Factorise $x^6+1$ in rational field vs real field Recently our class has gone through complex numbers, but I’m still confused about the differences between the rational field $\Bbb Q$ and the real field $\Bbb R$. There were some questions where we have to factorise equations over these fields.
For instance, “factorise $x^6+1$.” (I feel like many people here will find this easy, but I've just started this topic two days ago...)
I could factorise it into $\left(x^2+1\right)\left(x^4-x^2+1\right)$, but does this belong to the real or rational field? And how do I factorise it in the remaining field?
 A: In the real field $x^4-x^2+1$ is further factorizable into
$$x^4-x^2+1=(x^2+1)^2-3x^2=(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1).$$
In the rational field $x^4-x^2+1$ is not factorizable. In the complex field, both $x^2\pm\sqrt{3}x+1$ and $x^2+1$ are factorizable into first-degree polynomials.
A: So far, your have factorized $x^6 + 1$ in both the rationals and the reals, though you haven't factorized it into irreducible polynomials. This are the ones that cannot be expressed as the product of two other non-constant polynomials with coefficients in the field. Since $x^2 + 1$ has roots $-i$ and $i$, it's factorization in $\mathbb{C}$ is $(x-i)(x+i)$. These are not polynomials with real or rational coefficients, so $x^2 +1$ is in fact irreducible in $\mathbb{Q}$ and $\mathbb{R}$. As for the other factor, you can check that:
$$
x^4-x^2+1=(x^2+1)^2-3x^2=(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)
$$
(If this feels like cheating, set $w = x^2$, solve a quadratic equation, and then "go back" to $x^2$). By calculating the discriminant, you can see that these two quadratic polynomials have no real roots and therefore cannot be further factorized in $\mathbb{R}$. This shows that the second factor can be further factorized in $\mathbb{R}$ but not in the rationals, since it decomposes into polynomials that don't have all rational coefficients. We're left with the following decompositions:
$(x^2+1)(x^4-x^2+1)$, in $\mathbb{Q}$
$(x^2+1)(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)$, in $\mathbb{R}$
