Let $\alpha \in \mathbb{C}$ a root of $f(x)=x^3-3x-1$. Prove $f$ is irreducible over $\mathbb{Q}$.

Question

I encountered the following claim:

Let $\alpha \in \mathbb{C}$ a root of $f(x)=x^3-3x-1$. Prove $f$ is irreducible over $\mathbb{Q}$.

The explanation included something of the form:

Since $f$ has degree $3$ then it must include either two polynomials of degree one, or one of degree one and one of degree two.

Say it is the latter: $f(x)=(x-\alpha)(x^2+ax+b)$. Then if there is a root $m/n \in \mathbb{Q}$ it must be either $1$ or $-1$, which clearly cannot satisfy $f(x)=0$.

I cannot see how the conclusion of it being in $\{-1,1\}$ was obtained (it mentioned something about $m$ dividing $-1$ and $n$ dividing the leading $1$).

Any help would be appreciated.

Answer 1

The rational root theorem is used.

It says that if $p/q$ is a rational root of a polynomial in one variable with integer coefficients such that $p$ and $q$ are coprime, then $p$ divides the constant term and $q$ divides the leading coefficient.