Quadratic subfield of cyclotomic field

Question

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.

Answer 1

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.]

Answer 2

Hint. The quadratic Gauss sum $S=\sum_{i-1}^{p-1}\left(\frac{i}{p}\right)\zeta_p^i$ satisfies $S^2=\left(\frac{-1}{p}\right)p$, where $\left(\frac{a}{p}\right)$ is the Legendre symbol.

Answer 3

Below is a not-so-arithmetic proof. It can be found, for example, in this Outline article, and is not my original idea.

First determine explicitly the galois group of the $p$-th cyclotomic field. Then find its unique subgroup of index $2$.Hence compute the fixed field of that subgroup. You shall end up with the quadratic gauss sum as in the answer by Warren Moore. So the task is now reduced to showing the proposed equality. Now let the notations be as in the answer by Moore.

Then we evaluate the gauss sum. Write
$$S^2=\sum_{i=1}^{p-1}\sum_{j=1}^{p-1}\left(\dfrac{ij}{p}\right)\zeta_p^{i+j}=\sum_i\sum_{j\equiv ik\pmod p}\left(\dfrac{i^2k}{p}\right)\zeta_p^{i+ik}$$
$$=\sum_i\sum_k\left(\dfrac{k}{p}\right)\zeta_p^{i(1+k)}=\sum_{k=1}^{p-1}\left(\dfrac{k}{p}\right)\sum_i\zeta_p^{i(1+k)}$$.

When $p\not\mid(1+k)$, the latter sum is -1; else the sum is $p-1$. So the gauss sum turns out to be

$$-\sum_{k=1}^{p-2}\left(\dfrac{k}{p}\right)+\left(\dfrac{-1}{p}\right)(p-1)=-\sum_{i=1}^{p-1}\left(\dfrac{k}{p}\right)+\left(\dfrac{-1}{p}\right)p$$.

As a last step, show that the final sum becomes the desired result.

In addition, one could avail, either of the theory of ramifications, in algebraic number theory, or of the class field theory, to give different proofs. But I suppose that they are not so appropriate to your question, right?
In any case, hope you enjoy the proof.
Inform me of any errors, thanks in advance.