Let $K \mid \mathbb{Q}_2$ be a finite unramified extension of fields of degree $f$, i.e., $2 \in \mathbb{Z}_2$ is a uniformizer in $\mathcal{O}_K$ and $\mathcal{O}_K/2 \mathcal{O}_K \cong \mathbb{F}_{2^f}$ is a field containing precisely $2^f$ elements.

What can be said about the structure of the unit groups $K^\times$ and $\mathcal{O}_K^\times$?

It is clear that $K^\times \cong \mathbb{Z} \times \mathcal{O}_K^\times$. Further, $\exp$ and $\log$ induce an isomorphism $$(\mathcal{O}_K,+) \cong (4 \mathcal{O}_K,+) \cong (1+4\mathcal{O}_K, \cdot) .$$ Now I would like to write $$\mathcal{O}_K^\times \cong G \times (1+4\mathcal{O}_K)$$ for some group $G$. It should be $$|G| = |\mathcal{O}_K/4 \mathcal{O}_K| = |(\mathcal{O}_K/2 \mathcal{O}_K)^\times| \cdot |\mathcal{O}_K/2 \mathcal{O}_K| = (2^f-1) \cdot 2^f.$$ The ''$(2^f-1)$-part'' of $G$ could be the subgroup $\mu_{2^f-1}(K)$ of $(2^f-1)$-th roots of unity in $K$.

My question now is whether such a group $G$ actually exists and how to construct the isomorphism $$\mathcal{O}_K^\times \cong G \times (1+4\mathcal{O}_K).$$ Does anyone have an idea?

  • 2
    $\begingroup$ You always have $\mathcal{O}_K^\times=(1+2\mathcal{O}_K)\times \mu_{2^f-1}$. I think the only torsion in the factor $1+2\mathcal{O}_K$ is $-1$. Odd order torsion elements project to non-zero elements of $\mathcal{O}_K/2\mathcal{O}_K$, and those come from $\mu_{2^f-1}$. Torsion of order a power of two is limited to $\mu_2$, because a fourth root of unity generates a ramified extension. $\endgroup$ – Jyrki Lahtonen Oct 29 '18 at 17:33
  • $\begingroup$ @JyrkiLahtonen: So this means that $G$ cannot exist? $\endgroup$ – Algebrus Oct 29 '18 at 17:35
  • $\begingroup$ I think that's the conclusion. The exception being the case $f=1$, when $G=\langle -1\rangle$ works as prescribed. $\endgroup$ – Jyrki Lahtonen Oct 29 '18 at 17:36
  • $\begingroup$ The general principle manifesting itself here is that a finite subgroup of the multiplicative group of any field is always cyclic. The restriction on the power of two order tells that maximum order of a finite subgroup here has $2(2^f-1)$ elements. It's a drag that the exponential/logarithm doesn't work for the whole thing. IIRC more can be said, but I don't have the time to think/recall this. We have users who know this. Let's wait. $\endgroup$ – Jyrki Lahtonen Oct 29 '18 at 17:47

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