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

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 $$\mathbb{Q}?$$

• It is an act unfriendly to aged mathematicians with failing eyesight to use both $a$ and $\alpha$ in a formula or proposition. – Lubin Apr 18 '17 at 23:03

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 $\mathbb Q[X]$.
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