Let $F/K$ be a finite extension of local fields which is totally ramified. The inertia subgroup is defined as

$$ I_{F/K} = \{ \sigma \in \operatorname{Gal}(F/K) : \sigma(x) \equiv x \mod \pi_F \: \forall x \in \mathcal{O}_F \}$$ where $\pi_F$ denotes a uniformizer of $F$ and

$$\mathcal{O}_F = \{ x \in F : |x|\leq 1 \} = \{ x \in F : v_F(x) \geq 0\}$$

denotes the ring of integers of $F$.

Question: Is $I_{F/K} = \operatorname{Gal}(F/K)$?

My own progress so far (on an example): Consider $F = \mathbb{Q}_5(\sqrt{5})$. $K = \mathbb{Q}_5$, then $F/K$ is totally ramified of index 2. A possible uniformizer of $F$ would be $\pi_F = \sqrt{5}$. So the Galois group consists of two automorphisms, namely the unique automorphisms which map $\sqrt{5}$ to $\sqrt{5}$ and $-\sqrt{5}$, respectively. As every element of $F$ can be written in the form $x= a + b \sqrt{5}$ with $a,b \in \mathbb{Q}_5$, for each $\sigma \in \operatorname{Gal}(F/K)$ we have $\sigma(x) = \sigma(a+b\sqrt{5}) = \sigma(a) + \sigma(b)\sigma(\sqrt{5}) = a+b\sigma(\sqrt{5}) \equiv a + b \sqrt{5} = x$.

So in this case, we have $I_{F/K} = \operatorname{Gal}(F/K) $ indeed.

Some ideas (that I maybe could use):

  • If we assume that $F/K$ is a simple extension (I don't know if this is always true though), then there exists an $\alpha \in F$ with $F = K(\alpha)$. Can we choose $\alpha = \pi_F$? Maybe we need to use the fact that $F/K$ is totally ramified here.

  • If we can, do we have $\sigma(\pi_F) \equiv \pi_F \mod \pi_F$?

Any help is really appreciated.

  • 1
    $\begingroup$ Your extension is Galois to start with? In a Galois extension, the quotient of the full group by the inertia group is of order $f$, the residue-field extension degree. Totally ramified means $f=1$, so I think that’s it. $\endgroup$ – Lubin Sep 11 '18 at 1:20

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