# Frobenius automorphism only defined on valuation ring?

Let $K$ be a local field and $\tilde{K}/K$ be the maximal unramified extension of $K$.

Why do authors define the Frobenius automorphism $\varphi$ of $\tilde{K}$ simply by saying \begin{equation} a^{\varphi}\equiv a^{q}\pmod{ \tilde{\mathfrak{p}}},\quad a\in\mathcal{O}_{\tilde{K}} \end{equation} where $\mathbb{F}_{q}$ is the residue field of $K$, $\mathcal{O}_{\tilde{K}}$ the valuation ring of $\tilde{K}$ and $\mathfrak{p}$ the maximal ideal of $\tilde{K}$?

Why is it enough to say what it does in the valuation ring? Thanks.

• Oh, seems right. Though we had more of a theoretical significance for this rather than just lack of expliciticity – Shoutre Nov 16 '16 at 17:15
• everything in the full field is a ratio of elements of the integer ring, so since we are dealing with a field homomorphism, it is enough to know it on the ring. – Adam Hughes Nov 16 '16 at 22:02

## 1 Answer

Since $\tilde K$ is the fraction field of $\mathcal O_{\tilde K}$, it is sufficient to define the Frobenius automorphism on the valuation ring.

More theoretically, since $\tilde K$ is unramified, the surjective map $$\mathrm{Gal}(\tilde K/K)\to \mathrm{Gal}(\overline{\mathbb F}_p/\mathbb F_p)$$ given by letting $\mathrm{Gal}(\tilde K/K)$ act on $\mathcal O_{\tilde K}/\overline{\mathfrak p}\cong\overline{\mathbb F}_p$ is also injective. The right hand side is generated by the Frobenius automorphism. The Frobenius automorphism of $\tilde K$ is exactly the element of $\mathrm{Gal}(\tilde K/K)$ it corresponds to. This explains why the Frobenius automorphism is defined by its action on $\mathcal O_{\tilde K}/\overline{\mathfrak p}$.