Finding a polynomial that proves $\sqrt6$ is an algebraic number

Question

So the full question is:

A number $\alpha \in \mathbb R$ is said to be algebraic if there exists a non-zero polynomial $p(x)$ with integer coefficients such that $p(\alpha) = 0$. Show that $\sqrt 6$ is an algebraic number.

My initial thoughts are that a polynomial like $p(x) = x^2 - 6$ would work since it would equal $0$. This seems too simple though (especially with regards to the other problems in the set). Am I missing something here?

Thank you.

Answer 1

No, you are missing nothing. Let $p(x)=x^2-6$. Then, since

• $p(x)\neq0$;
• the coefficients of $p(x)$ are integers;
• $\sqrt6$ is a root of $p(x)$,

$\sqrt6$ is an algebraic number. Yes, it's as simple as that.

Answer 2

Sometimes things in math seem too simple only because they are phrased in counter-intuitive ways.

First, what is the zero polynomial? According to Mathworld,

The constant polynomial $P(x) = 0$ whose coefficients are all equal to $0$. The corresponding polynomial function is the constant function with value $0$, also called the zero map. The zero polynomial is the additive identity of the additive group of polynomials.

Then $p(x) = 0x^3 + 0x^2 + 0x + 0$ is a manifestation of the zero polynomial, while $p(x) = x^2 - 6$ is a non-zero polynomial. The coefficients of the former are all $0$, while the coefficients of the latter are $0$ infinitely many times then $1, 6$.

As you have already verified, the principal root of $x^2 - 6$ is $x = \alpha = \sqrt{6}$ (the other is $-\alpha$). Then $\alpha$ is indeed an algebraic integer.

Try these in Wolfram Alpha: Exponent[x^2 - 6, x] and Exponent[0x^3 + 0x^2 + 0x + 0, x]. The former should evaluate to the meaningful $2$, the latter to the not so meaninful $-\infty$.