# Calculate the degree of $[\mathbb Q(a):\mathbb Q]$

How to determine $[\mathbb Q(a):\mathbb Q]$ for $a\in\overline{\mathbb Q}\setminus\mathbb Q$ with $a^p=1$ with $p$ prime?

So $a$ is an algebraic, complex number which cannot be written as a quotient. It is algebraic and so it is the root of a polynomial $\in\mathbb Q[X]$, for example $f(X)=X-1$ as $a^p=1$. The degree can be determined by finding the minimal polynomial and the degree of the minimal polynomial will be the degree of the field extension. But what is the minimal polynomial?

If $\alpha$ is a primitive root of the unit then it is root of $$f(X) = X^{p-1} + X^{p-2} + \ldots + X + 1$$
which is irreducible over $\mathbb{Q}$. Can you take it from here?
• So it is the minimal polynomial and the degree of the field extension is $p-1$? Don't we have to use $a\in\overline{\mathbb Q}\setminus\mathbb Q$ somewhere? – user149868 Nov 30 '14 at 13:49
• Well, given $\alpha$ as it is we'd have, say, $\alpha = \cos \frac{2\pi}{p} + i\sin \frac{2\pi}{p} \neq 1$. And yes, the degree would be $p-1$. – Aaron Maroja Nov 30 '14 at 13:51