Homomorphism from inertia group to $(\mathcal{O}_L/Q)^*$

Question

(This is exercise 21.b in Chapter 4 of Marcus' Number Fields)

Let $L/K$ be a normal extension and let $Q$ be a prime ideal of $\mathcal{O}_L$.

Fixing $\pi \in Q - Q^2$ and considering an automorphism $\sigma$ in the inertia group $E(Q)$, we can prove that there is some $\alpha_\sigma \in \mathcal{O}_L$ (unique modulo $Q$) such that $\sigma \pi \equiv \alpha_\sigma \pi \pmod{Q^2}$.

Now we want to prove that the map $\sigma \mapsto \alpha_\sigma$ is an homomorphism from $E$ to $(\mathcal{O}_L/Q)^*$. So we need to prove that for $\sigma, \tau \in E$, $\sigma \tau \pi \equiv \alpha_\sigma \alpha_\tau \pi \pmod{Q^2}$.

If we had $\sigma (\tau \pi) \equiv \alpha_\sigma (\tau \pi) \pmod{Q^2}$ then we'd be done. I'm guessing this is true since it'd be nicer if the map $\sigma \mapsto \alpha_\sigma$ did not depend on $\pi$. But I'm not sure how to prove this from the definition of $\alpha_\sigma$.

Answer 1

From part (a) of the exercise, $\tau(\pi) \equiv \alpha_{\tau} \pi\ (Q^2)$; therefore $\sigma(\tau(\pi)) \equiv \sigma(\alpha_{\tau}) \sigma(\pi) \equiv \sigma(\alpha_{\tau}) \alpha_{\sigma}\pi\ (Q^2)$.

Since $\sigma \in E$, $\sigma(\alpha_{\tau}) \equiv \alpha_{\tau}\ (Q)$, therefore $\sigma(\alpha_{\tau})\pi \equiv \alpha_{\tau}\pi\ (Q^2)$. Combined with the above, this gives $\sigma(\tau(\pi)) \equiv \alpha_{\sigma}\alpha_{\tau}\pi\ (Q^2)$; combined with $(\sigma\tau)(\pi) \equiv \alpha_{\sigma\tau}\pi\ (Q^{2})$ this shows $\alpha_{\sigma}\alpha_{\tau} \equiv \alpha_{\sigma\tau}\ (Q)$.