Factoring a polynomial

The key here is to notice that $x^4 - 10x^2 + 1$ is the minimal polynomial of $\alpha = \sqrt{2} + \sqrt{3}$ over $\mathbf Q$ , which has four conjugates in $\mathbf Q(\sqrt{2}, \sqrt{3})$ by Galois theory. Thus, it is irreducible. To examine the situation modulo primes, we note that $\alpha$ has degree 2 over all intermediate fields, which are $\mathbf Q(\sqrt{2}), \mathbf Q(\sqrt{3}), \mathbf Q(\sqrt{6})$ . Being an algebraic integer, its minimal polynomial over all of these fields has algebraic integer coefficients, therefore they can be reduced modulo primes. We may explicitly compute the factorizations as follows:

$x^4 - 10x^2 + 1 = \prod (x \pm \sqrt{2} \pm \sqrt{3}) = (x^2 - (\sqrt{2} + \sqrt{3})^2)(x^2 - (\sqrt{2} - \sqrt{3})^2) = (x^2 - 5 - 2\sqrt{6})(x^2 - 5 + 2\sqrt{6})$ $x^4 - 10x^2 + 1 = ( (x + \sqrt{2})^2 - 3 )( (x - \sqrt{2})^2 - 3)$ $x^4 - 10x^2 + 1 = ( (x + \sqrt{3})^2 - 2)( (x - \sqrt{3})^2 - 2)$

Now, the final observation is that modulo a prime, one of 2, 3 or 6 must be a quadratic residue. This follows from the multiplicativity of the quadratic character: if 2 and 3 are not residues, then $\chi(6) = \chi(2)\chi(3) = (-1)(-1) = 1$ , so 6 is a residue. Therefore, at least one of these factorizations can be reduced modulo any prime, therefore our polynomial $x^4 - 10x^2 + 1$ is reducible modulo every prime, whereas it is irreducible over $\mathbf Z$ .