TITLE: Quadratic subfield of cyclotomic field
QUESTION [12 upvotes]: Let $p$ be prime and let $\zeta_p$ be a primitive $p$th root of unity. Consider the quadratic subfield of $\mathbb{Q}(\zeta_p)$. For instance, for $p=5$ we get the quadratic subfield to be $\mathbb{Q}(\sqrt5)$ obtained by the relation between the golden ratio and $\cos(2\pi/5)$ (the minimal polynomial of the golden ratio $\phi$ is $x^2-x-1$ and so the extension is quadratic). 
Also $\cos(\pi/5)=\phi/2\implies \cos(2\pi/5)=\phi^2/2-1$ by double angle identity. 

Is it true that in general the quadratic extension is $\mathbb{Q}(\sqrt{\pm p})$ where $\pm$ depends on whether $p$ is $1$ or $-1$ $\pmod 4$? 

Please prove or disprove.

REPLY [13 votes]: Here is another approach, of a more algebraic number-theoretic nature.
A basic fact is that any finite extension $K$ of $\mathbb Q$ has an invariant
attached to it, a number, called its discriminant.  As the name suggests, 
it is related to discriminants of polynomials.
It is defined as follows: if $\alpha \in K$, we say $\alpha$ is an algebraic integer of $K$ if the minimal polynomial of $\alpha$ (i.e. the unique irreducible monic polynomial in $\mathbb Q(x)$ that has $\alpha$ as a root) has integer coefficients.  (So e.g. $i$ is an algebraic integer in $\mathbb Q(i)$, but $i/2$ isn't.  Algebraic integers are to $K$ as usual integers are to $\mathbb Q$.)
If $\alpha \in K$ is an algebraic integer, then the discriminant of its minimal polynomial is an integer, which we refer to as the discriminant of $\alpha$.  We define the discriminant of $K$ to be the g.c.d of
the discriminants of the $\alpha$, where $\alpha$ runs over all algebraic integers for which $K = \mathbb Q(\alpha)$.  (If one wants to be subtle, one
can include a sign in the discriminant, but I won't worry about that here,
so everywhere that I talk about the discriminant of a polynomial, I should really
be talking about its absolute value.)
E.g. if $\zeta_p$ is a $p$th root of $1$, so that $L := \mathbb Q(\zeta_p)$ is the $p$th cyclotomic field, then it turns out that the discriminant of $L$ is actually equal to the discriminant of $\zeta_p$, which is equal to $p^{p-2}$.
E.g. the discriminants of quadratic extensions of $\mathbb Q$ are easy to compute, and if $p$ is an odd prime, then there is a unique quadratic field
whose discriminant is divisible by no prime but $p$, namely $\mathbb Q(\sqrt{\pm p})$, where the sign is chosen so that $\pm p \equiv 1 \bmod 4$.  (And in fact
the discriminant is then exactly equal to $p$.)
One property of discriminants is that if $K$ is a subfield of $L$,
then the discriminant of $L$ is divisible by the discriminant of $K$ raised to
the power of the degree $[L:K].$
Combining this fact with the preceding two examples, we see that if
$K$ is a quadratic subfield of $\mathbb Q(\zeta_p)$, then its discriminant is divisible by
only prime, namely $p$, and  this uniquely determines $K$ to be $\mathbb Q(\sqrt{\pm p})$ (where $\pm p \equiv 1 \bmod 4$).
[If you know no algebraic number theory, this will seem like hocus-pocus; but 
in fact it doesn't take more than the most basic algebraic number theory to fill in all the details.]