# Prove $X^4-2X^2+4$ is irreducible in $\mathbb{Q}[X]$

For some problem from my Galois Theory course, I need to prove that the polynomial $$X^4-2X^2+4$$ is irreducible in $$\mathbb{Q}[X]$$.

I know it has no roots in $$\mathbb{Q}$$ (by rational root theorem), but I can't conclude it's irreducible with just that fact because it's degree is $$4$$, and tried several changes to try to use Eisenstein Criterion but I had no luck.

How can I prove it? Is it really irreducible in $$\mathbb{Q}[X]$$? Thanks in advance, any help will be appreciated.

If it was reducible, you would be able to write it as $$(X^2+aX+b)(X^2-aX+c)$$. But$$(X^2+aX+b)(X^2-aX+c)=X^4+(-a^2+b+c)X^2+a(c-b)X+bc.$$So, you would have$$\left\{\begin{array}{l}-a^2+b+c=-2\\a(c-b)=0\\bc=4.\end{array}\right.$$So, $$a=0$$ or $$b=c$$. If $$a=0$$, then $$b+c=-2$$ and $$bc=4$$, and you can easuly check that this system has no rational solutions. And if $$b=c$$, you get $$-a^2+2b=-2$$ and $$b^2=4$$. Again, you can check that this system has no rational solutions.

If the polynomial isn’t reducible, then there is an algebraic number $$\alpha$$ with minimal polynomial of degree $$2$$ over $$\mathbb{Q}$$ such that $$\alpha^4-2\alpha^2+4=0$$. This means $$(\alpha^2-1)^2=-3$$, so $$\alpha^2=1 \pm i\sqrt{3}$$. In particular, we can write $$\alpha=x+iy\sqrt{3}$$ with $$x,y$$ rationals and thus $$x^2-3y^2=1, 2xy=\pm 1$$. So in any case, $$x^2-3/(4x^2)=1$$, so $$4x^4-4x^2-3=0$$. So if $$x=p/q$$, by the rational root theorem, $$p^2|3$$ and $$q^2|4$$, so that $$x^2=1$$ or $$x^2=1/4$$. But that’s impossible since $$y \neq 0$$ so $$x^2=1+3y^2 > 1$$.

Let $$\alpha=\sqrt{\frac{2+\sqrt{-12}}{2}}$$ and $$\beta=\sqrt{\frac{2+\sqrt{-12}}{2}}$$ (more precisely I choose a square root of each complex number).

Then $$P=X^4-2X^2+4=(X-\alpha)(X-\beta)(X+\beta)(X-\beta)$$.

Hence if $$P$$ has a monic factor of degree $$2$$ with rational coefficients, it is a product of two linear factors amongst these four factors. Looking at the constant terms, this means that $$-\alpha^2,-\beta^2$$ or $$\alpha\beta$$ lies in $$\mathbb{Q}$$ . This would implies in turn that $$\sqrt{-10}$$ or $$\sqrt{2}$$ are rational, which is not the case.

Over the real numbers, we can factorize: $$x^4 - 2x^2 + 4 = x^4 + 4x^2 + 4 - 6x^2 = (x^2+2)^2 - (\sqrt 6 x)^2 = (x^2 + \sqrt 6 x + 2) (x^2 - \sqrt6 x + 2)$$

The discriminant of the factors are $$(\pm \sqrt 6)^2 - 4 (1)(2) = -2 < 0$$, so they are irreducible.

Any irreducible rational factor of $$x^4 - 2x^2 + 4$$ has to be a product of the real factors, so it can only be $$x^4 - 2x^2 + 4$$ itself.