# Relationship between Frobenius homomorphism and legendre symbol for $K/\mathbb{Q}$ a quadratic extension

Let $K/\mathbb{Q}$ be a quadratic extension and d its discriminant. Show that $Frob_p(\sqrt{d}) = \Big( \frac{d}{p}\Big)\sqrt{d}$. I am not sure what the Frobenius homomorphism means. I was told that it is the preimage of the usual frobenius homomorphism in the reduction homomorphism $D_{\beta}\to Gal(\frac{O_K/\beta}{\mathbb{F}_p})$ where $D_\beta$ is the decomposition group of an ideal that contains p. I don't really understand how you actually calculate this homomorphism.

## 1 Answer

The homomorphism $f:D_\beta\to \mathrm{Gal}((\mathscr{O}_K/\beta)/\mathbb{F}_p)$ is calculated as follows. For any $\overline{a}\in \mathscr{O}_K/\beta$, let $a\in \mathscr{O}_K$ be any representative of $\overline{a}$, then for $\tau\in \mathrm{Gal}(K/\mathbb{Q})$, we simply let $f_\tau:=f(\tau)$ act by $f_\tau(\overline{a})=\tau(a)$, which is well-defined from the fact that $\tau(\beta)=\beta$ for all $\tau\in D(\beta)$. In other words, the action in on residue field $\mathscr{O}_K/\beta$ is just determined by looking at the action on $\mathscr{O}_K$ and reducing everything mod $\beta$.

When $p$ is unramified, this $f$ is an isomorphism, so we can speak explicitly of the element $f^{-1}(\mathrm{Frob}_\beta)\in D(\beta)$, which is what is being used in this exercise. When $p$ is ramified, the map $f$ has nontrivial kernel, and fits into a short exact sequence $$0\to I(\beta)\to D(\beta)\xrightarrow{f} \mathrm{Gal}((\mathscr{O}_K/\beta)/\mathbb{F}_p)\to 0$$ where $I(\beta)$ is called the inertia group of $\beta$, so pulling back the Frobenius element only describes a coset of $I(\beta)$, so there is some ambiguity. For this reason, when one speaks of "the" Frobenius element, it assumes the unramified case (this is obvious here: if $p$ is ramified we know $p\mid d$, so $\left(\frac{p}{d}\right)=0$, and the equality you ask to prove would be nonsense). From this point, the solution to your problem follows quickly if you think about the reduction of $\sqrt{d}$ mod $\beta$, and you should stop reading here and try to find it yourself.

Now let $p$ be unramified, and from the way we have defined the homomorphism $f$, we can calculate the action of $\mathrm{Frob}_p$ on $\sqrt{d}$ by reducing mod $\beta$. Denote by $\overline{\sqrt{d}}$ the reduction of $\sqrt{d}$ mod $\beta$. Then by definition, $f^{-1}(\mathrm{Frob}_p)\in \mathrm{Gal}(K/\mathbb{Q})$ acts nontrivially (i.e. by negation) on $\sqrt{d}$ iff $\mathrm{Frob}_\beta\in \mathrm{Gal}((\mathscr{O}_K/\beta)/\mathbb{F}_p)$ acts nontrivially on $\overline{\sqrt{d}}$.

We know that $\mathrm{Frob}_\beta$ generates $\mathrm{Gal}((\mathscr{O}_K/\beta)/\mathbb{F}_p)$, so the condition that $\mathrm{Frob}_\beta$ act trivially on $\overline{\sqrt{d}}$ is exactly the condition that $\overline{\sqrt{d}}\in \mathbb{F}_p$. But this must mean $d$ is a square mod $p$, so $\left(\frac{p}{d}\right)=1$. Likewise, if $\overline{\sqrt{d}}\not\in \mathbb{F}_p$, then $\mathrm{Frob}_\beta$ acts nontrivially on $\overline{\sqrt{d}}$, so $f^{-1}(\mathrm{Frob}_\beta)$ acts nontrivially on $\sqrt{d}$.