prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$

Question

Prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$.
Let $\theta$ be a root of $f(x)$. Compute $\frac{1}{\theta}$ and $(2 + \theta^2)^{-1} $ in $\mathbb Q[\theta ]$.

\begin{array}{l} f\left( \theta \right) = \theta ^3 + 3\theta - 1 = 0 \\ \Leftrightarrow \theta \left( {3 + \theta ^2 } \right) = 1 \\ \Leftrightarrow \frac{1}{\theta } = \left( {3 + \theta ^2 } \right);\left( {\theta \ne 0} \right) \\ \end{array}

\begin{array}{l} \frac{1}{\theta } = 3 + \theta ^2 ;\left( {\theta \ne 0} \right) \\ \Leftrightarrow \frac{1}{\theta } - 1 = 2 + \theta ^2 \\ \Leftrightarrow \left( {\frac{1}{\theta } - 1} \right)^{ - 1} = \left( {2 + \theta ^2 } \right)^{ - 1} \quad ;\left( { \pm \sqrt 2 \notin Q} \right) \\ \Leftrightarrow \left( {2 + \theta ^2 } \right)^{ - 1} = \frac{\theta }{{1 - \theta }}\quad ;\left( {\theta \ne 1} \right) \\ \end{array}

But I can't show that $f$ is irreducible.

Answer 1

Let $$g(x)=f(x+1)=x^3+3x^2+3x+1+3x+3-1=x^3+3x^2+6x+3$$

Now we can apply Eisenstein's Criterion (with $p=3$) to find that $g(x)$ is irreducible in $\mathbb{Q}[x]$, so $f(x)$ is also irreducible in $\mathbb{Q}[x]$.

Answer 2

Another idea is that if $f(x)$ is factorable over $\Bbb Q[x]$ it must have at least one rational zero. However, by the rational zero theorem, $\pm 1$ are the only possible rational zeros, but neither one is a zero.

Answer 3

For polynomials of deg $\leq 3$ it is enough to check that there are no rational roots (i.e a linear factor). Applying the Rational Roots Theorem the set of possible roots are $\pm 1$, which clearly $g(\pm 1) \not = 0$.