Why is $x^2-2ux+1$, where $u = \cos(\frac{2\pi}{n})$, irreducible in $\mathbb Q(u)$?

Question

My textbook states that $x^2-2ux+1$, where $u = \cos(\frac{2\pi}{n})$ for some $n \in \mathbb N$ is clearly irreducible in $\mathbb Q(u)$.

Is this obvious?

I tried to write it as a product of linear factors, but I found this difficult because I do not know what an element in $\mathbb Q(u)$ looks like.

Any insights?

Answer 1

If we in fact have $\;u=\cos\frac{2\pi}n\;$ , then the quadratic's discriminant is

$$\Delta:=b^2-4ac=\cos^2\frac{2\pi}n-4<0$$

and we're done...