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

Question

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.

Answer 1

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.

Answer 2

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 $x^2=1+3y^2 > 1$.

Answer 3

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.

Answer 4

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.