Show that a polynomial is irreducible using Galois

Question

I need to show that the polynomial $X^4-3X^2+4$ is irreducible in $\mathbb{Q}[X]$. As Eisenstein's criterion fails and I hardly try to avoid something such as “Let's assume $X^4-3X^2+4$ is reducible, then $(X^2+aX+b)(X^2+cX+d) = X^4-3X^2+4$”.

My idea is that its roots $\pm\left(\sqrt{\frac{1}{2}(3\pm i \sqrt 7)}\right)$ and squared $\frac{1}{2}(3\pm i \sqrt 7)$ are obviously $\notin \mathbb{Q}$ (because if it was reducible, the degree $[\mathbb{Q}(\sqrt{\frac{1}{2}(3 + i \sqrt 7}): \mathbb{Q}]=2$, and no polynomial of degree $2$ then has $\sqrt{\frac{1}{2}(3 + i \sqrt 7)}$ as a root. My question is simply if this argument is valid.

Answer 1

We see that our polynomial has no rational roots.

Now, since coefficients before $x^3$ and before $x$ they are $0$, we have two cases only: $$x^4-3x^2+4=(x^2+px+2)(x^2-px+2)$$ or $$x^4-3x^2+4=(x^2+px-2)(x^2-px-2)$$ for $p\in\mathbb Z$ and easy to check that they are impossible even for a rational $p$.

Answer 2

Your argument is not valid, I'm afraid. The fact that the big number has not degree 2 is exactly equivalent to the given polynomial being irreducible.

The polynomial is reducible over $\mathbb{R}$ as $$ X^4-3X^2+4=X^4+4X^2+4-7X^4=(X^2+2)^2-7X^2= (X^2-\sqrt{7}X+2)(X^2+\sqrt{7}X+2) $$ Both these factors are irreducible over $\mathbb{R}$.

A factorization over $\mathbb{Q}$ is also a factorization over $\mathbb{R}$. Does $\sqrt{7}\in\mathbb{Q}$?