals.
  • There are no rational roots (using the Rational Root Theorem).
  • The polynomial has a degree of 3, which is not decomposable into lower degree polynomials with rational coefficients.
This confirms \( x^3 - 2 \) is irreducible over \( \mathbb{Q} \).
Thus, it qualifies as the minimal polynomial for \( \sqrt[3]{2} \).
Roots and Coefficients
Understanding the relationship between the roots and coefficients of a polynomial is crucial.

For a polynomial \( ax^3 + bx^2 + cx + d \),
  • The sum of the roots is \( -\frac{b}{a} \).

  • The sum of the product of the roots taken two at a time is \( \frac{c}{a} \).

  • The product of the roots is \( -\frac{d}{a} \) if the degree is odd and \( \frac{d}{a} \) if even.


Applying this to the polynomial \( x^3 - 2 \), we see:
  • Sum of roots: 0

  • Sum of products of roots taken two at a time: 0

  • Product of the roots: 2

These relationships help verify the calculations and give insights into polynomial properties, ensuring the minimal polynomial is correctly determined.

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Most popular questions from this chapter

Let \(\alpha\) be an algebraic number. Show that there is an integer \(n\) such that \(n \alpha\) is an algebraic integer.

Show that \(\left|\sum_{i=m}^{n}(t / p)\right|<\sqrt{p} \log p .\) The inequality holds for the sum over any range. This remarkable inequality is associated with the names of Polya and Vinogradov. [Hint: Use the relation \((t / p) g=g_{1}\) and sum. The inequality \(\sin x \geq(2 / \pi) x\) for any acute angle \(x\) will be useful.]

If \(\alpha\) and \(\beta\) are algebraic integers, prove that any solution to \(x^{2}+\alpha x+\beta=0\) is an algebraic integer. Generalize this result.

Let \(x^{2}+m x+n \in \mathbb{Z}[x]\) be irreducible and \(\alpha\) be a root. Show that \(\mathbb{O}[\alpha]=\) \(\\{r+s x \mid r, s \in Q\\}\) is a ring (in fact, it is a field). Let \(m^{2}-4 n=D_{0}^{2} D\), where \(D\) is square-free. Show that \(Q[\alpha]=Q[\sqrt{D}]\).

A polynomial \(f(x) \in \mathbb{Z}[x]\) is said to be primitive if the greatest common divisor of its coefficients is 1 . Prove that the product of primitive polynomials is again primitive. [fint: Let \(f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n}\) and \(g(x)=b_{0} x^{m}+b_{1} x^{n-1}+\cdots+b_{m}\) be primitive. If \(p\) is a prime, let \(a_{i}\) and \(b_{j}\) be the coefficients with the smallest subscripts such that \(p X a_{i}\) and \(\left.p\right\\} b_{j}\). Show that the coefficient of \(x^{i+j}\) in \(f(x) g(x)\) is not divisible by \(p\).] This is one of the many results known as Gauss" lemma.
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