$x^3-2x-2$ is irreducible over $\mathbb{Q}$

Question

I tried doing this by Eisenstein's criterion:

$2$ is prime in $\mathbb{Q}$ and I then proceeded to write that it divides $-2$, $-2$ and $0$ but doesn't divide $1$ and also $2^2=4$ doesn't divide $2$. I then noticed these last two are erroneous since we're in $\mathbb{Q}$ and we have that $(1/2)(2)=1$ and thus $2$ does divide $1$; also $(4)(1/2)=2$ so $4$ divides $2$. How could I have done it instead?

Answer 1

What you're missing here is the result known as "Gauss' lemma". The result says the following.

Theorem (Gauss' lemma)

Let $p(X) \in \mathbb{Z}[X]$ be a non-constant polynomial. Then $p(X)$ is irreducible over $\mathbb{Z}[X]$ if and only if it is irreducible over $\mathbb{Q}[X]$ and primitive in $\mathbb{Z}[X]$.

To be primitive just means that the gcd of its coefficients is equal to 1. Thus, one can state a version of Gauss' lemma as follows.

Theorem (Gauss' lemma version 2)

Let $p(X) \in \mathbb{Z}[X]$ be a non-constant polynomial and assume that $p(X)$ is primitive. Then $p(X)$ is irreducible over $\mathbb{Z}[X]$ if and only if it is irreducible over $\mathbb{Q}[X]$.

In your case, your polynomial $p(X) = X^3 - 2X - 2 \in \mathbb{Z}[X]$ is clearly primitive, so you can deduce irreducibility over $\mathbb{Q}$ from irreducibility over $\mathbb{Z}$ by applying Gauss' lemma along with your preferred irreducibility criterion over $\mathbb{Z}$.

Answer 2

You solve it by noting that the polynomial is monic, which means that a factorisation over the rationals gives a factorisation over the integers (or contrapositively: of the is no non-trivial factorisation over the integers, then there are none over the rationals). And over the integers you can use Eisenstein without any issues.

Answer 3

Suppose that $p/q\in\mathbb Q$ is a root of this polynomial, with $p,q$ coprime integers. Then you have that $$\frac{p^3}{q^3}-2\frac{p}{q}-2=0$$ or, by doing a bit of algebra, $$p^3=2q^2 (p+q)$$ from which you can derive a contradiction.