$f(x)=x^4+ax^2+b\in \mathbb{Z}[x]$ is irreducible iff $\alpha ^2, \alpha \pm\beta$ are not elements of $\mathbb{Q}.$

Question

Let $\pm \alpha, \pm \beta$ denote the roots of the polynomial $f(x)=x^4+ax^2+b\in \mathbb{Z}[x]$. Prove that $f(x)$ is irreducible over $\mathbb{Q}$ if and only if $\alpha ^2, \alpha \pm\beta$ are not elements of $\mathbb{Q}.$

$\Rightarrow$

Suppose $f(x)$ is irreducible. $f(x)=(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)=(x^2-\alpha^2)(x^2-\beta^2)$. So, $\alpha^2$ is not in $\mathbb{Q}$. $f(x)=(x-\alpha)(x+\alpha)(x-\beta)(x+\beta)=(x^2+(\alpha-\beta)x-\alpha\beta)(x^2-(\alpha-\beta)x-\alpha\beta)$. $\alpha\beta$ is in $\mathbb{Q}$ if $b$ is a square.

How to show $\alpha\pm\beta$ are not in $\mathbb{Q}?$

Answer 1

Since your polynomial has rational coefficients you have $$2(\alpha^2+\beta^2)=x_1^2+x_2^2+x_3^2+x_4^2=(x_1+x_2+x_3+x_4)^2-2(x_1x_2+...+x_3x_4)\in \mathbb Q$$

Now, if $\alpha \pm \beta \in \mathbb Q$ then

$$(\alpha \pm \beta)^2=\alpha^2+\beta^2\pm 2\alpha \beta \in \mathbb Q$$ and hence $$\alpha \beta \in \mathbb Q$$

This shows that at least one of the factors in your factorisation is in $\mathbb Q[X]$.

Answer 2

$$f(x)=(x^2-\alpha^2)(x^2-\beta^2)\implies a=-\alpha^2-\beta^2.$$ Suppose that $\alpha+\beta\in \mathbb{Q}$, then $$\alpha\beta=\frac{(\alpha+\beta)^2+a}{2}$$ is rational, which gives $$f(x)=(x^2-(\alpha+\beta)x+\alpha\beta)(x^2+(\alpha+\beta)x+\alpha\beta),$$ a contradiction. Similar for $\alpha-\beta\in\mathbb{Q}$.