Question

Let \(f \in \mathbb{Z}[x]\) be monic, and \(\alpha \in \mathbb{Q}\) be a root of \(f\). Show that \(\alpha \in \mathbb{Z}\).

Short answer

The root \(\alpha\) is an integer because the denominator must divide 1.

Step-by-step solution

Understanding the Problem

We need to show that if a monic polynomial \( f(x) \) with integer coefficients has a rational root \( \alpha \), then this root must actually be an integer.

Expressing the Root

Since \( \alpha \in \mathbb{Q} \), we can write \( \alpha = \frac{p}{q} \) where \( p \) and \( q \) are integers with \( q eq 0 \) and \( \gcd(p, q) = 1 \).

Applying the Root Theorem

Substitute \( \alpha = \frac{p}{q} \) into the polynomial \( f(x) \). Since \( f(x) \) is monic, the leading coefficient is 1. The polynomial has the form \( f(x) = a_n x^n + a_{n-1} x^{n-1} + \, ... \, + a_1 x + a_0 \) with \( a_n = 1 \).

Clearing the Denominator

We have \( f\left(\frac{p}{q}\right) = 0 \). Multiply through by \( q^n \) to clear the denominator, resulting in an equation of the form \( p^n + a_{n-1} p^{n-1} q + \, ... \, + a_0 q^n = 0 \).

Using Gauss's Lemma

Gauss's Lemma tells us that if a polynomial with integer coefficients has a rational root \( \frac{p}{q} \), then \( q \) must divide the leading coefficient of the polynomial. Since the polynomial is monic, its leading coefficient is \( 1 \). Hence, \( q \mid 1 \).

Concluding the Argument

Since \( q \mid 1 \), we have \( q = 1 \). Therefore, \( \alpha = \frac{p}{1} = p \). So, \( \alpha \) is an integer.

Key concepts

Rational Root Theorem

The Rational Root Theorem is a fundamental tool in algebra that helps determine potential rational roots of a polynomial with integer coefficients. According to this theorem, if a polynomial has a rational root \(\alpha = \frac{p}{q}\) (where \(p\) and \(q\) are integers with no common factors other than 1), then \(p\) should evenly divide the constant term of the polynomial and \(q\) should divide the leading coefficient. This makes it a fantastic resource for swiftly eliminating many impossible options when seeking rational solutions.
In the case of a monic polynomial, where the leading coefficient is 1, if \(q\) must divide the leading coefficient, it means \(q\) can only be 1 or -1. Therefore, any rational root \(\alpha\) must necessarily be an integer. The theorem simplifies the quest for roots by reducing the problem to considering only integer divisors of the constant term.

Gauss's Lemma

Gauss's Lemma is a pivotal concept in understanding polynomials with integer coefficients and their rational roots. It asserts that if a polynomial with integer coefficients has a root that can be expressed as a rational number \(\frac{p}{q}\), then \(q\) must divide the leading coefficient of the polynomial. This lemma reminds us that rational roots don't exist independently—they are closely tied to the integer coefficients of the polynomial.
When applied to monic polynomials, where the leading coefficient is always 1, Gauss's Lemma implies that the denominator \(q\) of a rational root \(\alpha\) must divide 1. Consequently, this forces \(q\) to be either 1 or -1, meaning any possible rational root \(\alpha\) must actually be an integer.

Integer Coefficient Polynomials

Understanding integer coefficient polynomials is critical because they simplify the search for roots in polynomial equations. These are polynomials where every coefficient of the terms is an integer. Such polynomials are pervasive in algebra and number theory due to their structured nature.
When dealing with integer coefficient polynomials, especially monic ones, the relationship between the coefficients and the roots is typically studied through the Rational Root Theorem and Gauss's Lemma. One of the important properties of these polynomials is how they allow us to leverage these theorems fully. By scrutinizing the possible roots to only integers, we can make more direct calculations and predictions with ease. Hence, finding the roots for integer coefficient polynomials is often more about basic integer arithmetic and less about complex algebraic manipulations.