product considering the period of index over a cyclotomic extension

Question

This is an exercise from Milne Galois Theory (Chapter 3 Exercise 13).

Let p be an odd prime, and let $\zeta$ be a primitive $p^{\text{th}}$ root of 1 in $\mathbb C$. Let $E = \mathbb Q[\zeta ]$, and let $G = Gal(E/Q)$; thus $G = (Z/(p))^×$. Let H be the subgroup of index 2 in G. Put

$$\alpha=\sum_{\sigma \in H}{\sigma(\zeta)}$$

$$\beta=\sum_{\sigma \in G-H}{\sigma(\zeta)}$$

Show:

• $\alpha$ and $\beta$ are fixed by H:
• if $\sigma \in G-H$, then $\sigma\alpha = \beta, \sigma \beta = \alpha$. Thus $\alpha$ and $\beta$ are roots of the polynomial $X^2 + X + \alpha\beta \in \mathbb Q [X]$.

Finally compute $\alpha \beta$ and show that the fixed field of $H$ is $\mathbb Q[\sqrt{p}]$ when $p \equiv 1 \mod 4$ and $\mathbb Q [\sqrt{−p}]$ when $p=3 \mod 4$.

What I did:

I proved the first and second parts of the exercise (I wrote these parts anyway because they are probably useful for the final part)

I don't know how to compute $\alpha \beta$. Does anyone know how to do it in the most simple way?

Answer 1

There may be something simpler out there, but this is what I got by rubbing those three grey cells together real hard. I started out by doing it by hand in the cases $p=5$ and $p=7$. I was paying attention to how many times each power of $\zeta$ occurs in the product. It may be a good idea that you do the same, and see if you have a light bulb moment. If not, then do the same with $p=11$ and $p=13$. You can probably form at least a conjecture of what happens. Read further, if you're stuck.

When we view $G$ as $\Bbb{Z}_p^*$, the elements of $H$ are the (non-zero) quadratic residues modulo $p$, and the elements of $G\setminus H$ are the quadratic non-residues. So all the $(p-1)^2/4$ terms in the product $\alpha\beta$ are of the form $\zeta^{i+j}$ where $i\in H$ and $j\notin H$.

The observation I want to make is that each $k\in\Bbb{Z}_p^*$ occurs as the sum $i+j$ as often as $1$. Let $S_k$ be the set of pairs $(i,j)$ of the prescribed type such that $i+j=k$. The claim is equivalent to stating that $|S(k)|=|S(1)|$ whenever $k\in\Bbb{Z}_p^*$.

• If $k\in H$, then $(i,j)\mapsto (ki,kj)$ is a bijection from $S(1)$ to $S(k)$.
• If $k\notin H$, i.e. $k$ is a quadratic non-residue, then $(i,j)\mapsto (kj,ki)$ is a bijection from $S(1)$ to $S(k)$.

Because the sum of all non-trivial powers of $\zeta$ is equal to $-1$, we can already deduce that $$ \alpha\beta=|S(0)|-|S(1)|. $$ What about $S(0)$? Here there will be a difference according to the residue class of $p$ modulo $4$. Undoubtedly you know that $-1$ is a quadratic residue if and only if $p\equiv1\pmod4$.

• If $p\equiv1\pmod4$, then $-1\in H$. But if $i+j=0$, then $j=-i=(-1)i$. This is a contradiction because the product of two QRs is another QR. Therefore $S(0)$ is the empty set in this case.
• If $p\equiv3\pmod4$, then $-1\notin H$. So to each $i\in H$ we have $-i\notin H$. Therefore $$S(0)=\{(i,-i)\mid i\in H\}$$ and consequently $|S(0)|=(p-1)/2$.

The total number of pairs $(i,j)$ is clearly $$ \frac{(p-1)^2}4=|S(0)|+(p-1)|S(1)|. $$

Therefore

• When $p\equiv1\pmod4$ we get that $|S(1)|=(p-1)/4$ and consequently $$\alpha\beta=-\frac{p-1}4.$$
• When $p\equiv3\pmod4$ we get that $|S(1)|=(p-3)/4$ and consequently $$\alpha\beta=\frac{p-1}2-\frac{p-3}4=\frac{p+1}4.$$

I'm sure you can solve the equation $X^2+X+\alpha\beta=0$, and see which quadratic field is generated by $\alpha$ (or $\beta$).