# Quadratic Fields and Gauss Sums

I am given the p-cyclotomic field as $$K_p = \frac{\mathbb{Q}[x]}{(\Phi_p(x))}$$ where $$\Phi_p(x) = x^{p-1} + x^{p-2} + ... + x + 1$$

Then, a quadratic subfield L is defined such that $$\mathbb{Q} \subset L \subset K_p$$ with $$[L : \mathbb{Q}] = 2$$.

The Gauss sum is then defined as:

$$\tau = \sum_{a\in (\mathbb{Z} / p \mathbb{Z})^\times} \left(\frac{a}{p}\right) \zeta^a$$

where the $$\left(\frac{a}{p}\right)$$ is the Legendre symbol.

I have shown that $$\tau^2 = \left(\frac{-1}{p}\right) p$$.

I am asked to show that this sum $$\tau$$ is contained in the quadratic subfield L. My main confusion comes from the concept of L. Not really sure what an element of L would look like. I have read about how L is simply $$\mathbb{Q}$$ extended with some squareroot of a square free number. However, I do not see why this number must be p.

In other words, why must $$L = \mathbb{Q}(\sqrt p)$$, and even if this is the case, how does this take into account for the fact that if $$\left(\frac{-1}{p}\right) = -1$$, then $$\tau = i\sqrt p$$.

Any help/info would be very much appreciated.

• Hint: The reason the subfield $L$ exists and is unique is Galois theory, there is a Galois theoretic description of the elements of $L$ in terms of the Galois group of $K_p$ (which is a cyclic group!). So you can check this condition on $\tau$. – Alex J Best Mar 19 at 3:10
• The hard part is the calculation of ${\tau}^2$ (which you did). The very definition of $\tau$ shows that it belongs to $K_p$, so $Q(\tau)$ is a quadratic field contained in $K_p$. Besides, it is known that $K_p/Q$ is Galois, with Galois group $(Z/pZ)^*$, which is cyclic, hence $K_p$ admits a unique quadratic field $L$. The uniqueness shows that $L=Q(\tau)$. – nguyen quang do Mar 20 at 8:21